ALG1_L12

Linear maps in ${\mathbb{R}}^n$ are matrices

Theorem 1(Theorem).

Any linear map $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$ is of the form $f_A(\mathbf{x})$ for some $q\times p$ matrix $A$.

Proof
Take $\mathbf{x}=(c_1,c_2,c_3,\dots ,c_p{)}^T=c_1{\mathbf{e}}_1+c_2{\mathbf{e}}_2+\cdots +c_p{\mathbf{e}}_p$.
Then, $T\left({\sum }_{i=1}^p{c}_i{\mathbf{e}}_i\right)={\sum }_{i=1}^p{c}_{i}T({\mathbf{e}}_i)$.
Define ${\mathbf{u}}_i\equiv T({\mathbf{e}}_i)$.
It follows that, $T(\mathbf{x})=\left[\begin{array}{cccc}{\mathbf{u}}_1& {\mathbf{u}}_2& \dots & {\mathbf{u}}_p\end{array}\right]\mathbf{x}$. ❏

Thus, the matrix $A$ corresponding to $T$ (the matrix $A$ such that $T=f_A$) is the matrix whose columns are the images of the standard basis of ${\mathbb{R}}^p$ under $T$ in the canonical order.

Note that every linear map $T:{\mathbb{R}}^p\to {\mathbb{R}}^q$ has a unique $q\times p$ matrix $A$ associated with it, i.e, there exists a bijection $\zeta :\mathcal{L}({\mathbb{R}}^p,{\mathbb{R}}^q)\to {\mathcal{M}}_{q\times p}(\mathbb{R})$, $\zeta (T)=\text{ matrix of }T$, where the domain of $\zeta$ is the set of all linear maps from ${\mathbb{R}}^p$ to ${\mathbb{R}}^q$ and the codomain is the set of all $q\times p$ matrices with real entries. We have seen that both are vector spaces. It can also be easily verified that $\zeta$ is itself a linear map. Thus, $\mathcal{L}({\mathbb{R}}^p,{\mathbb{R}}^q)\cong {\mathcal{M}}_{q\times p}(\mathbb{R})$.

Examples

Observe that the following maps are linear for geometric reasons. Find their matrices (In the standard basis, of course).

1. ${\mathbb{R}}^2\to {\mathbb{R}}^2$, rotation by $\theta$.

$$\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$

2. ${\mathbb{R}}^3\to {\mathbb{R}}^3$, rotation by $\theta$ around $x,y,z$ axis.

$$R_z=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]$$ $$R_y=\left[\begin{array}{ccc}\cos \theta & 0& -\sin \theta \\ 0& 1& 0\\ \sin \theta & 0& \cos \theta \end{array}\right]$$ $$R_x=\left[\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right]$$

3. ${\mathbb{R}}^2\to {\mathbb{R}}^2$, reflection about the $y=x$ line.

$$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$$

4. ${\mathbb{R}}^3\to {\mathbb{R}}^3$, reflection about the $XY$ plane.

$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& -1\end{array}\right]$$

5. Composition of 1 followed by 3, i.e, rotation by $\theta$ followed by reflection about $y=x$.

$$\left[\begin{array}{cc}\sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{array}\right]$$

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Matrices of linear maps in fdvsp

In lecture 8, we saw how any fdvsp $V$ of dimension $p$ is isomorphic to ${\mathbb{R}}^p$. We also saw what an isomorphism between $V$ and ${\mathbb{R}}^p$ looked like, given a basis of $V$. Now, consider a linear map $T:V\to W$, where $V$ and $W$ are finite dimensional vector spaces of dimensions $p$ and $q$ respectively. If we fix bases $B_V$ for $V$ and $B_W$ for $W$, we can define an isomorphism which maps elements of $V$ to their coordinate vectors in ${\mathbb{R}}^p$ (ditto for $W$):

$$\begin{array}{rl}\psi :V\to {\mathbb{R}}^p,& V\cong {\mathbb{R}}^p\\ \varphi :W\to {\mathbb{R}}^q,& W\cong {\mathbb{R}}^q\end{array}$$

With representations of elements of $V$ and $W$ in ${\mathbb{R}}^n$, we can construct a matrix for $T$. To do this, we first need to find a linear map which maps the coordinate vector of $\mathbf{v}\in V$ in ${\mathbb{R}}^p$ to the coordinate vector of $T(\mathbf{v})$ in ${\mathbb{R}}^q$. Then, we can express this map as a matrix as described in the first section.

$T$ takes in an element of $V$ and spits out an element of $W$. So, given a coordinate vector of $\mathbf{v}\in V$, we have to map it to $\mathbf{v}$, then apply $T$, then map the result to its coordinate vector in ${\mathbb{R}}^q$. Thus, the transformation we seek is $L=\varphi T{\psi }^{-1}$.

Finally, the matrix of $T$ in bases $B_V$ and $B_W$ (denoted ${\mathcal{M}}_{B_V,B_W}(T)$) is $\left[\begin{array}{cccc}L({\mathbf{e}}_1)& L({\mathbf{e}}_2)& \dots & L({\mathbf{e}}_p)\end{array}\right]$. It is a $q\times p$ matrix.

Change of basis

Let $B_V$ and $B_V^{\prime }$ be two bases for $V$. Similarly, let $B_W$ and $B_W^{\prime }$ be two bases for $W$. Let $T:V\to W$. Given ${\mathcal{M}}_{B_V,B_W}(T)$, how do we find ${\mathcal{M}}_{B_V^{\prime },B_W^{\prime }}(T)$?

Just as we did previously, all we need to do is change the “interface” of ${\mathcal{M}}_{B_V,B_W}(T)$. We can define the following isomorphisms:

$$\begin{array}{ll}\psi :V\to {\mathbb{R}}^p,& \psi (\mathbf{v})=\text{ coordinate vector of }\mathbf{v}\text{ in }{B}_V\\ {\psi }^{\prime }:V\to {\mathbb{R}}^p,& {\psi }^{\prime }(\mathbf{v})=\text{ coordinate vector of }\mathbf{v}\text{ in }{B}_V^{\prime }\\ & \\ \varphi :W\to {\mathbb{R}}^q,& \varphi (\mathbf{w})=\text{ coordinate vector of }\mathbf{w}\text{ in }{B}_W\\ {\varphi }^{\prime }:W\to {\mathbb{R}}^q,& {\varphi }^{\prime }(\mathbf{w})=\text{ coordinate vector of }\mathbf{w}\text{ in }{B}_W^{\prime }\end{array}$$

Then, ${\mathcal{M}}_{B_V^{\prime },B_W^{\prime }}(T)={\varphi }^{\prime }{\varphi }^{-1}{\mathcal{M}}_{B_V,B_W}(T)\psi {\psi }^{\prime -1}$.

$\psi {\psi }^{\prime -1}$ is a map from ${\mathbb{R}}^p\to {\mathbb{R}}^p$, so it must have a unique matrix representation, say $A$. Similarly, let $B$ be the matrix of $\varphi {\varphi }^{\prime -1}$. Note that $A$ is a $p\times p$ matrix, and $B$ is a $q\times q$ matrix. $A$ is called the change of basis matrix in $V$, and $B$ is called the change of basis matrix in $W$. Observe:

$$\begin{array}{rl}& (\text{coordinate vector of }\mathbf{v}\text{ in }{B}_V)=A(\text{coordinate vector of }\mathbf{v}\text{ in }{B}_V^{\prime })\\ & (\text{coordinate vector of }\mathbf{w}\text{ in }{B}_W)=B(\text{coordinate vector of }\mathbf{w}\text{ in }{B}_W^{\prime })\end{array}$$

Observe that the columns of $A$ are the coordinate vectors of the elements of $B_V^{\prime }$ in $B_V$. Similarly, the columns of $B$ are the coordinate vectors of the elements of $B_W^{\prime }$ in $B_W$.

Finally, we have ${\mathcal{M}}_{B_V^{\prime },B_W^{\prime }}(T)=B^{-1}{\mathcal{M}}_{B_V,B_W}(T)A$.