Linear Mappings and Bases

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The examples of linear mappings from $\R ^n\to \R ^m$ that we introduced in Section ?? were matrix mappings. More precisely, let $A$ be an $m\times n$ matrix. Then $L_A(x)=Ax$ defines the linear mapping $L_A:\R ^n\to \R ^m$ . Recall that $Ae_j$ is the $j^{th}$ column of $A$ (see Chapter ??, Lemma ??); it follows that $A$ can be reconstructed from the vectors $Ae_1,\ldots ,Ae_n$ . This remark implies (Chapter ??, Lemma ??) that linear mappings of $\R ^n$ to $\R ^m$ are determined by their values on the standard basis $e_1, \ldots , e_n$ . Next we show that this result is valid in greater generality. We begin by defining what we mean for a mapping between vector spaces to be linear.

Let $V$ and $W$ be vector spaces and let $L:V\to W$ be a mapping. The map $L$ is linear if

$\begin{eqnarray*} L(u+v) & = & L(u) + L(v) \\ L(cv) & = & cL(v) \end{eqnarray*}$

for all $u,v\in V$ and $c\in \R$ .

Examples of Linear Mappings

(a) Let $v\in \R ^n$ be a fixed vector. Use the dot product to define the mapping $L:\R ^n\to \R$ by $L(x)= x\cdot v.$ Then $L$ is linear. Just check that $L(x+y) = (x+y)\cdot v = x\cdot v + y\cdot v = L(x) + L(y)$ for every vector $x$ and $y$ in $\R ^n$ and $L(cx) = (cx)\cdot v = c(x\cdot v) = cL(x)$ for every scalar $c\in \R$ .

(b) The map $L:\CCone \to \R$ defined by $L(f) = f'(2)$ is linear. Indeed, $L(f+g) = (f+g)'(2) = f'(2) + g'(2) = L(f) + L(g).$ Similarly, $L(cf)=cL(f)$ .

(c) The map $L:\CCone \to \CCone$ defined by $L(f)(t)=f(t-1)$ is linear. Indeed, $L(f+g)(t) = (f+g)(t-1) = f(t-1) + g(t-1) = L(f)(t) + L(g)(t).$ Similarly, $L(cf)=cL(f)$ . It may be helpful to compute $L(f)(t)$ when $f(t)=t^2-t+1$ . That is, $L(f)(t) = (t-1)^2-(t-1)+1 = t^2-2t+1-t+1+1 = t^2-3t+3.$

Constructing Linear Mappings from Bases

Let $V$ and $W$ be vector spaces. Let $\{v_1,\ldots ,v_n\}$ be a basis for $V$ and let $\{w_1,\ldots ,w_n\}$ be $n$ vectors in $W$ . Then there exists a unique linear map $L:V\to W$ such that $L(v_i)=w_i$ .

Proof

Let $v\in V$ be a vector. Since $\Span \{v_1,\ldots ,v_n\}=V$ , we may write $v$ as $v = \alpha _1v_1 + \cdots + \alpha _nv_n,$ where $\alpha _1,\ldots ,\alpha _n$ in $\R$ . Moreover, $v_1,\ldots ,v_n$ are linearly independent, these scalars are uniquely defined. More precisely, if $\alpha _1v_1 + \cdots + \alpha _nv_n = \beta _1v_1 + \cdots + \beta _nv_n,$ then $(\alpha _1-\beta _1)v_1 + \cdots + (\alpha _n-\beta _n)v_n = 0.$ Linear independence implies that $\alpha _j-\beta _j=0$ ; that is $\alpha _j=\beta _j$ . We can now define $\begin{equation} \label{e:v-coord} L(v) = \alpha_1 w_1+\cdots+\alpha_n w_n. \end{equation}$

We claim that $L$ is linear. Let $\hat {v}\in V$ be another vector and let $\hat {v} = \beta _1v_1+\cdots +\beta _nv_n.$ It follows that $v+\hat {v} = (\alpha _1+\beta _1)v_1+\cdots +(\alpha _n+\beta _n)v_n,$ and hence by (??) that

$\begin{eqnarray*} L(v+\hat{v}) & = & (\alpha_1+\beta_1)w_1+\cdots+(\alpha_n+\beta_n)w_n\\ & = & (\alpha_1w_1+\cdots+\alpha_nw_n) + (\beta_1w_1+\cdots+\beta_nw_n) \\ & = & L(v) + L(\hat{v}). \end{eqnarray*}$

Similarly

$\begin{eqnarray*} L(cv) & = & L( (c\alpha_1)v_1+\cdots +(c\alpha_n)v_n)\\ & = & c(\alpha_1w_1+\cdots+\alpha_nw_n)\\ & = & cL(v). \end{eqnarray*}$

Thus $L$ is linear.

Let $M:V\to W$ be another linear mapping such that $M(v_i)=w_i$ . Then

$\begin{eqnarray*} L(v) & = & L(\alpha_1v_1+\ldots +\alpha_nv_n)\\ & = & \alpha_1w_1+\cdots+\alpha_nw_n \\ & = & \alpha_1M(v_1) + \cdots +\alpha_nM(v_n)\\ & = & M(\alpha_1v_1 + \cdots +\alpha_nv_n)\\ & = & M(v). \end{eqnarray*}$

Thus $L=M$ and the linear mapping is uniquely defined. $\blacksquare$

There are two assertions made in Theorem ??. The first is that a linear map exists mapping $v_i$ to $w_i$ . The second is that there is only one linear mapping that accomplishes this task. If we drop the constraint that the map be linear, then many mappings may satisfy these conditions. For example, find a linear map from $\R \to \R$ that maps $1$ to $4$ . There is only one: $y=4x$ . However there are many nonlinear maps that send $1$ to $4$ . Examples are $y=x+3$ and $y=4x^2$ .

Finding the Matrix of a Linear Map from $\R ^n\to \R ^m$ Given by Theorem ??

Suppose that $V=\R ^n$ and $W=\R ^m$ . We know that every linear map $L:\R ^n\to \R ^m$ can be defined as multiplication by an $m\times n$ matrix. The question that we next address is: How can we find the matrix whose existence is guaranteed by Theorem ???

More precisely, let $v_1,\ldots ,v_n$ be a basis for $\R ^n$ and let $w_1,\ldots ,w_n$ be vectors in $\R ^m$ . We suppose that all of these vectors are row vectors. Then we need to find an $m\times n$ matrix $A$ such that $Av_i^t=w_i^t$ for all $i$ . We find $A$ as follows. Let $v\in \R ^n$ be a row vector. Since the $v_i$ form a basis, there exist scalars $\alpha _i$ such that $v=\alpha _1 v_1 + \cdots + \alpha _n v_n.$ In coordinates

$\begin{equation} \label{e:v^t} v^t = (v_1^t|\cdots|v_n^t)\vect{\alpha}{n}, \end{equation}$ where $(v_1^t|\cdots |v_n^t)$ is an $n\times n$ invertible matrix. By definition (see (??)) $L(v) = \alpha _1 w_1 + \cdots + \alpha _n w_n.$ Thus the matrix $A$ must satisfy $Av^t = (w_1^t|\cdots |w_n^t)\vect {\alpha }{n},$ where $(w_1^t|\cdots |w_n^t)$ is an $m\times n$ matrix. Using (??) we see that $Av^t = (w_1^t|\cdots |w_n^t)(v_1^t|\cdots |v_n^t)\inv v^t,$ and $\begin{equation} \label{e:defA} A = (w_1^t|\cdots|w_n^t)(v_1^t|\cdots|v_n^t)\inv \end{equation}$ is the desired $m\times n$ matrix.

An Example of a Linear Map from $\R ^3$ to $\R ^2$

As an example we illustrate Theorem ?? and (??) by defining a linear mapping from $\R ^3$ to $\R ^2$ by its action on a basis. Let $v_1=(1,4,1)\quad v_2=(-1,1,1) \quad v_3=(0,1,0).$ We claim that $\{v_1,v_2,v_3\}$ is a basis of $\R ^3$ and that there is a unique linear map for which $L(v_i)=w_i$ where $w_1=(2,0) \quad w_2=(1,1) \quad w_3=(1,-1).$

We can verify that $\{v_1,v_2,v_3\}$ is a basis of $\R ^3$ by showing that the matrix $(v_1^t|v_2^t|v_3^t) = \left (\begin {array}{rrr} 1 & -1 & 0 \\ 4 & 1 & 1 \\ 1 & 1 & 0 \end {array}\right )$ is invertible. This can either be done in MATLAB using the inv command or by hand by row reducing the matrix

$\left (\begin {array}{rrr|ccc} 1 & -1 & 0 & 1 & 0 & 0 \\ 4 & 1 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 \end {array}\right )$ to obtain $(v_1^t|v_2^t|v_3^t)\inv = \frac {1}{2}\left (\begin {array}{rrr} 1 & 0 & 1 \\ -1 & 0 & 1 \\ -3 & 2 & -5 \end {array}\right ).$ Now apply (??) to obtain $A = \frac {1}{2} \left (\begin {array}{rrr} 2 & 1 & 1\\ 0 & 1 & -1 \end {array}\right ) \left (\begin {array}{rrr} 1 & 0 & 1 \\ -1 & 0 & 1 \\ -3 & 2 & -5 \end {array}\right ) = \left (\begin {array}{rrr} -1 & 1 & -1 \\ 1 & -1 & 3 \end {array}\right ).$ As a check, verify by matrix multiplication that $Av_i=w_i$ , as claimed.

Properties of Linear Mappings

Let $U,V,W$ be vector spaces and $L:V\to W$ and $M:U\to V$ be linear maps. Then $L\compose M :U\to W$ is linear.

Proof: The proof of Lemma ?? is identical to that of Chapter ??, Lemma ??. $\blacksquare$

A linear map $L:V\to W$ is invertible if there exists a linear map $M:W\to V$ such that $L\compose M:W\to W$ is the identity map on $W$ and $M\compose L:V\to V$ is the identity map on $V$ .

Let $V$ and $W$ be finite dimensional vector spaces and let $v_1,\ldots ,v_n$ be a basis for $V$ . Let $L:V\to W$ be a linear map. Then $L$ is invertible if and only if $w_1,\ldots ,w_n$ is a basis for $W$ where $w_j=L(v_j)$ .

Proof

If $w_1,\ldots ,w_n$ is a basis for $W$ , then use Theorem ?? to define a linear map $M:W\to V$ by $M(w_j)=v_j$ . Note that $L\compose M(w_j)= L(v_j) =w_j.$ It follows by linearity (using the uniqueness part of Theorem ??) that $L\compose M$ is the identity of $W$ . Similarly, $M\compose L$ is the identity map on $V$ , and $L$ is invertible.

Conversely, suppose that $L\compose M$ and $M\compose L$ are identity maps and that $w_j=L(v_j)$ . We must show that $w_1,\ldots ,w_n$ is a basis. We use Theorem ?? and verify separately that $w_1,\ldots ,w_n$ are linearly independent and span $W$ .

If there exist scalars $\alpha _1,\ldots ,\alpha _n$ such that $\alpha _1w_1+\cdots +\alpha _nw_n = 0,$ then apply $M$ to both sides of this equation to obtain $0=M(\alpha _1w_1+\cdots +\alpha _nw_n)=\alpha _1v_1+\cdots +\alpha _nv_n.$ But the $v_j$ are linearly independent. Therefore, $\alpha _j=0$ and the $w_j$ are linearly independent.

To show that the $w_j$ span $W$ , let $w$ be a vector in $W$ . Since the $v_j$ are a basis for $V$ , there exist scalars $\beta _1,\ldots ,\beta _n$ such that $M(w) = \beta _1v_1+\cdots +\beta _nv_n.$ Applying $L$ to both sides of this equation yields $w = L\compose M(w) = \beta _1w_1+\cdots +\beta _nw_n.$ Therefore, the $w_j$ span $W$ . $\blacksquare$

Let $V$ and $W$ be finite dimensional vector spaces. Then there exists an invertible linear map $L:V\to W$ if and only if $\dim (V)=\dim (W)$ .

Proof: Suppose that $L:V\to W$ is an invertible linear map. Let $v_1,\ldots ,v_n$ be a basis for $V$ where $n=\dim (V)$ . Then Theorem ?? implies that $L(v_1),\ldots ,L(v_n)$ is a basis for $W$ and $\dim (W)=n=\dim (V)$ .
Conversely, suppose that $\dim (V)=\dim (W)=n$ . Let $v_1,\ldots ,v_n$ be a basis for $V$ and let $w_1,\ldots ,w_n$ be a basis for $W$ . Using Theorem ?? define the linear map $L:V\to W$ by $L(v_j)=w_j$ . Theorem ?? states that $L$ is invertible. $\blacksquare$

Exercises

Use the method described above to construct a linear mapping $L$ from $\R ^3$ to $\R ^2$ with $L(v_i)=w_i$ , $i=1,2,3$ , where $v_1=(1,0,2)\quad v_2=(2,-1,1) \quad v_3=(-2,1,0)$ and $w_1=(-1,0) \quad w_2=(0,1) \quad w_3=(3,1).$

Let ${\cal P}_n$ be the vector space of polynomials $p(t)$ of degree less than or equal to $n$ . Show that $\{1,t,t^2,\ldots ,t^n\}$ is a basis for ${\cal P}_n$ .

Show that $\frac {d}{dt}:{\cal P}_3\to {\cal P}_2$ is a linear mapping.

Show that $L(p) = \int _0^tp(s)ds$ is a linear mapping of ${\cal P}_2\to {\cal P}_3$ .

Use Exercises ??, ?? and Theorem ?? to show that $\frac {d}{dt}\compose L:{\cal P}_2\to {\cal P}_2$ is the identity map.

Let $\C$ denote the set of complex numbers. Verify that $\C$ is a two-dimensional vector space. Show that $L:\C \to \C$ defined by $L(z) = \lambda z,$ where $\lambda =\sigma +i\tau$ is a linear mapping.

Let ${\cal M}(n)$ denote the vector space of $n\times n$ matrices and let $A$ be an $n\times n$ matrix. Let $L:{\cal M}(n)\to {\cal M}(n)$ be the mapping defined by $L(X)=AX-XA$ where $X\in {\cal M}(n)$ . Verify that $L$ is a linear mapping. Show that the null space of $L$ , $\{X\in {\cal M}:L(X)=0\}$ , is a subspace consisting of all matrices that commute with $A$ .

Let $L:\CCone \to \R$ be defined by $L(f) = \int _0^{2\pi }f(t)\cos (t)dt$ for $f\in \CCone$ . Verify that $L$ is a linear mapping.

Let $\cal P$ be the vector space of polynomials in one variable $x$ . Define $L:{\cal P}\to {\cal P}$ by $L(p)(x)=\int _0^x(t-1)p(t)dt$ . Verify that $L$ is a linear mapping.