4.1 Vector Spaces and Subspaces

Vector Space

A nonempty set $V$of vectors satisfying the following conditions for $u,v \in V$:

Example: The set of polynomials $P_n$ of degree $n\geq 0$

This set has the zero polynomial

It is closed under vector addition

and scalar multiplication

Subspace

$H$ is a subspace of $V$ if it satisfies:

• The zero vector of $V$ is in $H$

• $H$ is closed under vector addition

• $H$ is closed under scalar multiplication

Example: $\{\vec 0 \} \sub V$

The set consisting of only the zero vector in a vector space $V$ is a subspace of $V$, called the zero subspace.

Theorem 1

If $v_1,\dots,v_p$ are in a vector space $V$, then $\text{Span}\{v_1,\dots,v_p\}$ is a subspace of $V$

Example: Let $H = \text{Span}\{v_1,v_2\}$

We know $H$ is a subspace of $V$ by Theorem 1.

Example: $\R^2$ is not a subspace of $\R^3$ because $\R^3$ is the set of $3$-dimensional vectors and $\R^2$ is not even a subset of $\R^3$.

Example: For what values of $h$ will $y= \begin{bmatrix} -4\\3\\h \end{bmatrix}$ be in the subspace of $\R^3$ spanned by the columns of $\begin{bmatrix} 1 & 5 & -3\\ -1 & -4 & 1\\ -2 & -7 & 0\\ \end{bmatrix}$?

This is equivalent to asking for what values of $h$ there exists a linear combination of the vectors that equals to $y$, which we know is equivalent to the existence of a solution $x$ to the linear equation $Ax=y$. So we just row reduce the augmented matrix $[A|b]$ and find that the system is consistent for $h=5$.

Questions

Let $W$ be the union of the first and third quadrants in the $xy$-plane. $W= \left\{ \begin{bmatrix} x\\ y \end{bmatrix} : xy \geq 0 \right\}$

a. Is $cu$ in $W$?

Yes.

b. Is $W$ a vector space?

No. It’s not closed under vector addition.

Is the following set a subspace of $P^n$: All polynomials of the form $p(t) = a +t^2$, $a\in\R$?

We know that a subspace $H\sub P_n$ must satisfy the following conditions:

• The zero vector of $P_n$ is in $H$

• Vector addition

• Scalar multiplication

We find that the zero vector $0 \in P_n$ is not in $H$ if $a \neq 0$.

Is the following set a subspace of $P_n$: All polynomials in $P_n$ such that $p(0)=0$?

This is the zero vector $\vec 0 \in P_n$, which we know to be a subset of $P_n$

Let $W$be the set of all vectors of the form $\begin{bmatrix} s +3t\\ s-t\\ 2s-t\\ 4t \end{bmatrix}$. Show that $W$ is a subspace of $\R^4$.

We can break this definition into the span of two vectors

By Theorem 1, we know that the span of vectors from $V$is a subspace of $V$

Is $W=\begin{bmatrix} -a + 1\\ a-6b\\ 2b+a \end{bmatrix}$ a vector space?

First, we express $W$ as the span of vectors

Now we check if this linear combination satisfies the three condition of vector spaces. Firstly, we find if the zero vector is in $W$. This is equivalent to asking if this transposed hyperplane goes through the origin such that $au+bv + (1,0,0) = \vec 0$. We find that the system $au + bv = -(1,0,0)$ does not have a solution, so the zero vector is not in $W$, thus it is not a vector space.

Let $F$ be a fixed $3\times 2$ matrix, and let $H$ be the set of all matrices $A$ in $M_{2\times 4}$ with the property that $FA= \bold 0 \in \R^{3\times 4}$. Determine if $H$ is a subspace of $M_{2\times 4}$.

$H$ is the null space of $F$, which makes it a subspace by Theorem 2.

4.2 Null Spaces, Column Spaces, Row Spaces, and Linear Transformations

Null Space (Kernel)

The null space of a matrix is the set of solutions $x$ to the homogenous system $Ax=\vec0$

Theorem 2

The null space of an $m \times n$ matrix $A$ is a subspace of $\R^n$ ($\dim{A}$).

Example: Let $H$ be the set of all vectors in $\R^4$ whose coordinates satisfy the equations $a - 2b +5c = d$ and $c-a = b$. Show that $H$ is a subspace of $\R^4$.

We can actually rearrange this system of equations like so:

Which is now the set of solutions to the homogeneous system $Ax=0$, which we know is a subspace of $n$ by Theorem 2.

Column Space (Range)

The column space of an $m \times n$ matrix $A$, $\text{Col}{A}$, is the set of all linear combinations of the columns of $A$, which is a subspace of $\R^m$

Theorem 3

The column space of an $m\times n$ matrix $A$ is a subspace of $\R^m$. The Span of $p$ vectors $v$ from $V$ is automatically a subspace of $V$.

Row Space

The set of all linear combinations of the row vectors is called the row space of $A$, $\text{Row}{A}$, which is a subspace of $\R^n$

Questions

Find an explicit definition of $\text{Nul}{A}$. $A = \begin{bmatrix}1 & -6 & 4 & 0\\0 & 0 & 2 & 0\end{bmatrix}$.

$\text{Nul}{A}$ is the set of solutions to the equation $Ax=0$. We can express this solution set in general form. First, we row reduce into RREF and get the following matrix:

Which gives us the following general solution in terms of the free variables $x_2$ and $x_4$

Thus, the null space is the span of the two vectors, and the nullity is $2$.

Either show that $W$ is a vector space, or prove the contrary.

We can rearrange the two equations into a homogeneous system:

$W$ is the set of all solutions to a homogeneous system, thus it is a subspace of $\R^4$ by Theorem 2.

Either show that $W$ is a vector space, or prove the contrary.

We can rewrite $W$ as a linear combination:

$W$ does not include the origin, thus it is not a subspace of $\R^4$.

4.3 Bases

Theorem 4

An indexed set $\{v_1,\dots,v_p\}$ of two or more vectors, with $v_q \neq 0$, is linearly dependent if and only if some $v_j$ is a linear combination of the preceding vectors.

Basis

A set of vectors $\mathcal{B}$ in $V$ is a basis for $H$ if the following two conditions are satisfied

• $\mathcal{B}$ is a linearly independent set

• $H = \text{Span}{\mathcal{B}}$

Theorem 5: The Spanning Set Theorem

Let $S=\{v_1,\dots,v_p\}$ be a set in a vector space $V$, and let $H=\text{Span}\{v_1,\dots,v_p\}$

• If $v_k$ in $S$ is a linear combination of the remaining vectors, $S-v_k$ still spans $H$

• If $H \neq \{0\}$, some subset of $S$ is a basis for $H$

Theorem 6

The pivot columns of a matrix $A$ form a basis for $\text{Col}{A}$.

Theorem 7

If two matrices $A$ and $B$ are row equivalent, then their row spaces are the same. If $B$ is in echelon form, the nonzero rows of $B$ form a basis for the row space of $A$ and $B$.

Example: Find a basis for the row space of $A$

We find the basis for $\text{Row}A$ by row-reducing to RREF. The pivot columns of $A$ form the basis for $\text{Row}A$.

Questions

Is the following set a basis for $\R^3$? $S = \left\{ \begin{bmatrix} 1\\2\\-3 \end{bmatrix}, \begin{bmatrix} -4\\-5\\6 \end{bmatrix} \right\}$

We know that a set $S$ must be linearly independent and span $H$ to be a basis for $H$. A set of two vectors obviously can’t span $\R^3$.

Example: Find bases for the null space and column space for $A = \begin{bmatrix} 1 & 0 & -5 & 1 & 4\\ -2 & 1 & 6 & -2 & -2\\ 0 & 2 & -8 & 1 & 9 \end{bmatrix}$

First, we row reduce to RREF:

Columns $1,2,4$ are the pivot columns thus

And we write $\text{Nul}A$ as the general form solution to $Ax=0$ as a linear combination of the free variables $x_3,x_5$

True or False: If $B$ is an echelon form of a matrix $A$, then the pivot columns of $B$ form a basis for $\text{Col}A$.

False. The pivot columns of the reduced matrix correspond to the pivot columns in the original set which form the basis

4.4 Coordinate Systems

Theorem 8: The Unique Representation Theorem

Let $\mathcal{B} = \{b_1,\dots,b_n\}$ be a basis for a vector space $V$. Then for each $x$ in $V$, there exists a unique set of scalars $c_1,\dots,c_n$ such that

where the coordinates of $x$ relative to basis $\mathcal{B}$

Using these coefficients $[x]_\mathcal{B}$, we can express the unique point $x$ as a linear combination of the basis vectors in $\mathcal{B}$:

Example: Let $\mathcal{B} =\{ \begin{bmatrix} 1\\0 \end{bmatrix}, \begin{bmatrix} -2\\3 \end{bmatrix} \}$ be a basis for $\R^2$, $[x]_\mathcal{B} = \begin{bmatrix} -2\\3 \end{bmatrix}$. Find $x$.

The vector $x$$x$$\mathcal{B}$$x$$[x]_\mathcal{B}$

Example: Let $\mathcal{B} = \left\{ \begin{bmatrix}2\\1\end{bmatrix}, \begin{bmatrix}-1\\1\end{bmatrix} \right\}$, $x=\begin{bmatrix}4\\5\end{bmatrix}$. Find the coordinate vector $[x]_\mathcal{B}$

We can solve the equation $x = \mathcal{B}[x]_\mathcal{B}$$[\mathcal{B}|x]$

Theorem 9

Let $\mathcal{B}=\{b_1,\dots,b_n\}$ be a basis for a vector space $V$. Then the coordinate mapping $x \mapsto [x]_\mathcal{B}$ is a one-to-one linear transformation from $V$ onto $\R^n$.

Isomorphism

Two vector spaces $V$ and $W$ are isomorphic when $\text{Col}V = \text{Col}W$. For example, $P_3$ is isomorphic to $\R^4$.

Questions

The set $\mathcal{B} = \{1-t^2,t-t^2,2-2t+t^2\}$ is a basis for $P_2$. Find the coordinate vector of $p(t) = 3+t-6t^2$ relative to $\mathcal{B}$.

Solve the system $x = \mathcal{B}[x]_\mathcal{B}$ for $[x]_\mathcal{B}$ by reducing the augmented matrix $[\mathcal{B}|x]$.

True or False? If $\mathcal{B}$ is the standard basis for $\R^n$, then the $\mathcal{B}$-coordinate vector of an $x$ in $\R^n$ is $x$ itself.

True. The standard basis for $\R^n$ is $[e_1,\dots,e_n]$, thus $\mathcal{B} = I$.

Let $p_1(t) = 1 + t^2, p_2(t) = t-3t^2, p_3(t) = 1 + t - 3t^2$.

a. Use coordinate vectors to show that these polynomials form a basis for $P_2$

Rewriting the given polynomials as vectors in $P_2$:

We find that these $3$ vectors are linearly independent and span $\R^3$, which is isomorphic to $P_2$. Thus, the vectors form a basis of $P_2$ by Theorem 13.

b. Consider the basis $\mathcal{B} = \{p_1,p_2,p_3\}$ for $P_2$. Find $q$ in $P_2$, given that $[q]_\mathcal{B} = \begin{bmatrix}-1\\1\\2\end{bmatrix}$

4.5 The Dimension of a Vector Space

Theorem 10

If a vector space $V$ has a basis $\mathcal{B}=\{b_1,\dots,b_n\}$, then any set in $V$ containing more than $n$ vectors must be linearly dependent

Theorem 11

If a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ must consist of exactly $n$ vectors.

Dimension

The dimension of a vector space, $\dim V$, is the number of vectors in a basis for $V$. $\dim\{\vec 0\}=0$

Example: Find the dimension of the subspace $H = \left\{ \begin{bmatrix} a-3b+6c\\ 5a+4d\\ b-2c-d\\ 5d \end{bmatrix} \right\}$

First, we rewrite this set of linear equations as a linear combination of vectors

Now we row-reduce to RREF and find that the pivot columns correspond to the basis vectors which define $\dim V$

Theorem 12

Let $H$ be a subspace of a finite-dimensional vector space $V$. Any linearly independent set in $H$ can be expanded to a basis for $H$, where $\dim{H} \leq \dim{V}$

Theorem 13: The Basis Theorem

Let $V$ be a $p$-dimensional vector space, $p\geq 1$. Any linearly independent set of exactly $p$ elements in $V$ is automatically a basis for $V$. Any set of exactly $p$ elements that spans $V$ is automatically a basis for $V$

Rank and Nullity

The rank of an $m \times n$ matrix $A$ is the dimension of the column space (number of pivot columns) and the nullity of $A$ is the dimension of the null space (number of free variables).

Theorem 14: The Rank Theorem

Thus, we can add the following properties to invertible matrices:

• The columns of $A$ form a basis for $\R^n$

• $\text{Col}A = \R^n$

• $\text{Rank}{A} =n$

• $\text{Nul}{A} = \{0\}$

Questions

Find a basis for $W$ and state its dimension

Firstly, we can express this set of linear equations as a linear combination of vectors

Now we row-reduce these columns as one matrix:

We find that there are $2$ pivot columns $1$ free variable. Thus the first $2$ columns form the basis.

Find a basis for $W$ and state its dimension.

$W$ is defined as one linear equation with $4$ variables. This corresponds to a homogeneous system with one row containing $1$ pivot column and $3$ free variables.

We can write the solution to this homogeneous equation in general form

The dimension of a homogeneous system is equivalent to the number of free variables (nullity).

Determine the dimensions of $\text{Nul}A, \text{Col}A, \text{Row}A$

Firstly, we reduce into RREF

Where we have $3$ pivot columns and $3$ free variables. We know that the dimension of a vector space is given by the number of vectors in its basis, given by the number of pivot columns and free variables.

True or False? The dimensions of the row space and the column space of $A$ are the same, even if $A$ is not square.

True. The dimension of the row and column space corresponds to the number of pivot entries, which is the same across columns and rows.

The first four Laguerre polynomials are $1,1-t,2-4t+t^2, 6-18t+9t^2-t^3$. Show that these polynomials form a basis of $P_3$.

First, we write these polynomials as vectors

Given that $P_3$ is isomorphic to $\R^4$, by the Basis Theorem we know that $n$ linearly independent vectors form a basis for $\R^n$. Thus these $4$ vectors automatically form a basis because they are linearly independent.

4.6 Change of Basis

Change-of-Coordinates Matrix

A change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$ is used to express the set of basis vectors in $\mathcal{B}$ as a linear combination of the basis vectors in $\mathcal{C}$ such that

Alternatively, this matrix can be used to take the coefficients $[x]_\mathcal{B}$ which represent the vector $x$ as a linear combination of the basis vectors in $\mathcal{B}$ and transform it into the coefficients $[x]_\mathcal{C}$.

Theorem 15

Let $\mathcal{B}=\{b_1,\dots,b_n\}$ and $\mathcal{C}=\{c_1,\dots,c_n\}$ be bases of $V$. Then there is a unique $n\times n$ change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}[x]_\mathcal{B}$ such that

where the change-of-coordinates matrix from $\mathcal{B}$ to $\mathcal{C}$ contains the coefficients to express each basis vector of $\mathcal{B}$ as a linear combination of the basis vectors in $\mathcal{C}$:

Think of the column vectors $[b_i]_\mathcal{C}$ in $P_{\mathcal{C} \leftarrow \mathcal{B}}$ as the coeffients that are used to express each basis vectors $b_i$ in $\mathcal{B}$ as a linear combination of basis vectors $c_i$ in $\mathcal{C}$.

We can say that applying change-of-coordinates mapping directly to the basis vectors of $\mathcal{C}$ transforms it into $\mathcal{B}$

This allows us to just directly solve for $P_{\mathcal{C} \leftarrow \mathcal{B}}$ via row reduction:

Example: Let $b_1 = \begin{bmatrix} -9\\1 \end{bmatrix}, b_2 = \begin{bmatrix} -5\\-1 \end{bmatrix}, c_1 = \begin{bmatrix} 1\\-4 \end{bmatrix}, c_2 = \begin{bmatrix} 3\\-5 \end{bmatrix}$. Find the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$

We solve the system $\mathcal{C}P_{\mathcal{C} \leftarrow \mathcal{B}} = \mathcal{B}$ via row reduction:

Thus:

Questions

Let $\mathcal{B}=\{b_1.b_2\}$ and $\mathcal{C}=\{c_1.c_2\}$ be bases for a vector space $V$, and let $b_1 = -c_1+4c_2, b_2=5c_1-3c_2$. Find the change-of-coordinates matrix $P_{\mathcal{C} \leftarrow \mathcal{B}}$. Find $[x]_\mathcal{C}$ for $x=5b_1+3b_2$