LEC ALG1 13

Recall

Let $T:V\to W$ is a linear map, with $B_V$ and $B_W$ being bases of $V$ and $W$ respectively. The matrix of $T$ in the given bases is the matrix whose columns are coordinate vectors of $T(B_V)$ with respect to the basis $B_W$.

Example 79.1(Example).

Let $T:V\to W$ be the derivative map where $V={\mathbb{P}}_3$ and $W={\mathbb{P}}_2$. Let the bases be

$$B_V=\{1,x,x^2,x^3\}{B}_W=\{1,x,x^2\}$$

Observe that:

$$T(1)=0T(x)=1T(x^2)=2xT(x^3)=3x^2$$

We can now write the matrix of $T$ in the bases $B_V$ and $B_W$ denoted by ${\mathcal{M}}_{B_V,B_W}(T)$ as (note the implicit establishment of isomorphisms to ${\mathbb{R}}^n$) :

$${\mathcal{M}}_{B_V,B_W}(T)=\left[\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right]$$

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Homomorphisms

Definition 79.2(Definition).

A homomorphism is a structure preserving map between two algebraic structures of the same type.

Contrast with isomorphism, which is a structure preserving map between two structures of the same type which can be reversed by an inverse mapping.

A linear map is a homomorphism of vector spaces.

$\mathrm{hom}(V,W)\equiv \{\text{all linear maps}T:V\to W\}$.

Relation to Matrices

For two vector spaces $V$ and $W$ with dimensions $p$ and $q$ and $T:V\to W$, ${\mathcal{M}}_{B_V,B_W}(T)$ is a $q\times p$ matrix, as we have seen. As every $T$ has a single matrix for a fixed set of bases, ${\mathcal{M}}_{B_V,B_W}$ is a function. That is,

$${\mathcal{M}}_{B_V,B_W}:\mathrm{hom}(V,W)\to \{q\times p\text{ matrices}\}.$$

Exercise: Prove that ${\mathcal{M}}_{B_V,B_W}$ is an isomorphism of vector spaces. Solution here
Thus, $\mathrm{hom}(V,W)\cong \{q\times p\text{matrices}\}$. Therefore $\dim \mathrm{hom}(V,W)=pq$.

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Change of basis

(Ref. 2.4 Hoffman)

Let $V$ be a vector space of dimension $n$. Now, let

$$\begin{array}{rl}\text{old basis}{B}_V& \equiv \{{\mathbf{v}}_1,{\mathbf{v}}_2,\dots {\mathbf{v}}_n\}\\ \text{new basis}{\tilde{B}}_V& \equiv \{{\mathbf{w}}_1,{\mathbf{w}}_2,\dots {\mathbf{w}}_n\}\end{array}$$

Each ${\mathbf{w}}_j$ permits representation as a linear combination of $B_V$.

$${\mathbf{w}}_j=\left[\begin{array}{cccc}{\mathbf{v}}_1& {\mathbf{v}}_2& \dots & {\mathbf{v}}_n\end{array}\right]\left[\begin{array}{c}{p}_{1,j}\\ p_{2,j}\\ ⋮\\ p_{n,j}\end{array}\right],$$

where the left “row vector of vectors” is called a hyper vector.
Using the rule for matrix multiplication, we have

$$\left[\begin{array}{cccc}{\mathbf{w}}_1& {\mathbf{w}}_2& \dots & {\mathbf{w}}_n\end{array}\right]=\left[\begin{array}{cccc}{\mathbf{v}}_1& {\mathbf{v}}_2& \dots & {\mathbf{v}}_n\end{array}\right]\left[\begin{array}{cccc}{p}_{1,1}& p_{1,2}& \dots & p_{1,n}\\ p_{2,1}& p_{2,2}& \dots & p_{2,n}\\ ⋮& ⋮& \ddots & ⋮\\ p_{n,1}& p_{n,2}& \dots & p_{n,n}\end{array}\right].$$

Denote the hyper vector of the ${\mathbf{v}}_i$‘s as $A$ and that of the ${\mathbf{w}}_i$‘s as $B$. The above will become $B=AP$. Fix an arbitrary vector $\mathbf{u}\in V$. $\mathbf{u}$ will have a coordinate vector in $B_V$ (call it $\mathbf{c}$) and one in $B_W$ (call it ${\mathbf{c}}^{\prime }$) such that $\mathbf{u}=A\mathbf{c}=B{\mathbf{c}}^{\prime }$.

$$\begin{array}{rl}\mathbf{u}=A\mathbf{c}& =B{\mathbf{c}}^{\prime }\\ A\mathbf{c}& =AP{\mathbf{c}}^{\prime }\\ \mathbf{c}& =P{\mathbf{c}}^{\prime }\end{array}$$

Thus,

$$\begin{array}{rl}\text{old coordinates}& =(\text{change of basis matrix})(\text{new coordinates}),\\ \text{new basis}& =(\text{old basis})(\text{change of basis matrix}).\end{array}$$

Example 79.3(Example).

Let $T:V\to W$ be a linear transformation. We aim to find ${\mathcal{M}}_{B_V^{\prime },B_W^{\prime }}(T)$ given that ${\mathcal{M}}_{B_V,B_W}(T)=A$.

$$\begin{array}{cccc}& V& & W\\ \text{old basis}& B_V& \stackrel{A}{⟶}& B_W\\ & P\uparrow & & Q\uparrow \\ \text{new basis}& B_V^{\prime }& & B_W^{\prime }\end{array}$$

Fix a vector $\mathbf{v}\in V$ and let $\mathbf{x}$ be its “new” (given) coordinates. Now

$$\begin{array}{rl}\text{New coordinates of v}& =\mathbf{x}\\ \text{Old coordinates of v}& =P\mathbf{x}\\ \text{Transformed coordinates in old basis}& =AP\mathbf{x}\\ \text{Transformed coordinates in new basis}& =Q^{-1}AP\mathbf{x}\end{array}$$

Therefore,

$${\mathcal{M}}_{B_V^{\prime },B_W^{\prime }}(T)=Q^{-1}{\mathcal{M}}_{B_V,B_W}(T)P$$

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Composition of linear maps in terms of matrices

$$\begin{array}{cccccc}& V& \stackrel{T}{⟶}& W& \stackrel{S}{⟶}& U\\ \text{dim}& p& & q& & r\\ \text{basis}& B_V& & B_W& & B_U\end{array}$$

Theorem 79.4(Theorem).

${\mathcal{M}}_{B_V,B_U}(S\circ T)={\mathcal{M}}_{B_W,B_U}(S){\mathcal{M}}_{B_V,B_W}(T)$

Proof.

Let

$$\begin{array}{rlr}{A}_{q\times p}& ={\mathcal{M}}_{B_V,B_W}(T)& =(a_{ij})\\ B_{r\times q}& ={\mathcal{M}}_{B_W,B_U}(S)& =(b_{ij})\\ C_{r\times p}& ={\mathcal{M}}_{B_V,B_U}(S\circ T)& =(c_{ij})\end{array}$$

Recall the representation of linear transformations as matrices.

$$c_{i,j}=\text{Coefficient of}{\mathbf{u}}_{\mathbf{i}}\text{in}(S\circ T)({\mathbf{v}}_{\mathbf{j}}).$$

Now,

$$\begin{array}{rl}(S\circ T)({\mathbf{v}}_{\mathbf{j}})& =S\left(\sum _{\nu =1}^q{a}_{\nu ,j}{\mathbf{w}}_{\nu }\right)\\ & =\sum _{\nu =1}^q{a}_{\nu ,j}S({\mathbf{w}}_{\nu })\\ & =\sum _{\nu =1}^q{a}_{\nu ,j}\sum _{\lambda =1}^r{b}_{\lambda ,\nu }{\mathbf{u}}_{\lambda }.\end{array}$$

As we want the coefficient of ${\mathbf{u}}_{\mathbf{i}}$, we will only take the $\lambda =i$ term from the second sum. That is

$$c_{ij}=\sum _{\nu =1}^q{a}_{\nu ,j}{b}_{i,\nu }=(AB{)}_{ij}.$$ □

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A “Good” basis for a linear map

Let $T:V\to W$ be a linear map with $\dim V=p$ and $\dim W=q$. We want to choose bases $B_V$ and $B_W$ such that ${\mathcal{M}}_{B_V,B_W}(T)$ is as simple as possible.

Let $\text{rank}T=r$. Now, take a basis of $\ker T$, $\{{\mathbf{k}}_1,\dots {\mathbf{k}}_{p-r}\}$ (Recall the rank nullity theorem). Extend to get a basis of $V$, $B_V=\{{\mathbf{k}}_1,\dots {\mathbf{k}}_{p-r},{\mathbf{v}}_1,\dots {\mathbf{v}}_r\}$. We have shown that $T({\mathbf{v}}_1)\dots T({\mathbf{v}}_r)$ are a basis of $\text{Im}T$ (See proof of Thm 77.2). Extend this to a basis of $W$. $B_W=\{T({\mathbf{k}}_1)\dots T({\mathbf{k}}_r),{\mathbf{w}}_1,\dots {\mathbf{w}}_{q-r}\}$. Now ${\mathcal{M}}_{B_V,B_W}(T)$ will be

$$\left[\begin{array}{cccccccccc}0& 0& 0& \dots & 0& 1& 0& 0& \dots & 0\\ 0& 0& 0& \dots & 0& 0& 1& 0& \dots & 0\\ 0& 0& 0& \dots & 0& 0& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 0& 0& 0& 0& \dots & 1\\ 0& 0& 0& \dots & 0& 0& 0& 0& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 0& 0& 0& 0& \dots & 0\end{array}\right].$$

More concisely, we can write

$$\left[\begin{array}{cc}{0}_{r\times (p-r)}& I_{r\times r}\\ 0_{(q-r)\times (p-r)}& 0_{(q-r)\times r}\end{array}\right].$$