27 - Operations on Linear Maps

Combining Linear Transformations (addition + scalar multiplication)

Let $V,W$ be vector spaces over a field $\mathbf{F}$
For $S,T:V\to W$ and $\lambda \in \mathbb{F}$ then linear maps $S+T$ and $\lambda S$ are defined by

$$\begin{array}{rlr}S+T:V\to W& \text{ by }(S+T)(v)=S(v)+T(v)& \text{ for }v\in V\\ \lambda S:V\to W& \text{ by }(\lambda S)(v)=\lambda S(v)& \text{ for }v\in V\end{array}$$

With these operations (including the Zero map) the set of linear transformations $V\to W$ forms a vector space denoted $\text{Hom}(V,W)$

Proof for Linear Maps

Showing that $S+T$ is a linear map

$$\begin{array}{rl}(S+T)({\alpha }_1{v}_1+{\alpha }_2{v}_2)& =S({\alpha }_1{v}_1+{\alpha }_2{v}_2)+T({\alpha }_1{v}_1+{\alpha }_2{v}_2)[\text{by definition}]\\ & ={\alpha }_{1}S(v_1)+{\alpha }_{2}S(v_2)+{\alpha }_{1}T(v_1)+{\alpha }_{2}T(v_2)[\text{by linearity}]\\ & ={\alpha }_1(S(v_1)+T(v_2))+{\alpha }_2(S(v_2)+T(v_2))[\text{rearranging}]\\ & ={\alpha }_1(S+T)(v_1)+{\alpha }_2(S+T)(v_2)[\text{by defintion}]\end{array}$$

Hence $S+T$ is linear
Showing that $\lambda S$ is a linear map

$$\begin{array}{rl}(\lambda S)({\alpha }_1{v}_1+{\alpha }_2{v}_2)& =\lambda S({\alpha }_1{v}_1+{\alpha }_2{v}_2)\\ & =\lambda ({\alpha }_{1}S(v_1)+{\alpha }_{2}S(v_2))\\ & ={\alpha }_1(\lambda S(v_1))+{\alpha }_2(\lambda S(v_2))\\ & ={\alpha }_1(\lambda S)(v_1)+{\alpha }_2(\lambda S)(v_2)\end{array}$$

Hence $\lambda S$ is linear

Combining Linear Transformations (composition)

Let $U,V,W$ be vector spaces over $\mathbf{F}$
Let $S:U\to V$ and $T:V\to W$ be linear then

$$T\circ S:U\to W\text{ by }(T\circ S)(v)=T(S(v))$$

Note on notation

Sometimes $T\circ S$ is written as $TS$ but $T\circ S$ is removes any ambiguity

Proof for Linearity

Take $u_1,u_2\in U$ and ${\lambda }_1,{\lambda }_2\in \mathbb{F}$ then

$$\begin{array}{rl}(T\circ S)({\lambda }_2{u}_1+{\lambda }_2{u}_2)& =T(S({\lambda }_1{u}_1+{\lambda }_2{u}_2))\\ & =T({\lambda }_{1}S(u_1)+{\lambda }_{2}S(u_2))\\ & ={\lambda }_{1}T(S(u_1))+{\lambda }_{2}T(S(u_2))\\ & ={\lambda }_1(T\circ S)(u_1)+{\lambda }_2(T\circ S)(u_2)\end{array}$$

Hence $T\circ S$ is linear

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Invertible Linear Maps

Let $V,W$ be vector spaces and let $T:V\to W$ be linear
$T$ is invertible if there is linear transformation $S:W\to V$ such that

$$ST=i{d}_V\text{ and }TS=i{d}_W$$

where $i{d}_V$ and $i{d}_W$ are the identity maps on $V,W$ respectively
Then $S$ is called the inverse of $T$ (typically written as $T^{-1}$)

Note that invertible linear maps are an isomorphism (refer to groups)

Property of Bijective Linear Maps

Let $V,W$ be vector spaces
Let $T:V\to W$ be linear

$$T\text{ is invertible }⟺T\text{ is bijective}$$

Proof

If $T$ is invertible, then it is certainly bijective (from Intro to Uni Maths)

Assume that $T$ is bijective
Then $T$ has an inverse function $S:W\to V$ (but we need to show that $S$ is linear)
For $w_1,w_2\in W$ and ${\lambda }_1,{\lambda }_2\in \mathbb{F}$
Let $v_1=S(w_1)$ and $v_2=S(w_2)$ then

$$T(v_1)=TS(w_1)=w_1\text{ and }T(v_2)=TS(w_2)=w_2$$

Hence

$$\begin{array}{rl}S({\lambda }_1{w}_1+{\lambda }_2{w}_2)& =S({\lambda }_{1}T(v_1)+{\lambda }_{2}T(V_2))\\ & =S(T({\lambda }_1{v}_1+{\lambda }_2{v}_2))[\text{since }T\text{ is linear}]\\ & ={\lambda }_1{v}_1+{\lambda }_2{v}_2[\text{as }S\text{ is inverse to }T]\\ & ={\lambda }_{1}S(W_1)+{\lambda }_{2}S(w_2)\end{array}$$

So $S$ is linear

Inverse of Compositive Linear Maps

Let $U,V,W$ be vector spaces
Let $S:U\to V$ and $T:V\to W$ be invertible linear transformations
Then

$$TS:U\to W\text{ is invertible}\text{ and}(TS{)}^{-1}=S^{-1}{T}^{-1}$$

Dimension Property of Invertible Linear Maps

1. Let $V,W$ be vector spaces with $V$ finite-dimensional
   If there is an invertible linear map $T:V\to W$ then $\dim V=\dim W$
2. Let $V,W$ be finite-dimsional vector spaces with $\dim V=\dim W$
   Then there is invertible linear map $T:V\to W$

Consequently $V$ and $W$ are isomorphic if and only if $\dim V=\dim W$

Proof

Let $v_1,\cdots ,v_n$ be a basis for $V$
For $v\in V$ it can be uniquely expressed as

$$v={\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n$$

for some ${\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_n$

Suppose $T(v_1),\cdots ,T(v_n)$ is a basis for $W$

As $T:V\to W$ is invertible then for $w\in W$ there exists $v\in V$ such that $T(v)=w$
So

$$w=T(v)=T({\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n)={\alpha }_{1}T(v_1)+\cdots +{\alpha }_{n}T(v_n)$$

Hence $T(v_1),\cdots ,T(v_n)$ is spanning

Finding a solution to ${\alpha }_{1}T(v_1)+\cdots +{\alpha }_{n}T(v_n)=0_W$

$$\begin{array}{rl}{\alpha }_{1}T(v_1)+\cdots +{\alpha }_{n}T(v_n)& =0_W\\ T({\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n)& =0_W\\ {\alpha }_1{v}_1+\cdots +{\alpha }_n{v}_n& =0_V\end{array}$$

Hence ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_n=0$ as $v_1,\cdots ,v_n$ is a basis for $V$
Hence $T(v_1),\cdots ,T(v_n)$ is linearly independent
So $T(v_1),\cdots ,T(v_n)$ is a basis for $W$

Hence $\dim W=n=\dim V$

Suppose $n=\dim V=\dim W$
Let $v_1,\cdots ,v_n$ be a basis for $V$ and $w_1,\cdots ,w_n$ a basis for $W$
We need to show that

$$T:V\to W\text{ given by }T\left(\sum a_i{v}_i\right)=\sum {\alpha }_i{w}_i$$

is a well-defined, invertible linear map
TODO =_=

Uniqueness of Inverses of linear maps corollary

Let $V$ be a finite-dimensional vector space
Let $T:V\to V$ be linear
Then any one-sided inverse of $T$ is a two-sided inverse and hence the inverse of $T$ is unique

Proof

Suppose $T$ has a right inverse $S:V\to V$ so $T\circ S=i{d}_V$
Since $i{d}_V$ is surjective, $T$ is surjective, so $\mathrm{rank}(T)=\dim V$
So by Equivalent Statements for Rank and Nullity then $T$ is invertible
Suppose $T$ has a two-sided inverse $S^{\prime }$
Then

$$S^{\prime }=S^{\prime }\circ i{d}_V=S^{\prime }\circ (T\circ S)=(S^{\prime }\circ T)\circ S=i{d}_V\circ S=S$$

Hence $S$ is the unique two sided inverse