LADR Exercises 3A

Problem 1

Suppose $b,c\in \mathbb{R}$. Define $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ by

$$T(x,y,z)=(2x-4y+3z+b,6x+cxyz)$$

Show that $T$ is linear if and only if $b=c=0$.

Recall that a linear map have additivity and homogeneity. Let’s say we have $x,y\in {\mathbb{R}}^3$ where $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3)$.

Then, for additivity, we would need

$$\begin{array}{rl}T(x+y)& =Tx+Ty\\ T(x_1+y_1,x_2+y_2,x_3+y_3)& =T(x_1,x_2,x_3)+T(y_1,y_2,y_3)\end{array}$$

The left side gives us

$$\begin{array}{rl}(2(x_1+y_1)& -4(x_2+x_2)+3(x_3+y_3)+b,6(x_1+y_1)+c(x_1+y_1)(x_2+y_2)(x_3+y_3))\\ & =2x_1+2y_1-4x_2+y_2+3x_3+3y_3+b,\\ & 6x_1+6y_1+c(x_1{x}_2{x}_3+x_1{x}_2{y}_3+x_1{y}_2{x}_3+\\ & x_1{y}_2{y}_3+y_1{x}_2{x}_3+y_1{x}_2{y}_3+y_1{y}_2{x}_3+y_1{y}_2{y}_3)\end{array}$$

The right side gives us:

$$\begin{array}{rl}& (2x_1-4x_2+3x_3+b,6x_1+c{x}_1{x}_2{x}_3)+(2y_1-4y_2+3y_3+b,6y_1+c{y}_1{y}_2{y}_3)\\ & =(2x_1+2y_1-4x_2-4y_2+3x_3+3y_3+2b,\\ & 6x_1+6y_1+c{x}_1{x}_2{x}_3+c{y}_1{y}_2{y}_3).\end{array}$$

For the two of them to be equal

$$\begin{array}{rl}2x_1+2y_1-4x_2+y_2+3x_3+3y_3+b& =2x_1+2y_1-4x_2-4y_2+3x_3+3y_3+2b\\ b& =2b\\ \therefore b& =0\end{array}$$

and

$$\begin{array}{rl}& 6x_1+6y_1+c(x_1{x}_2{x}_3+x_1{x}_2{y}_3+x_1{y}_2{x}_3+x_1{y}_2{y}_3+y_1{x}_2{x}_3+y_1{x}_2{y}_3+y_1{y}_2{x}_3+y_1{y}_2{y}_3)\\ & =6x_1+6y_1+c{x}_1{x}_2{x}_3+c{y}_1{y}_2{y}_3\end{array}$$

For this to be true, will also need $c=0$. Thus, additivity needs $b=c=0$.

To confirm homogeneity, we compare $T(\lambda v)$ and $\lambda (Tv)$ for all $\lambda \in \mathbb{F}$ and $v\in {\mathbb{R}}^3$.

$$\begin{array}{rl}T(\lambda v)=T(\lambda x,\lambda y,\lambda z)& =(2\lambda x-4\lambda y+3\lambda z+b,6\lambda x+c\lambda x\lambda y\lambda z)\end{array}$$

and

$$\begin{array}{rl}\lambda (Tv)=\lambda (T(x,y,z))& =\lambda (2x-4y+3z+b,6x+cxyz)\\ & =2\lambda x-4\lambda y+3\lambda z+b\lambda ,6\lambda x+\lambda cxyz\end{array}$$

Thus, for $T(\lambda v)=\lambda (Tv)$, we must have

$$\begin{array}{rl}2\lambda x-4\lambda y+3\lambda z+b& =2\lambda x-4\lambda y+3\lambda z+b\lambda \\ b& =b\lambda \\ b& =0\text{ if }\lambda \ne 1\end{array}$$

and

$$\begin{array}{r}c\lambda x\lambda y\lambda z=\lambda cxyz\\ c{\lambda }^{3}xyz=c\lambda xyz\end{array}$$

which also requires $c=0$ if $\lambda \ne 1$.

Thus, for both additivity and homogeneity, we need $b=c=0$.

Problem 2

Suppose $b,c\in \mathbb{R}$. Define $T:\mathcal{P}(\mathbb{R})\to {\mathbb{R}}^2$ by

$$Tp=\left(3p(4)+5p^{\prime }(6)+bp(1)p(2),{\int }_{-1}^2{x}^{3}p(x)dx+c\sin p(0)\right)$$

Show that $T$ is linear if and only if $b=c=0$.

Let us have two polynomials $f,g\in \mathcal{P}(\mathbb{R})$. For additivity, we need to have

$$T(f+g)=Tf+Tg$$

The left side gives:

$$\begin{array}{rl}(& 3(f+g)(4)+5(f^{\prime }+g^{\prime })(6)+b(f(1)+g(1))(f(2)+g(2)),\\ & {\int }_{-1}^2{x}^3(f(x)+g(x))dx+c\sin ((f+g)(0)))\\ =(& 3f(4)+3g(4)+5f^{\prime }(6)+5g^{\prime }(6)+b(f(1)+g(1))(f(2)+g(2)),\\ & {\int }_{-1}^2{x}^3(f(x)+g(x))dx+c\sin ((f+g)(0)))\end{array}$$

The right side gives:

$$\begin{array}{rl}& \left(3f(4)+5f^{\prime }(6)+bf(1)f(2),{\int }_{-1}^2{x}^{3}f(x)dx+c\sin f(0)\right)\\ & +\left(3g(4)+5g^{\prime }(6)+bg(1)g(2),{\int }_{-1}^2{x}^{3}g(x)dx+c\sin g(0)\right)\\ & =(3f(4)+3g(4)+5f^{\prime }(6)+5g^{\prime }(6)+bf(1)f(2)+bg(1)g(2),\\ & {\int }_{-1}^2{x}^{3}f(x)dx+{\int }_{-1}^2{x}^{3}f(x)dx+c\sin f(0)+c\sin g(0))\end{array}$$

For the left and right sides to be equal, we need

$$\begin{array}{rl}b(f(1)+g(1))(f(2)+g(2))& =bf(1)f(2)+bg(1)g(2)\\ bf(1)f(2)+bf(1)g(2)+bg(1)f(2)+bg(1)g(2)& =bf(1)f(2)+bg(1)g(2)\\ \therefore b& =0\end{array}$$

We also need

$$\begin{array}{rl}c\sin ((f+g)(0))& =c\sin f(0)+c\sin g(0)\\ c\sin (f(0)+g(0))& =c\sin f(0)+c\sin g(0)\\ \therefore c& =0\text{ is the only universal solution}\end{array}$$

We can also check homogeneity

$$T(\lambda f)=\lambda (Tf)$$

The left side gives:

$$\left(3\lambda p(4)+5\lambda p^{\prime }(6)+b\lambda p(1)\lambda p(2),{\int }_{-1}^2{x}^3\lambda p(x)dx+c\sin \lambda p(0)\right)$$

The right side gives:

$$\lambda \left(3p(4)+5p^{\prime }(6)+bp(1)p(2),{\int }_{-1}^2{x}^{3}p(x)dx+c\sin p(0)\right)$$

Then, we have

$$\begin{array}{rl}b\lambda p(1)\lambda p(2)& =\lambda bp(1)p(2)\\ b{\lambda }^{2}p(1)p(2)& =\lambda bp(1)p(2)\end{array}$$

so the only solution is for all $\lambda$ is $b=0$.

And we also have

$$c\sin \lambda p(0)=\lambda c\sin p(0)$$

which is also only true for all $\lambda$ is $c=0$.

Problem 3

Suppose that $T\in \mathcal{L}({\mathbb{F}}^n,{\mathbb{F}}^m)$. Show that there exist scalars $A_{j,k}\in \mathbb{F}$ for $j=1,\dots ,m$ and $k=1,\dots ,n$ such that

$$T(x_1,\dots ,x_n)=(A_{1,1}{x}_1+\cdots +A_{1,n}{x}_n,\dots ,A_{m,1}{x}_1+\cdots +A_{m,n}{x}_n)$$

for every $(x_1,\dots ,x_n)\in {\mathbb{F}}^n$.

Note that this problem shows the 2nd last example from Linear Map.

We first determine $T$ using its action on the standard basis vectors of ${\mathbb{F}}^n$, denoted as $e_1,e_2,\dots ,e_n$, where $e_k$ is the vector with a $1$ in the $k$-th position and $0$ elsewhere. For each $k\in \{1,,\dots ,n\}$, the transformation of $e_k$ by $T$ is a vector in ${\mathbb{F}}^m$, denoted as

$$T(e_k)=(A_{1,k},A_{2,k},\dots ,A_{m,k})\in {\mathbb{F}}^m$$

where $A_{j,k}\in \mathbb{F}$ for each $j=1,\dots ,m$.

We can write an arbitrary vector $(x_1,\dots ,x_n)\in {\mathbb{F}}^n$ as a linear combination of the basis vectors $e_1,\dots ,e_n$:

$$(x_1,\dots ,x_n)=x_1{e}_1+\cdots +x_n{e}_n$$

By linearity of $T$, we have

$$T(x_1{e}_1+\cdots +x_n{e}_n)=x_{1}T(e_1)+\cdots +x_{n}T(e_n)$$

Substituting $T(e_k)=(A_{1,k},A_{2,k},\dots ,A_{m,k})$, we have

$$T(x_1,\dots ,x_n)=x_1(A_{1,1},\dots ,A_{m,1})+\cdots +x_n(A_{1,n},\dots ,A_{m,n})$$

If we consider the $j$-th component of $T(x_1,\dots ,x_n)$, we have

$$x_1{A}_{j,1}+x_2{A}_{j,2}+\cdots +x_n{A}_{j,n}$$

Thus, we have

$$T(x_1,\dots ,x_n)=(A_{1,1}{x}_1+\cdots +A_{1,n}{x}_n,\dots ,A_{m,1}{x}_1+\cdots +A_{m,n}{x}_n)$$

We can look at an example of $T:{\mathbb{F}}^3\to {\mathbb{F}}^2$ so that $T$ takes vectors from ${\mathbb{F}}^3$ of the form $(x_1,x_2,x_3)$ and maps them to vectors from ${\mathbb{F}}^2$ of the form $(y_1,y_2)$.

We’ll define $T$ by specifying how it acts on the basis standard vectors of ${\mathbb{F}}^3$, $e_1=(1,0,0)$, $e_2=(0,1,0)$, and $e_3=(0,0,1)$:

$$\begin{array}{r}T(e_1)=(1,2)\\ T(e_2)=(3,4)\\ T(e_3)=(5,6)\end{array}$$

For a general vector $(x_1,x_2,x_3)\in {\mathbb{F}}^3$, we can have:

$$\begin{array}{rl}(x_1,x_2,x_3)& =x_1{e}_1+x_2{e}_2+x_3{e}_3\\ T(x_1,x_2,x_3)& =x_{1}T(e_1)+x_{2}T(e_2)+x_{3}T(e_3)\\ & =x_1(1,2)+x_2(3,4)+x_3(5,6)\end{array}$$

Now we can compute the two components corresponding to ${\mathbb{F}}^2$

• First component:

$$x_1(1)+x_2(3)+x_3(5)=x_1+3x_2+5x_3$$

• Second component:

$$x_1(2)+x_2(4)+x_3(6)=2x_1+4x_2+6x_3$$

Here, the scalars $A_{j,k}$ are

• $A_{1,1}=1,A_{1,2}=3,A_{1,3}=5$
• $A_{2,1}=2,A_{2,2}=4,A_{2,3}=6$

So:

$$\begin{array}{rl}T(x_1,x_2,x_3)& =(A_{1,1}{x}_1+A_{1,2}{x}_2+A_{1,3}{x}_3,A_{2,1}{x}_1+A_{2,2}{x}_2+A_{2,3}{x}_3)\\ T(1,2,3)& =(1(1)+3(2)+5(3),2(1)+4(2)+6(3))=(22,32)\end{array}$$

Problem 4

Suppose $T\in \mathcal{L}(V,W)$ and $v_1,\dots ,v_m$ is a list of vectors in $V$ such that $T{v}_1,\dots ,T{v}_m$ is a linearly independent list in $W$. Prove that $v_1,\dots ,v_m$ is linearly independent.

If $v_1,\dots ,v_m$ are not linearly independent, there exist some $a_1,\dots ,a_m$ that are not all $0$, such that

$$a_1{v}_1+\cdots +a_m{v}_m=0$$

Applying the linear map to both sides:

$$\begin{array}{rl}T(a_1{v}_1+\cdots +a_m{v}_m)& =T(0)\\ T(a_1{v}_1+\cdots +a_m{v}_m)& =0\\ T(a_1{v}_1)+\cdots +T(a_m{v}_m)& =0\\ a_{1}T(v_1)+\cdots +a_{m}T(v_m)& =0\end{array}$$

Note that $T(0)=0$ because linear maps take 0 to 0.

The last line implies that $T{v}_1,\dots ,T{v}_m$ is not linearly independent, since we are using some $a_1,\dots ,a_m$ that are not all $0$. This presents a contradiction. Thus, $v_1,\dots ,v_m$ must be linearly independent.

This basically says that, by the linearity of the transformation $T$, if $v_1,\dots ,v_m$ are linearly independent, so must $T{v}_1,\dots ,T{v}_m$ and vice versa.

Problem 5

Prove that $\mathcal{L}(V,W)$ is a vector space.

Recall that for a vector space, we need to show a zero element, closure under addition, and closure under scalar multiplication. Specifically, for these three operations, we verify that the result is still a linear map (for example, the addition of two linear maps must still be a linear map).

We can define a zero map $0:V\to W$ by:

$$0(v)=0_W\forall v\in V$$

where $0_W$ is the zero vector in $W$. The zero map is linear because is satisfies:

• Additivity: $0(u+w)=0(u)+0(v)=0_W+0_W=0_W$
• Homogeneity: $0(cv)=0_W=c0(v)=c\cdot 0_W=c\cdot 0(v)$

For closure under addition, let $T_1,T_2\in \mathcal{L}(V,W)$. Define $(T_1+T_2):V\to W$ by:

$$(T_1+T_2)(v)=T_1(v)+T_2(v),\forall v\in V$$

Checking the linearity of $T_1+T_2$:

• Additivity:

$$\begin{array}{rl}(T_1+T_2)(u+v)& =T_1(u+v)+T_2(u+v)\\ & =T_1(u)+T_1(v)+T_2(u)+T_2(v)\\ & =(T_1+T_2)u+(T_1+T_2)v\end{array}$$

• Homogeneity:

$$\begin{array}{rl}(T_1+T_2)(\lambda v)& =T_1(\lambda v)+T_2(\lambda v)\\ & =\lambda T_1(v)+\lambda T_2(v)\\ & =\lambda (T_1(v)+T_2(v))\\ & =\lambda (T_1+T_2)(v)\end{array}$$

Therefore, $T_1+T_2\in \mathcal{L}(V,W)$, showing closure under addition.

For closure under scalar multiplication, let $T\in \mathcal{L}(V,W)$, and define $\lambda T$ by:

$$(\lambda T)(v)=\lambda T(v),\forall v\in V,c\in \mathbb{F}$$

Checking for linearity:

• Additivity:

$$\begin{array}{rl}(\lambda T)(u+v)& =\lambda T(u+v)\\ & =\lambda (T(u)+T(v))\\ & =\lambda T(u)+\lambda T(v)\\ & =(\lambda T)(u)+(\lambda T)(v)\end{array}$$

• Homogeneity:

$$\begin{array}{r}(\lambda T)(\alpha v)=\lambda T(\alpha v)=\lambda (\alpha T(v))=(\lambda \alpha )T(v)=\alpha (\lambda T)(v)\end{array}$$

Thus, $\lambda T\in \mathcal{L}(V,W)$, confirming closure under scalar multiplication.

Since we’ve shown zero element, closure under addition, and closure under scalar multiplication, we’ve shown that $\mathcal{L}(V,W)$ is a vector space.

Problem 6

Prove that multiplication of linear maps has the associative, identity, and distributive properties.

Associativity:

$$\begin{array}{rl}((T_1{T}_2)T_3)(v)& =(T_1{T}_2)(T_3(v))\\ & =T_1(T_2(T_3(v)))\\ & =T_1((T_2{T}_3)(v))\\ & =(T_1(T_2{T}_3))(v)\end{array}$$

Identity:

$$(IT)(v)=I(T(v))=T(v)$$

and

$$(TI)(v)=T(I(v))=T(v)$$

Distributive properties:

$$\begin{array}{rl}((S_1+S_2)T)(v)& =(S_1+S_2)(T(v))\\ & =S_1(T(v))+S_2(T(v))\\ & =(S_{1}T)(v)+(S_{2}T)(v)\\ & =(S_{1}T+S_{2}T)(v)\end{array}$$

and

$$\begin{array}{rl}(S(T_1+T_2))(v)& =S((T_1+T_2)(v))\\ & =S(T_1(v)+T_2(v))\\ & =S(T_1(v))+S(T_2(v))\\ & =(S{T}_1)(v)+(S{T}_2)(v)\\ & =(S{T}_1+S{T}_2)(v)\end{array}$$

Problem 7

Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if $\dim V=1$ and $T\in \mathcal{L}(V)$, then there exists $\lambda \in \mathbb{F}$ such that $Tv=\lambda v$ for all $v\in V$.

Since $\dim V=1$, there is some $v_1\in V$ that is a basis of $V$. Because $T{v}_1\in V$, there exists some $\lambda \in \mathbb{F}$ such that $T{v}_1=\lambda v_1$.

Since $v_1$ is a basis every $v\in V$, there exists some $a\in \mathbb{F}$ such that $v=a{v}_1$. This implies that

$$Tv=T(a{v}_1)=aT{v}_1=a\lambda v_1=\lambda a{v}_1=\lambda v$$

Problem 8

Give an example of a function $\varphi :{\mathbb{R}}^2\to \mathbb{R}$ such that

$$\varphi (av)=a\varphi (v)$$

for all $a\in \mathbb{R}$ and all $v\in {\mathbb{R}}^2$ but $\varphi$ is not linear.

For $v=(v_1,v_2)\in {\mathbb{R}}^2$, let

$$\varphi (v)=\{\begin{array}{ll}\frac{v_1^2}{v_2},& v_2\ne 0\\ 0,& v_2=0\end{array}$$

Clearly, we have homogeneity in $\varphi (av)=a\varphi (v)$ for every $a\in \mathbb{R}$ and $v\in {\mathbb{R}}^2$.

But, there obviously exists $u,v\in {\mathbb{R}}^2$ where $\varphi (u+v)\ne \varphi (u)+\varphi (v)$, because

$$\frac{(u_1+v_1{)}^2}{u_2+v_2}\ne \frac{u_1^2}{u_2}+\frac{v_1^2}{v_2}$$

Thus, $\varphi$ is not linear.

Problem 8 and 9 show that neither homogeneity nor additivity alone is enough to imply that a function is a linear map.

Problem 9

Give an example of a function $\varphi :\mathbb{C}\to \mathbb{C}$ such that

$$\varphi (w+z)=\varphi (w)+\varphi (z)$$

for all $w,z\in \mathbb{C}$ but $\varphi$ is not linear. (Here $\mathbb{C}$ is thought of as a complex vector space.)

Let $\mathbb{C}=\{a+bi:a,b\in \mathbb{R}\}$. For every $z=a+bi\in \mathbb{C}$, we define $\varphi (z)=a$, so the function $\varphi$ just returns the real part.

Hence, $\forall w=a+bi,z=c+di\in \mathbb{C}$, we have

$$\varphi (w+z)=\varphi ((a+c)+(b+d)i)=a+c$$

and

$$\varphi (w)+\varphi (z)=a+c$$

So we have $\varphi (w+z)=\varphi (w)+\varphi (z)$.

However, $\varphi (i(1+i))=\varphi (i-1)=-1$ and $i\varphi (1+i)=i\cdot 1=i$. Then, we have

$$\varphi (i(1+i))\ne i\varphi (1+i)$$

Thus, in this case, for some $a\in \mathbb{C}$ and $w\in \mathbb{C}$, we don’t have $\varphi (aw)=a\varphi (w)$. Therefore, $\varphi$ is not linear.

Problem 10

Prove or give a counterexample: If $q\in \mathcal{P}(\mathbb{R})$ and $T:\mathcal{P}(\mathbb{R})\to \mathcal{P}(\mathbb{R})$ is defined by $Tp=q\circ p$, then $T$ is a linear map.

• This differs from the composition example in Examples because the order of functions in the composition is different.

For additivity, we need $T(p_1+p_2)=T{p}_1+T{p}_2$. Expanding these, we have:

$$\begin{array}{rl}T(p_1+p_2)& =q\circ (p_1+p_2)\\ T{p}_1+T{p}_2& =(q\circ p_1)+(q\circ p_2)\end{array}$$

As a counterexample, let $q(x)=x^2,p_1(x)=x,p_2(x)=-x$. Then:

$$T(p_1+p_2)=T(0)=q\circ 0=q(0)=(0{)}^2=0$$

However,

$$T{p}_1+T{p}_2=(q\circ p_1)+(q\circ p_2)=q(x)+q(-x)=x^2+(-x{)}^2=2x^2$$

For homogeneity, we need $T(\lambda p)=\lambda \cdot Tp$. Expanding these, we have

$$\begin{array}{rl}T(\lambda p)& =q\circ \lambda p\\ \lambda \cdot Tp& =\lambda (q\circ p)\end{array}$$

As a counterexample, let $q(x)=x^2,p(x)=x,\lambda =2$. Then:

$$T(\lambda p)=q\circ 2x=(2x{)}^2=4x^2$$

However:

$$\lambda \cdot Tp=2(q(p(x)))=2(q(x))=2x^2$$

Problem 11

Suppose $V$ is finite-dimensional and $T\in \mathcal{L}(V)$. Prove that $T$ is a scalar multiple of the identity if and only if $ST=TS$ for every $S\in \mathcal{L}(V)$.

First we prove it in one direction, by showing that if $T=aI$, for some scalar $a\in \mathbb{F}$, we have $ST=TS$.

We have:

$$(ST)(x)=S(T(x))=S(aI(x))=S(ax)=aS(x)$$

and

$$(TS)(x)=T(S(x))=aI(S(x))=aS(x)$$

so $ST=TS$.

Next, we prove it in the opposite direction: if $ST=TS$ for every $S\in \mathcal{L}(V)$, we have $T=aI$. First, we show the effect of $T$ on basis vectors, and show that it is diagonal on a chosen basis.

Suppose $\dim V=n$ and $v_1,\dots ,v_n$ is a basis of $V$. Define $S_i\in \mathcal{L}(V)$:

$$S_i(a_1{v}_1+\cdots +a_n{v}_n)=a_i{v}_i\text{where}{a}_i\in \mathbb{F},i=1,\dots ,n$$

Thus, $S_i$ projects any vector onto the basis vector $v_i$, scaled by a corresponding coefficient.

From the assumption that $ST=TS$ for all $S\in \mathcal{L}(V)$, consider:

$$\begin{array}{rl}{S}_{i}T(v_i)& =T{S}_i(v_i)\\ S_i(T(v_i))& =T(S_i(v_i))\end{array}$$

Since $S_i(v_i)=v_i$, this reduces to

$$S_i(T(v_i))=T(v_i)$$

By definition of $S_i$, the transformation $S_i(T(v_i))$ retains only the $v_i$-component of $T(v_i)$. Therefore, $T(v_i)$ must be a scalar multiple of $v_i$:

$$T(v_i)={\lambda }_i{v}_i,\text{where }{\lambda }_i\in \mathbb{F}$$

Now, we want to show that $T$ cannot treat different basis vectors differently; it must scale them all by the same scalar. Specifically, we want ${\lambda }_i={\lambda }_j$ for all $i,j$ using the linear operator $S_{ij}$ and the assumption and that $ST=TS$ for all $S\in \mathcal{L}(V)$.

Let our $S_{ij}$ be defined to swap the coefficients of $v_i$ and $v_j$ in any vector, such that

$$S_{ij}(\cdots +a_i{v}_i+\cdots +a_j{v}_j+\dots )=\cdots +a_j{v}_i+\cdots +a_i{v}_j+\dots$$

For example, $S_{12}(a_1{v}_1+a_2{v}_2+a_3{v}_3)=a_2{v}_1+a_1{v}_2+a_3{v}_3$.

From the assumption that $ST=TS$, we have

$$S_{ij}T=T{S}_{ij}$$

Applying the left side to $a_i{v}_i+a_j{v}_j$, we have

$$T(a_i{v}_i+a_j{v}_j)={\lambda }_i{a}_i{v}_i+{\lambda }_j{a}_j{v}_j$$

and then applying $S_{ij}$, we have

$$S_{ij}({\lambda }_i{a}_i{v}_i+{\lambda }_j{a}_j{v}_j)={\lambda }_j{a}_i{v}_i+{\lambda }_i{a}_j{v}_j$$

Applying the right side, we have

$$S_{ij}(a_i{v}_i+a_j{v}_j)=a_j{v}_i+a_i{v}_j$$

and then applying $T$ gives

$$T(a_j{v}_i+a_i{v}_j)={\lambda }_i{a}_j{v}_i+{\lambda }_j{a}_i{v}_j$$

Equating the two sides gives

$${\lambda }_j{a}_j{v}_i+{\lambda }_i{a}_i{v}_j={\lambda }_i{a}_j{v}_i+{\lambda }_j{a}_i{v}_j$$

which means that ${\lambda }_i={\lambda }_j$ for all $i,j$. Thus, $T=\lambda I$, where $\lambda \in \mathbb{F}$.

Problem 12

Suppose $U$ is a subspace of $V$ with $U\ne V$. Suppose $S\in \mathcal{L}(U,W)$ and $S\ne 0$ (which means that $Su\ne 0$ for some $u\in U$). Define $T:V\to W$ by

$$Tv=\{\begin{array}{ll}Sv& \text{if }v\in U\\ 0& \text{if }v\in V\text{ and }v\notin U\end{array}$$

Prove that $T$ is not a linear map on $V$.

Let $u\in U$ such that $Su\ne 0$; such as $u$ exists because $S\ne 0$. Let $v\in V\setminus U$ such that $v\notin U$; such a $v$ exists because $U\ne V$.

Since $u\in U$ and $v\notin U$, their sum $u+v$ cannot lie in $U$. Otherwise, we could write $v=(u+v)+(-u)\in U$, which is a contradiction.

Therefore, $u+v\in V\setminus U$, and by the definition of $T$, we have $T(u+v)=0$. However, additivity requires $T(u)+T(v)=Su+0=Su$, implying $T(u+v)\ne Tu+Tv$.

Therefore, $T$ is not a linear map on $V$.

Problem 13

Suppose $V$ is finite-dimensional. Prove that every linear map on a subspace of $V$ can be extended to a linear map on $V$. In other words, show that if $U$ is a subspace of $V$ and $S\in \mathcal{L}(U,W)$, then there exists $T\in \mathcal{L}(V,W)$ such that $Tu=Su$ for all $u\in U$.

There exists a subspace $M$ of $V$ such that $V=U\oplus M$. For all $v\in V$, there exist $u\in U$ and $m\in M$ such that $v=u+m$.

Suppose $R\in \mathcal{L}(M,W)$. For all $v=u+m\in V$, define $Tv=Su+Rm$. We have

$$\begin{array}{rl}T(\lambda v)& =T(\lambda u+\lambda m)\\ & =S(\lambda u)+R(\lambda m)\\ & =\lambda Su+\lambda Rm\\ & =\lambda (Su+Rn)\\ & =\lambda Tv\end{array}$$

We also have:

$$\begin{array}{rl}T(v_1+v_2)& =T((u_1+m_1)+(u_2+m_2))\\ & =S(u_1+u_2)+R(m_1+m_2)\\ & =S{u}_1+S{u}_2+R{m}_1+R{m}_2\\ & =(S{u}_1+R{m}_1)+(S{u}_2+R{m}_2)\\ & =T{v}_1+T{v}_2\end{array}$$

Hence, $T$ is linear.

Problem 14

Suppose $V$ is finite-dimensional with $\dim V>0$, and suppose $W$ is infinite- dimensional. Prove that $\mathcal{L}(V,W)$ is infinite-dimensional.

Suppose $\dim V=n$ and $v_1,\dots ,v_n$ is a basis of $V$. As $W$ is infinite-dimensional, we can find a sequence of elements in $W$, denoted by $w_1,\dots ,w_m$, that is linearly independent for any positive integer $m$.

For every $v=a_1{v}_1+\cdots +a_n{v}_n\in V$, where $a_i\in \mathbb{F}$, define $T_k\in \mathcal{L}(V,W)$ such that

$$T_k(v)=a_1{w}_k$$

To prove $T_1,\dots ,T_m$ is linearly independent for any $m$, assume that we have a linear combination of these maps that equals the zero map, which must exist since $\mathcal{L}(V,W)$ is a vector space.

$${\lambda }_1{T}_1+\cdots +{\lambda }_m{T}_m=0$$

This means, for every $v\in V$:

$$\begin{array}{r}({\lambda }_1{T}_1+\cdots +{\lambda }_m{T}_m)(v)=0\\ a_1({\lambda }_1{w}_1+\cdots +{\lambda }_m{w}_m)=0\end{array}$$

For this to hold for all $a_1\in \mathbb{F}$, it must be true that ${\lambda }_1{w}_1+\cdots +{\lambda }_m{w}_m=0$. Since $w_1,\dots ,w_m$ are linearly independent by construction, the only solution is ${\lambda }_1,\dots ,{\lambda }_m=0$.

Hence, we find a sequence of elements in $\mathcal{L}(V,W)$, $T_1,\dots ,T_m$ that is linearly independent for any positive $m$. Therefore, $\mathcal{L}(V,W)$ is infinite-dimensional.

Problem 15

Suppose $v_1,\dots ,v_m$ is a linearly dependent list of vectors in $V$. Suppose also that $W\ne \{0\}$. Prove that there exist $w_1,\dots ,w_m\in W$ such that no $T\in \mathcal{L}(V,W)$ satisfies $T{v}_k=w_k$ for each $k=1,\dots ,m$.

$v_1,\dots ,v_m$ being linearly dependent implies that there exist some $a_1,\dots ,a_m\in \mathbb{F}$, not all $0$, such that $a_1{v}_1+\cdots +a_m{v}_m=0$.

As $W\ne \{0\}$ and $a_1,\dots ,a_m$ are not all $0$, we can find $w_1,\dots ,w_m$ such that $a_1{w}_1+\cdots +a_m{w}_m\ne 0$.

Now if we use this list $w_1,\dots ,w_m$ with $T\in \mathcal{L}(V,W)$ such that $T{v}_k=w_k$, we have

$$\begin{array}{rl}T(0)& =T(a_1{v}_1+\cdots +a_m{v}_m)\\ & =a_{1}T{v}_1+\cdots +a_{m}T{v}_m\\ & =a_1{w}_1+\cdots +a_m{w}_m\\ & \ne 0\end{array}$$

This is impossible. Hence, the result is proved.

Problem 16

Suppose $V$ is finite-dimensional with $\dim V>1$. Prove that there exist $S,T\in \mathcal{L}(V)$ such that $ST\ne TS$.

Suppose $v_1,\dots ,v_m$ is a basis of $V$, such that every element of $V$ can be written as some $a_1{v}_1+\cdots +a_m{v}_m$. Since $\dim V>1$, $m\ge 2$.

Let us define $S\in \mathcal{L}(V)$ to such that

$$S(a_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m)=a_2{v}_1+a_1{v}_2+\cdots +a_m{v}_m$$

where the coefficients of the $v_1$ and $v_2$ terms are switched.

Let us define $T\in \mathcal{L}(V)$ such that

$$S(a_1{v}_1+a_2{v}_2+\cdots +a_m{v}_m)=a_1{v}_1$$

Then:

$$ST(v_1+2v_2)=S(T(v_1+2v_2))=S(v_1)=0v_1+1v_2=v_2$$

and

$$TS(v_1+2v_2)=T(2v_1+v_2)=2v_1$$

Thus, $ST\ne TS$.

Problem 17

Suppose $V$ is finite-dimensional. Show that the only two-sided ideals of $\mathcal{L}(V)$ are $\{0\}$ and $\mathcal{L}(V)$.

A subspace $\mathcal{E}$ of $\mathcal{L}(V)$ is called a two-sided ideal of $\mathcal{L}(V)$ if $TE\in \mathcal{E}$ and $ET\in \mathcal{E}$ for all $E\in \mathcal{E}$ and all $T\in \mathcal{L}(V)$.

It’s easy to verify that $\{0\}$ and $\mathcal{L}(V)$ are two-sided ideals of $\mathcal{L}(V)$.

Suppose for the sake of contradiction that another $\mathcal{E}$ exists. Let $e_1,\dots ,e_m$ be its basis and let $e_1,\dots ,e_m,e_{m+1},\dots ,e_n$ be the basis of $V$.

Define $T(v)$ as

$$T(v)=T\left(\sum _{i=1}^n{a}_i{e}_i\right)=\sum _{i=1}^n{a}_i{e}_{n-i}$$

to be a linear map in $V$. Essentially, this reverses the coefficients of the basis vectors.

Define $E_j(u)$ as

$$E_j(u)=E_j\left(\sum _{k=1}^m{b}_k{e}_k\right)=b_j{e}_j$$

This maps acts a projection onto the basis vector $e_j$. Note that $E_j\in \mathcal{E}$ because $\mathcal{E}$ is spanned by $e_1,\dots ,e_m$.

We have reached the contradicting the example such that

$$T{E}_j(v)=a_j{e}_{n-j}\in V\setminus \mathcal{E}$$