3. VectorSpaces Properties
Subspaces, Bases and Dimensions, Coordinates, Row-equivalence summary, Computations Concerning Subspaces.
0. Index.
- Introduction.
- Subspaces.
- 2.1. Definition and characterization.
- 2.2. Example of subspaces.
- 2.3. Combination of vector spaces.
- 2.3.1. Intersection of subspaces.
- 2.3.2. Spanned subspace by a set.
- 2.3.3. Sum of subsets of vector spaces.
- 2.4 Example of combinated vector spaces.
- 2.4.1. Subspace defined by equations.
- 2.4.2. Matrix.
- 2.4.3. Row-spaces.
- 2.4.4. Caracterization of the polynomial vectorspace.
-
2.4.5. Subspaces exercises.
1. Introduction.
In the past post; Vector Spaces, we've introduced that a vector space $(K,V,·)$ is the algebraic structure resulting from using a field $K$ to weigh the compositions of an abelian group $V$, through a field action $·$. Meaning that a vector space is a triple $(K,V, ·)$ where:
- $K$ is a field.
- $V$ is an abelian (commutative) group.
- $· : K \times V \to V$ is a field action, which acts over $V$ using $K$'s elements to scale vectors, making families of proportional vectors.
The resulting compositions are what we call linear combinations, independent contributions of the group's elements mediated by the field's scalars:
\[\alpha v + \beta u : \alpha ,\beta \in \mathbb{R}, v,u \in V\]We presented some examples of sets which, with proper operations involved, are examples of vector spaces:
- The $n$-tuples space: $F^n$
- The space of $m \times n$ matrices: $M_{m \times n}(F)$
- The space of functions from a set $S$ to a field $K$: $K^S$
-
The space of polynomial functions over a field K
We also developed, surreptitiously, the affine space structure, which is the way in which we use vectors to study a non-empty set in a simply transitive way, and we defined the arrow concept as the segment that connects two points of the affine space with magnitude, direction and orientation, which captures the displacement from the point at the base of the segment to the one lying at the end of the arrow.
We also demonstrated that if you fix one point $O$ and consider the family of all possible arrows with base at $O$, then that family has the structure of a vector space, which leaves us with a geometric intuition of what a vector is.
Then, with this information as a starting point, let's develop the fundamental properties of vector spaces.
2. Subspaces.
2.1. Definition and characterization.
In this section we shall introduce some of the basic concepts in the study of vector spaces.
Let $V$ be a vector space over the field $F$. A subspace of $V$ is a non-empty subset $W$ of $V$ which is itself a vector space over $F$ with the operations of vector addition and scalar multiplication on $V$.
Let's observe that, from the axioms of vector spaces, if $W$ is a vector space such $W \subset V$, then:
-
Closure relative to linear combinations:
\[v,u \in W \implies (\alpha v + \beta u \in V \ \forall \alpha, \beta \in F)\] -
Contains zero vector: $0_V \in W$, or $0_W = 0_V$. Remember that $V$ being a vector space means that $(V,+)$ is an abelian group, so $(W,+)$ is an abelian subgroup of $V$ and it inherits $0_V$ from $V$ by the uniqueness of this element. (Observe that this also extends to associativity and inverses).
Let's take a characterization for any non-empty subset $W \subset V$ to be a vector space. If $V$ is a $K$-vector space, then:
\[W \text{ is a vector space } \iff (\alpha u + v \in W \quad \forall u,v \in W, \alpha \in K)\]Let's reconstruct the structure:
-
$(W,+)$ is an abelian subgroup of $(V,+)$.
Since $W \neq \varnothing$ (by the premise), we can consider $u,v \in W$, then: $(-1)u + v = v - u \in W$, so it is a subgroup of $V$. Let's see that $W$ also inherits commutativity from $V$ so it is an abelian subgroup.
-
$· |_W: K \times W \to W$ is a field action:
-
Observe that $-1 u + u = 0 \in W$, thus $1 u + 0 = u \in W$ so the unit in $F$ doesn't change the vector and it falls inside $W$.
-
Take some $v \in W \implies \alpha v + 0 = \alpha v \in W \implies \beta (\alpha v) + 0 = \beta (\alpha v) \in W$. For the same reason $(\alpha \beta)v \in W$. Then observe that in $V$, by associativity, is $\beta (\alpha v) = (\beta \alpha)v$ so both elements are equal in $V$ and the same in $W$ (by unicity of the inverse), thus $\beta (\alpha v) = (\beta \alpha)v$
-
Take $(\alpha + \beta)v + 0 = (\alpha + \beta)v \in W \subset V$, then in $V$ is $(\alpha + \beta)v = \alpha v + \beta v$ so, for the same argument as above, is $(\alpha + \beta)v = \alpha v + \beta v$ in $W$.
-
Take $\alpha v + 0 \in W \implies \exists w = \alpha v \in W$, take now other $u \in W$, then $\alpha u + w \in W$, again in $V$ is $\alpha u + w = \alpha u + \alpha v = \alpha(u + v)$ and again is $ \alpha(u + v) = \alpha u + \alpha v$ in $W$.
-
2.2. Example of subspaces.
Let's consider some well-known examples of subspaces. Consider some $V$ a $K$-vector space, then:
- $V$ is a subspace of $V$.
-
$\Set{0} \subset V$ is the zero subspace of $V$; $\alpha 0 + 0 = 0 \in \Set{0} \quad \forall \alpha \in K$ so $\Set{0}$ is a subspace of $V$.
-
$A:= \Set{x \in K^n \mid x_1 = 0}$ is a subspace of $K^n$, check that always $\alpha x + y \in A$, trivially.
But let's observe that $B := \Set{x \in K^n \mid x_1 = 1 + x_2}$ (the solution of a non-homogeneous equation) does not satisfy the rule, take some $\alpha$ and observe that:
\[\alpha x + y = (\alpha (1 + x_2) + 1 + y_2), \alpha x_2 + y_2,..., \alpha x_n + y_n)\]Then, $\alpha + 1 + \alpha x_2 + y_2 = 1 + \alpha x_2 + y_2 \iff \alpha=0$.
So we have that $\forall \alpha \neq 0 \quad (x,y \in B \implies \alpha x + y \notin B)$
-
The space of polynomial functions over $K$;
\[\operatorname{Pol}(K, K) := \left\{\, f \in K^K \ \middle|\ \exists n \in \mathbb{N}_0,\ \exists (\alpha_0, \ldots, \alpha_n) \in K^{n+1} : \forall s \in K,\ f(s) = \sum_{i=0}^{n} \alpha_i\, s^i \,\right\}\]Is a subspace of $K^K$.
Observe that, if we get $f,g \in Pol(K,K)$, then is: $ f(s) = \sum_{i=0}^{n} \alpha_i\, s^i$, $g(s) = \sum_{i=0}^{m} \beta_i\, s^i$, consider is $n \leq m$, then we can expand $f(s)$ to $\sum_{i=0}^{m} \alpha_i\, s^i = \sum_{i=0}^{n} \alpha_i\, s^i +\sum_{i=n+1}^{m} 0\, s^i$ and then is:
\[(\gamma f + g)(s) = \gamma f(s) + g(s) = \gamma \sum_{i=0}^{m} \alpha_i\, s^i + \sum_{i=0}^{m} \beta_i\, s^i =\] \[\sum_{i=0}^{m} \gamma\alpha_i\, s^i + \sum_{i=0}^{m} \beta_i\, s^i = \sum_{i=0}^{m} (\gamma\alpha_i + \beta_i) s^i = \sum_{i=0}^m \varphi_i \, s^i \in Pol(K,K)\] -
Consider $S:=\Set{(a_{ij}){ij} \in M_n(K) \mid (a{ij}){ij} = (a{ij})_{ji}}$, the set of all symmetric matrices form a subspace of $M_n(K)$
Take, $A,B \in S$, then let's see that $\alpha A + B = \alpha (a_{ij}){ij} + (b{ij}){ij} = (\alpha a{ij} + b_{ij})_{ij}$, then let's see that this matrix is in $S$ by observing that:
\[A \in S \implies a_{ij} = a_{ji} \quad \forall i,j \implies \alpha a_{ij} = \alpha a_{ji} \quad \forall i,j \implies \alpha A \in S\]Now, observe that:
\[A,B \in S \implies \begin{cases}a_{ij} = a_{ji} \quad \forall i,j \\ b_{ij} = b_{ji} \quad \forall i,j \end{cases} \implies a_{ij} + b_{ij} = a_{ji} + b_{ji} \quad \forall i,j \implies A+B \in S\]Ultimately concluding $\alpha A + B \in S$, so $S$ is a subspace of $M_n(K)$
-
Consider $M_n(\mathbb{C})$, then we say that $A \in M_n(\mathbb{C})$ is Hermitian or self-adjoint if $(a_{ij}){ij} = \overline{(a{ij})_{ji}}$, meaning that is equal to the transverse of the conjugate.
Then, the set $H := \Set{A \in M_n(\mathbb{C}) \mid (a_{ij}){ij} = \overline{(a{ij}){ji}} }$ is not a subspace of $M_n(\mathbb{C})$ check that the diagonal elements impose $(a{ii}){ii} = \overline{(a{ii})_{ii}}$, since the entries are complex numbers, a complex number is equal with his conjugate when is real.
Consider some $A \in H$ and observe that $iA \notin H$, since the diagonal is not real and is not an Hermitic matrix, so our criteria $\alpha A +B \in H \quad \forall A,B \in H, \alpha \in \mathbb{C}$ doesn't apply.
How ever it does applies when $\alpha \in \mathbb{R}$, observe that $(\mathbb{R},(H,+), ·)$ is a $\mathbb{R}$-vector space.
-
The solution space of a system of homogeneous linear equations over $K$
Let be $A \in M_{m\times n}(K)$, then consider $W := \Set{X \in M_{n \times 1}(K) \mid AX = 0}$ that's not empty, let's remember that:
\[AX = \begin{pmatrix} a_{11} \cdots a_{1n} \\ \vdots \quad \quad \quad \vdots \\ a_{m1} \cdots a_{mn} \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n\end{pmatrix} = \begin{pmatrix} \sum_{i=1}^n a_{1i}x_i \\ \vdots \\ \sum_{i=1}^n a_{mi}x_i \end{pmatrix}\]Then, $X \in W \implies \sum_{i=1}^n a_{ji}x_i = 0\quad \forall j \leq m$
Let's now consider $\alpha X + Y$, then:
\[A[\alpha X + Y] = \begin{pmatrix} a_{11} \cdots a_{1n} \\ \vdots \quad \quad \quad \vdots \\ a_{m1} \cdots a_{mn} \end{pmatrix} \begin{pmatrix} \alpha x_1 + y_1 \\ \vdots \\ \alpha x_n + y_n\end{pmatrix} = \begin{pmatrix} \sum_{i=1}^n a_{1i}(\alpha x_i + y_i) \\ \vdots \\ \sum_{i=1}^n a_{mi}(\alpha x_i + y_i) \end{pmatrix}\]Observe that, for being $X,Y \in W$:
\[\sum_{i=1}^n a_{mi}(\alpha x_i + y_i) = \alpha \sum_{i=1}^n a_{ji} x_i + \sum_{i=1}^n a_{ji}y_i = 0 \quad \forall j \leq m\]So, $\alpha X +Y \in W$.
2.3. Combination of vector spaces.
Let's now see that the intersection of vector subspaces is a subspace, the spanned vector space of a subset of vectors is at the same time the intersection of all the subspaces that contains the subset and the set of all linear combinations of the vectors of the subset.
The the addition or sum of subspaces is a subspace
2.3.1. Intersection of subspaces.
Let $V$ be a vector space over the field $K$. The intersection of any collection of subspaces of $V$ is a subspace of $V$.
Consider $V$ a $K$-vector space, then $\Set{W_i}$ a family of subspaces of $V$, consider $\bigcap_i W_i$. Then, observe that for being each $W_i \leq V \implies 0 \in W_i \quad \forall i \implies 0 \in \bigcap_i W_i \neq \varnothing$. Then, consider:
\[u,v \in \bigcap_i W_i \implies u,v \in W_i \leq V \quad \forall i \implies \alpha u +v \in W_i \quad \forall i \implies \alpha u + v \in \bigcap_i W_i\]And thus, $\bigcap_i W_i \leq V$
This result gains its importance from the fact that the subspace generated by a subset $S \subset V$ is the intersection of all the subspaces of $V$ that contains $S$, this is in fact the smallest subspace that contains $S$.
2.3.2. Spanned subspace by a set.
Spanned subspace definition
Consider again a $K$-space, $V$, and a subset $S$ of $V$.
Then, the subspace spanned by $S$ is the intersection of all those subspaces of $V$ which contains $S$, which, as we see above, is a vector space.
When $S$ is a finite set of vectors, $S:=\Set{\alpha_1,…, \alpha_n}$, we shall simply call $W$ the subspace spanned by the vectors $\alpha_1, \alpha_2, …, \alpha_n$,
Spanned subspace caracterization
Be $S$ a non-empty subset of a $K$-vector space $V$, then the spanned subspace of $S$, $W$ is the set of all the linear combinations of $S$'s vectors.
-
First, let's see that obviously $W \neq \varnothing$: $S \neq \varnothing \implies \exists u \in S$, then $u = 1u \in W \implies W \neq \varnothing$. Lastly, just consider $u,v \in W$, then $\alpha u + v \in W$ so $W \leq V$.
Let's make an observation, $\varnothing$ vacuosly satisfies the closure on linear combinations, there are no $u,v$ to test so is a null antecedent implication. The fact which separates $W$ from this degenerated case is the fact that $W \neq \varnothing$, meaning that for a vector space is unconditional to be non-empty, linear combination closure is not enough.
-
Let now be $\Set{W_i \leq V \mid S \subseteq W_i}_{i \in I}$ the family of subspaces of $V$ that contains $S$ let's see that in fact is $W = \bigcap_i W_i$ by demonstrating each set contains the other.
-
$W \subset \cap_i W_i$
If $w \in W \implies \exists u,v \in S : \alpha u + \beta v = w$. Since $S \subset \cap_i W_i \implies u,v \in \cap_i W_i$ and, since each $W_i \leq V$ and the intersection of vector spaces is a vector space then any linear combination is in $\cap_i W_i$, meaning $w = \alpha u + \beta v \in \cap_i W_i$
-
$\cap_i W_i \subset W$
Let's observe $W$ is the subspace of all the linear combinations of $S$ elements, is contained in each $W_i$ because every vector space contains the linear combinations of the elements of $S$ for being vector spaces.
Thus, observe that $W$ is the minimum element of the collection which, coincides by definition with his ínfimum which is by definition the spanned vector space this is, the intersection $\cap_i W_i$.
-
2.3.3. Sum of subsets of vector spaces.
If $S_1,S_2,\cdots, S_k$ are subsets of a vectror space $V$, then the set of all sums:
\[\sum_{i=1}^k S_i = S_1 + S_2 \cdots + S_k = \Set{\sum_{i=1}^k \alpha_i \mid \alpha_i \in S_i}\]Now consider: $W_i \leq V : i = 1,2…,k$, then the set $\sum_{i=1}^k W_i$ is immediately a subspace. Observe that this spaces is the space spanned by the set $\bigcup_i W_i$.
2.4 Example of combinated vector spaces.
2.4.1. Subspace defined by equations.
Let $K$ be a subfield of the field $C$ of complex numbers. Then suppose:
\[\begin{cases} \alpha_1 = (1,2,0,3,0) \\ \alpha_2 = (0,0,1,4,0) \\ \alpha_3 = (0,0,0,0,1) \end{cases}\]Let's observe that the spanned vector space by $\Set{\alpha_1,\alpha_2, \alpha_3}$, $W$, is the set of al the lineal combinations of $\alpha_1, \alpha_2$ and $\alpha_3$. Hence, a vector
\[\alpha \in W \iff \exists c_1,c_2,c_3, \in K : \alpha = \sum_{i=1}^5 c_i \alpha_i = (c_1,2c_1,c_2,3c_1 + 4c_2,c_3)\]Observe that this subspace $W$ can be described using the equations as the solution set of the linear equation system:
\[M := \begin{cases} x_2 = 2x_1 \\ x_4 = 3x_1 + 4x_3\end{cases}\]2.4.2. Matrix.
Let be $K$ a subfield of the field $\mathbb{C}$ of complex numb ers, and let $V$ be the vector space of all $2 \times 2$ matrices over $K$.
Let consider be the subset of $V$ consisting of all matrices of the form:
\[W_1 := \Set{\begin{pmatrix} x & y \\ z & 0\end{pmatrix}: x,y,z \in K}\] \[W_2 := \Set{\begin{pmatrix} x & 0 \\ 0 & y\end{pmatrix}: x,y \in K}\]Observe that: $V = W_1 + W_2$, or, in other terms, the linear combinations of the elements of the two spaces can generate any other element in $V$:
\[\begin{pmatrix} x & y \\ z & t\end{pmatrix} = \begin{pmatrix} x & y \\ z & 0\end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 0 & t\end{pmatrix} \in V\]Also, the intersection subspace is:
\[W_1 \cap W_2 = \Set{\begin{pmatrix} x & 0 \\ 0 & 0\end{pmatrix}: x \in K}\]2.4.3. Row-spaces.
Let $A \in M_{m \times n}(K)$. Then, the row vectors of $A$ are vectors $\alpha \in K^n$; $A := (\alpha_1, \ldots, \alpha_n)$.
In this context we call the row-space of $A$ to the spanned vector space, $W \leq K^n$ of the elements $\Set{\alpha_i \in K^n \mid A = (\alpha_1,\ldots, \alpha_n)}$.
2.4.4. Caracterization of the polynomial vectorspace.
Consider:
\[Pol(K,K) := \Set{f \in K^K \mid \exists n \in \mathbb{N}, \exists (\alpha_1,\ldots, \alpha_n) \in K^n : f(x) = \sum_{i=1}^n \alpha_i x^i \ \forall x \in K}\]Then, consider now $ S \subset Pol(K,K)$ defined as:
\[S := \Set{f \in K^K \mid \exists i \in \mathbb{N} :f(x)=x^i \ \forall x \in K}\]Note that the spanned vector space of $S$ is $Pol(K,K)$.
2.4.5. Subspaces exercises.
2.4.5.1. Subspaces of $\mathbb{R}^n$.
-
$\Set{\alpha \in \mathbb{R^n} \mid a_1 \geq 0}$
Naturally isn't a vector space. Observe that trivially $A \neq \varnothing$ and $-1·\alpha + 0 = -\alpha \notin A$, since $a_1 \geq 0 \implies -a_1 \leq 0$.
-
$\Set{\alpha \in \mathbb{R^n} \mid a_1 + 3a_2 =a_3}$
We already covered this set in $2.2-3/7$, the solution set of any homogeneous linear equation system is a vector space.
-
$\Set{\alpha \in \mathbb{R^n} \mid a_2 =a^2_1}$
Let's see that this equation is not linear, thus, it do not respects linear combinations and is not a subspace.
Take for example $\alpha = (1,1,0)$ and consider the linear combination $\alpha + \alpha$ which not satisfies the condition.
-
$\Set{\alpha \in \mathbb{R^n} \mid a_1a_2=0}$
Clearly is not a subspace, consider again: $\alpha = (1,0,1)$ and $\beta = (0,1,0)$ and his linear combination $\alpha + \beta$ which, again, do not satisfies the condition.
-
$\Set{\alpha \in \mathbb{R^n} \mid a_2 \in \mathbb{Q}}$
Is not a subspace, take $\alpha = (0,1,0)$, then $\sqrt{2}\alpha = (0,\sqrt{2},0)$ do not satisfies the condtion.
2.4.5.2. Function subspaces.
Consider again $\mathbb{R}^{\mathbb{R}}:=\Set{f \in \mathcal{P}(\mathbb{R} \times \mathbb{R}) \mid \forall x \in \mathbb{R} \ \exists ! y \in \mathbb{R}: (x,y) \in f}$, with the operations:
\[(f+g)(x) = f(x) + g(x)\] \[(\alpha f)(x) = \alpha (f(x)) : \alpha \in \mathbb{R}\]Then, which of the following sets of functions are subspaces of $\mathbb{R}^{\mathbb{R}}$
-
$\Set{f \mid f(x^2) = f(x)^2}$
Let's observe immediately that $(\alpha f)(x^2) = \alpha f(x^2)= \alpha f(x)^2 \neq (\alpha f)(x)^2 = \alpha^2 f(x)^2$ for any $\alpha \neq 0$, so is not a vector subspace.
-
$\Set{f \mid f(0) = f(1)}$
Check that $(\alpha f + g)(0) = \alpha f(0) + g(0) = \alpha f(1) + g(1) = (\alpha f + g)(1)$, so is a subspace.
-
$\Set{f \mid f(3) = 1 + f(-5)}$
Check that
\[(\alpha f + g)(3) = \alpha f(3) + g(3) = \alpha (1 + f(-5))+ 1 + g(-5) = (\alpha f + g)(-5) + 1 + \alpha\]Thus, the condition is not meeted for any $\alpha \neq 0$.
-
$\Set{f \mid f(-1) = 0}$
Check that $(\alpha f + g)(-1) = \alpha f(-1) + g(-1) = 0$, so is a subspace.
-
$\Set{f \mid f \text{ is continuous}}$
Naturally yes.
2.4.5.3. Spanned vector.
Consider the vector $\alpha = (3,-1,0,-1)$, is spanned by the vectors $u = (2, -1, 3, 2), v = (-1, 1, 1, -3), w = (1, 1, 9, -5)$?
Consider $W$ the vector subspace spanned by the set $S := \Set{u,v,w}$. Then, the statement ask if $\alpha \in W$.
By $2.3.2$, we do know that the spanned vector subspace, $W$ is the intersection of all the subspaces that contains $S$ which coincides with the set of all the linear combinations of the vectors of $S$ which is a subspace it self.
Then, $\alpha$ is an element of $W$ only if is a linear combination of $u,v,w$, formally:
\[\alpha \in W \iff \exists x,y,z \in \mathbb{R}: \alpha = xu + yv + zw\]Taking coordinate to coordinate, we can form the following equation system on $\mathbb{R}$:
\[\begin{cases} 3 = 2x -y +z \\ -1 = -x + y + z \\0 = 3x + y +9z \\ -1 = 2x -3y -5z \end{cases}\]2.4.5.4. Set of spanned subspace.
Consider $W \leq \mathbb{R}^5$ of those $(x_1,x_2,x_3,x_4,x_5) \in \mathbb{R}^5$ satisfying the following equation system:
\[M:=\begin{cases} 2x_1 -x_2 + \frac{4}{3}x_3 - x_4 = 0 \\ x_1 + \frac{2}{3} x_3 - x_5 = 0 \\ 9x_1 - 3x_2 + 6x_3 - 3x_4 - 3x_5 = 0 \end{cases}\]Find a finite set of vector that spans $W$.
We have to find some set $S$ such the set of the linear combinations of his elements coincides with $W$. Leveraging the exercise before, giving some $\alpha \in \mathbb{R}^5$, we have to find a finite some of vectors $u_1,u_2, \ldots, u_n$ such the equation system on the escalars $t_1,\ldots, t_n \in \mathbb{R} : \alpha = \sum_{i=1}^n t_i u_i$ is equivalent to the one gived by the exercise.
First, let's consider the $RREM$ form of the matrix previous matrix:
\[R := \begin{pmatrix} 1 & 0 & \frac{2}{3} & 0 & -1 \\ 0 & 1 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}\]Which leave us with the system:
\[M' := \begin{cases} x_1 + \frac{2}{3}x_3 - x_5 = 0 \\ x_2 + x_4 -2x_5 = 0 \end{cases}\]Since $M$ and $M'$ has row-equivalent asociated matrix, we do know that both are equivalents; $M \equiv M'$, so both shares the same solution set, $W$.
Ultimately, let's solve it, the solution of $M'$ is:
\[x_1 = -\tfrac{2}{3}x_3 + x_5, \quad x_2 = -x_4 + 2x_5\]Calling; $x_3 = \alpha,\ x_4 = \beta,\ x_5 = \gamma$, then:
\[W := \Set{(\gamma -\frac{2}{3}\alpha,2\gamma - \beta ,\alpha, \beta, \gamma) : \alpha, \beta, \gamma \in \mathbb{R}}\]Let's observe that:
\[\begin{pmatrix} \gamma -\frac{2}{3}\alpha \\ 2\gamma - \beta \\ \alpha \\ \beta \\ \gamma \end{pmatrix} = \alpha\begin{pmatrix} -\tfrac{2}{3} \\ 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ -1 \\ 0 \\ 1 \\ 0 \end{pmatrix} +\gamma \begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}\]Hence, any vector of $W$ is a linear combination of $(−2/3,0,1,0,0), (0,−1,0,1,0), (1,2,0,0,1)$
In summary, the solution set of a homogeneous linear system is a vector subspace. Writing the solution in its parametrized form yields the vectors that span the subspace.
2.4.5.5. Matrix subspaces.
Let be $M_{n \times n} (K) : n \geq 2$, which of the following sets $\Phi_n$ are subspaces?
-
$ \Set{A \in M_{n \times n}(K) \mid \exists A^{-1}}$
Let's observe that if we consider $A \in \Phi_n \implies -A \in \Phi_n$ but $A + -A = 0 \notin \Phi_n$, hence is not a subspace.
Observe that, we could quicker check that isn't a subspace because is not a vector space since it not contains $0$.
-
$\Set{A \in M_{n \times n}(K) \mid \nexists A^{-1}}$
Let observe that we can consider:
\[A = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}, \quad B = \begin{pmatrix}0 & 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\]Both of them aren't invertible since they are already their $RREM$ which not coincides with $I_3$ but $A+B = I_3 \in \Phi_n$
-
$\Set{A \in M_{n \times n}(K) \mid AB = BA}$, for some fixed $B$.
Observe that taking $A, C \in \Phi_n$ then:
\[(\alpha A + C)B = \alpha AB + CB = B(\alpha A) + BC = B(\alpha A + C)\]Thus, it conforms a subspace
-
$\Set{A \in M_{n \times n}(K) \mid A^2 = A}$
Observe that if $A,B \in \Phi_n$, then:
\[(\alpha A + B)^2 = \alpha^2 A^2 + B^2 + \alpha AB + B \alpha A = \alpha^2A + B + \alpha AB + \alpha BA \neq \alpha A + B\]
2.4.5.6. Subspaces of $\mathbb{R}$.
-
Prove that the only subspaces of $\mathbb{R}$ are $\mathbb{R}$ and the zero subspace.
That $\mathbb{R} \leq \mathbb{R}$ and $\Set{0} \leq \mathbb{R}$ comes immediately from $2.2.$
Note directly that if there is some subspace $V \leq \mathbb{R}$, then $V$ contains all the linear combinations of himself, so consider some $a \in V \implies 1 = (a/a) \in V$, then since we can multiply $1$ by any scalar, we can reach any element of $\mathbb{R}$ inside of $V$ so $V = \mathbb{R}$.
-
Prove that a subspace of $\mathbb{R}$ is $\mathbb{R}^2$, or the zero subspace, or consists of all scalar multiples of some fixed vector in $\mathbb{R}^2$.
First, again that $\mathbb{R} \leq \mathbb{R}$ and $\Set{0} \leq \mathbb{R}$ comes immediately from $2.2.$
Next, the lines on $\mathbb{R}^2$ are subspaces is also clear, take some not null $v \in R^2$ then any scaled element of $v$ is a linear combination of $v$ so the line contains all the linear combinations of his elements.
Lastly, let's suppose two elements $u,v \in \mathbb{R}^2: \nexists \alpha \in \mathbb{R} : u = \alpha v$. Then, this two elements doesn't belong to the same line. Suppose $u = (a,b), v = (c,d)$, hence observe that for some $\alpha, \beta \in \mathbb{R}$ is $\alpha u + \beta v = (\alpha a+ \beta c,0) + (0, \alpha b+ \beta d)$, observe that we can form the following equation system:
\[\begin{cases} \alpha a + \beta b = 1 \\ \alpha c + \beta d = 1\end{cases}\]Solving it we get:
\[\alpha=\dfrac{d-c}{ad-bc},\qquad \beta=\dfrac{a-b}{ad-bc}\]Observe that the solution has sense, since $u,v$ are no proportionals, $a,c$ and $b,d$ can't be zero at the same time so $ad - bc \neq 0$
And:
\[\dfrac{d-c}{ad-bc} u + \dfrac{a-b}{ad-bc} v = (1,0) + (0,1) = e_1 + e_2\]Observe that (without the need to invocate orthogonality or basis), we can ensure that, being $w = (x,y) \in \mathbb{R}^2$, then is:
\[w = \dfrac{x(d-c)}{ad-bc} u + \dfrac{y(a-b)}{ad-bc} v \quad \forall w \in \mathbb{R}^2\]Meaning, with two non-proportional vectors, we can reach any other element in $\mathbb{R}^2$ thus the spanned subspace $V$ of the subset $S := \Set{u,v \mid \nexists \alpha \in \mathbb{R} : u = \alpha v} \subset \mathbb{R}^2$ is $\mathbb{R}^2$ it self.
Thus, observe that for any $W \leq V$, then if $W$ contain at least two non-proportional vectors then $W = \mathbb{R}^2$, since a vector space contain all the linear combination of his vectors, if do not then is either a line or the zero subspace.
-
Can you describe the subspaces of R3?
Extending the argumentation in $2.$ is the zero subspace, those subspaces for which exists at least three independant vectors (one vector which is not linear combination of other two) which spans $\mathbb{R}^3$, then those subspaces with at least two non-proportional vectors which are planes and those with all his vectors proportionals which are lines.
2.4.5.7. Union of subspaces.
Let $W_1, W_2 \leq V$ be subspaces of a vector space such that the set-theoretic union of $W_1$ and $W_2$ is also a subspace. Prove that one of the spaces $W_i$ is contained in the other.
Let's consider some interesting visual approach about the union of subspaces. Consider $\mathbb{R}^2$, then consider two non proportional vectors and the spanned subspaces of each of them, which are two distinct lines crossing in the origin. Is easy to see that a linear combination of the vectors of each line can fall out of each line since the subspace containing two non-proportional vectors coincides with $\mathbb{R}^2$. This means that the union of the two lines doesn't contains all his linear combinations so is not a subspace. For the union to be a subspace, each linear combination should end up as vector of one of the lines wich is pretty much to say that is proportional to one of the vectors of the linear combination but observe that this also implies that the third vector is also proportional so to the two lines are the same.
Then, generalizing:
\[\bigcup_i W_i \leq V \implies w = u + v \in \bigcup_i W_i \quad \forall u \in W_1, \forall v \in W_2\]Then, suppose $w \in W_i \implies v = w - u \in W_i$, observe that this means that we can't consider simulatenously $u \in W_i \setminus W_j$ and $v \in W_j \setminus W_i$ since the last afirmation would be false.
2.4.5.8. Even and odd function subspaces. Direct Sum.
Consider $\mathbb{R}^\mathbb{R}$ and the following subsets:
\(A := \Set{f \in \mathbb{R}^\mathbb{R} \mid f(-x) = f(x)}\) \(B := \Set{f \in \mathbb{R}^\mathbb{R} \mid f(-x) = -f(x)}\)
Then, prove that:
-
Both subsets are subspaces of $\mathbb{R}^\mathbb{R}$
Observe that:
\[(\alpha f + g)(-x) = \alpha f(-x) + g(-x) = \begin{cases} \alpha f(x) + g(x) = (\alpha f + g)(x) \quad \forall f,g \in A \\ -\alpha f(x) - g(x) = -(\alpha f + g)(x) \quad \forall f,g \in B \end{cases}\]So both contains the linear combinations of his own elements so their are subspaces.
-
$A + B = \mathbb{R}^\mathbb{R}$
Let's suppose that $f \in \mathbb{R}^\mathbb{R}$, then let's suppose also $g \in A$ and $h \in B$ such $f = g + h$, observe that this functions verifies at the same time:
\[\begin{cases}f(x) = g(x) + h(x) \\ f(-x) = g(-x) + h(-x) = g(x) - h(x) \end{cases}\]In this context is, combining both equatlities we get:
- $g(x) = \frac{1}{2}[f(x) + f(-x)]$
- $h(x) = \frac{1}{2}[f(x) - f(-x)]$
So is:
\[f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)]\] -
$A \cap B = \Set{0}$
Observe that this is almost immediate, suppose some common function from $A \cap B$, then:
\[f(x) = \frac{1}{2}(f(-x) - f(-x)) = 0 \quad \forall x \in \mathbb{R}\]
Observe that we have demonstrated that two disjoint subspaces $A,B \leq \mathbb{R}^\mathbb{R}$ sum the total space:
\[A,B \leq \mathbb{R}^\mathbb{R} : A \oplus B \iff (A \cap B =\Set{0} \wedge A + B = \mathbb{R}^\mathbb{R})\]Observe that this allows to dispose every vector of $\mathbb{R}^\mathbb{R}$ in terms of two componentes of $A$ and $B$.
2.4.5.9. Property of direct sum.
Let $W_1, W_2$ subspace such are direct sum of $V$. Prove that for each vector $\alpha \in V$ there are unique vectors $\alpha_1 \in W_1$ and $\alpha_2 \in W_2$ such that $a = \alpha_1 + \alpha_2$.
Take $a \in V$ and suppose that is $a = \alpha_1 + \alpha_2 = \beta_1 + \beta_2 : \alpha_i, \beta_i \in W_i$ with $i = 1,2$, then observe that it cannot be $\alpha_i - \beta_j = 0$ since that would imply $\alpha_i \in W_j$ or $\beta_j \in W_i$ contradicting the premise that both subspaces do not share not-null vectors, so it can only be $\alpha_i - \beta_i = 0 : i = 1,2$, so both vectors are the same exact vector.
3. Bases and Dimension.
3.1. Conceptal introduction.
Let's first introduce what a basis is, a basis is the minimum set of a vector space which contains all the information the vector space is capable to express. Meaning that is the minimum set of vectors that spans the vector space; any vector is reachable trough a linear combination of the elements of a basis and at the same time this set is minimum, any subset of a basis is uncapable to span the vector space.
The dimension of the vector space is the number of "degrees of freedom" that a vector space has; the number of independent vector required to specify an arbitrary vector materialized in the cardinal number of the basis set.
This two concepts has a fundamental importance; the dimension is the fundamental invariant that clasifies vector spaces, and the basis what transform linear algebra (operation with vectors) in matrix computations.
3.2. Linearly dependent/independent.
Let first introduce the concept of dependant and independant linearity, which basically determines wheter a vector is owned by the spanned vector space of a set of vectors.
Let be $V$ a $K$-vector space and $S \subset V$. Then:
-
$S$ is said to be linear dependant if and only if:
\[\exists \alpha_1,\ldots, \alpha_n \in S \text{ distinct}, \exists c_1,\ldots, c_n \in K \text{ not all 0} : \sum_{i=1}^n c_i\alpha_i = 0\] -
$S$ is said to be linear independant if and only if is not dependant, negating the above we get
\[\forall \alpha_1,\ldots, \alpha_n \in S \text{ distinct}, \forall c_1,\ldots, c_n \in K \text{ not all 0} : \sum_{i=1}^n c_i\alpha_i \neq 0\]Observe that we can convert the statement by quitting the "not all 0" obtaining:
\[\forall \alpha_1,\ldots, \alpha_n \in S \text{ distinct}, \forall c_1,\ldots, c_n \in K : \sum_{i=1}^n c_i\alpha_i = 0 \iff c_i = 0 \quad \forall i \in [n]\]
Let's observe this more carefully, check something interesting, a subset of vectors of a vector space is said to be dependant when it can reference the zero vector, $0$ through a finite, non-trivial, linear combination.
Take for example two vectors, $u,v \neq 0 \in V$, then if $\exists \alpha,\beta\neq 0 \in K : \alpha u + \beta v = 0 \iff u = \frac{-\beta}{\alpha}v$, both are proportionals and we can retrieve one from the other without involve any other vector, so we can say they contains the same information, both are the same type of vector so to speak. Thus, let's suppose that $u,v$ can't reference null vector and consider $w = tu + lv \neq 0 : t,l \neq 0 \in K$, then observe that this three vectors can again reference the zero vector, $tu + lv -w = 0$ through a finite non-trivial linear combination, qwe can think in $w$ as a proportional vector to a combination of $u,v$, thus $w$ and $u,v$ contains the same information despite scaling and observe that the fact that three vectors can reference through a non-trivial linear combination states the same, each of them is proportional to a combination of the other ones.
Thus, information redundancy is the essense of linear dependence, a set of vectors is said to be linear dependent when at least one of them carries complete redundant information, in the sense that the whole set can express a part and the counterpart leading to a cancelation on the zero vector. Linear independence is just the negation of this statement, a set of vectors is said to be linear independent when all vectors contribute with some non-replicable information, or, in other terms, substract any vector in the set implies to loose information.
Let's see some easy consequences of the definition.
- Any set which contains a linearly dependent set is linearly dependent.
- Any subset of a linearly independent set is linearly independent.
-
Any set which contains the 0 vector is linearly dependent; for $1 · 0 = 0$.
3.2.1. Examples.
Consider $K$ a field, then in $K^3$ the vectors:
\[\begin{cases} \alpha_1 = (3,0,-3) \\ \alpha_2 = (-1,1,2) \\ \alpha_3 = (4,2,-2) \\ \alpha_4 = (2,1,1)\end{cases}\]Satisfies:
\[2\alpha_1 + 2\alpha_2 - \alpha_3 + 0 · \alpha_4 = 0\]So are linear dependant. Note that $\Set{\alpha_1, \alpha_2, \alpha_3}$ are linear dependant as well, hence, any set containing this vectors is also linear dependant.
The vectors:
\[\begin{cases} \alpha_1 = (1,0,0) \\ \alpha_2 = (0,1,0) \\ \alpha_3 = (0,0,1)\end{cases}\]Are linear independant.
3.3. Basis definition and examples.
Let $V$ be a vector space. A basis, $\mathcal{B}$, of $V$ is a linearly independent set of vectors in $V$ which spans the space $V$. The space $V$ is finitedimensional if it has finite basis.
Let's observe here that we are giving continuity to the dependence/independence frame but we are adding generators sets in the process. A basis is nothing more that the combination of two objects:
-
A generator set, this is; a set which contains all the information that the vector space can express.
-
A independent set, a set from which you can't free any element without loose information.
This two objects gives us a compress notion; a basis is the minimal generator set of a vector space; substracting a vector make a piece of information to get lost, so the span doesn't hold. This will be explain in a later theorem.
3.3.1. Standard basis of $K^n$.
Let $K$ be a field and in $K^n$, consider the subset:
\[\mathcal{B}: = \begin{cases} \alpha_1 = (1,0,0,\ldots, 0) \\ \alpha_2 = (0,1,0,\ldots,0) \\ \alpha_3 = (0,0,1,\ldots,0) \\ \quad \vdots \\ \alpha_n = (0,0,0,\ldots,1) \end{cases}\]At first, observe that for any $x = (x_1,x_2,\ldots,x_n) \in K^n$ is:
\[x = \sum_{i=1}^n \alpha_i x_i\]So $\mathcal{B}$ spans $K^n$ and also is linear independent as we see in the particular case with $K^3$. So $\mathcal{B}$ is a basis of $K^n$, particularly called standard basis.
3.3.2. Invertible matrix and basis.
Let's consider some $P \in M_n(K)$ invertible. We do know that $P \simeq I_n$, which means that no row in $P$ can be zeroed through a linear combination of the other rows in $P$ so the set of the rows in $P$ forms a linear independent set. Observe that, in more simple terms, be $X \in K^{n\times 1}$, then, by $6.3.3$ in Linear Equations:
\[\exists P^{-1} \iff (PX = 0 \iff X = 0)\]Thus, $PX = \sum_{i=1}^n x_iP_i = 0 \iff x_i = 0 \quad \forall i \in [n]$, meaning that $\Set{P_1, \ldots, P_n}$ is a linear independent set.
And also, as we see above, it spans any column on $K^{n \times 1}$ so $\Set{P_1, \ldots, P_n} \subset K^{n \times 1}$ is in fact a basis of $K^{n \times 1}$.
Let's observe that we are saying that in $K^n$, a basis and an invertible matrix is literally the same object. Is the naturall continuity to the conception of matrix seen as linear-information codified packeges viewed in Linear Equations.
The columns of any invertible matrix become a basis of the same matrix-dimension tuple vector space, in a way that the coordinates of any vector in terms of that basis becomes the solution of a linear equation system. Again from $6.3.3$ in the post above the solution is unique so:
\[\forall Y \in K^{n \times 1} \ \exists ! X \in K^{n \times 1} : \quad (PX = Y \iff X = P^{-1} Y)\]3.3.3. Basis of non-squared matrix vector spaces.
Let $A \in M_{m \times n}(K)$ and let $S := \Set{X \mid AX = 0}$ the solution on the homogeneous system. Take the $RREM$, $R \simeq A$, then, $RX=0$ share the space solution.
Observe that since $R$ is $RREM$, it has $r$ non-zero rows, which allow to clear $r$ unknowns in terms of $n-r$ unknowns, be $J$ the set of the index of the uncleared unknowns, then $RX = 0$ is equivalent to the system:
\[\begin{cases} x_1 = \displaystyle\sum_{i = 1}^J \alpha_{1i} x_i \\ \quad \vdots \\ x_r = \displaystyle\sum_{i = 1}^J \alpha_{ri} x_i\end{cases}\]All solution are retrieved by giving values to the dependent unknowns assigning arbitrary values to the independent unknows, those with and index in $J$.
Observe then that, be $j \in J$ then consider $E_j$ the solution by giving $x_j=1$ and $x_i=0$ for any other $i \in J$. Observe since the solutions are indeed linear combinations of the independent unknowns, then any arbitrary solution are is a linear combination of the family $\Set{E_j}_{j \in J}$ as described above and also is clear that are linearly independant so is a basis.
3.3.4. Infinite Basis.
Let's note first that an infinite basis do not introduce infinite sums, each vector gets obtained through a finite linear combination. Which is infinite is the generator engine, the basis it self which, algebraicly speacking, means that there is only no finite basis at all. The richness of infinite-dimensional theory (functional analysis, operators) lies not in the vector space structure itself, but in the additional structure superimposed upon it.
Take some subfield $W \subset \mathbb{C}$, and consider $Pol(W,W)$, remember:
\[Pol(W,W) := \Set{f \in W^W \mid \exists n \in \mathbb{N}:(\exists \alpha \in \mathbb{C}^n : f(x) = \sum_{i=0}^n \alpha_i x^i \quad \forall x \in W )}\]Note then that, for any $f \in Pol(W,W)$, it takes the form:
\[f(x) = \alpha_0 + \alpha_1 x + \ldots + \alpha_n x^n\]Let's call $f_n(x) = x^n$, then the family $\Set{f_n}_{n \in \mathbb{N}}$ is a basis over $Pol(W,W)$.
First, is clear that it can generate any element of $Pol(W,W)$, but let's also see that they are independent. We do know that a set is independent if any linear combination of his elements is $0$, it would be sufficent to proove that for any $n$, the finite family $\Set{f_0,f_1,\ldots,f_n}$ is independent, then, for some $n$ is:
\[\sum_{i=0}^n \alpha_i f_i = \alpha_0 + \alpha_1 x + \cdots + \alpha_n x^n = 0\]We assume that the reader knows that a polynomial of degree $n$ with complex coefficients cannot have more than $n$ distinct roots. It follows that $\alpha_0 = \cdots = \alpha_n = 0$.