LADR Exercises 3B

Problem 1

Give an example of a linear map $T$ with $\dim \text{null }T=3$ and $\dim \text{range }T=2$.

Consider

$$T\in \mathcal{L}({\mathbb{F}}^5,{\mathbb{F}}^2):T(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2)$$

Then the null space is all vectors of the form $(0,0,x_3,x_4,x_5)$, giving $\dim \text{null }T=3$. The range is all vectors of the form $(x_1,x_2)$, which clearly gives $\dim \text{range }T=2$. Note that this also fulfills the fundamental theorem of linear maps, since $3+2=5$.

Problem 2

Suppose $S,T\in \mathcal{L}(V)$ are such that $\text{range }S\subseteq \text{null }T$. Prove that $(ST{)}^2=0$.

We have:

$$(ST{)}^2(v)=STST(v)=S(T(S(Tv)))$$

As $\text{range }S\subseteq \text{null }T$, we have

$$T(S(u))=0$$

for any $u$ in $V$.

This means that $T(S(Tv))=0$ for all $v$. The final operation would be $S(0)$, and since linear maps take 0 to 0, we have $(ST{)}^2=0$.

Problem 3

Suppose $v_1,\dots ,v_m$ is a list of vectors in $V$. Define $T\in \mathcal{L}({\mathbb{F}}^m,V)$ by

$$T(z_1,\dots ,z_m)=z_1{v}_1+\cdots +z_m{v}_m$$

• (a) What property of $T$ corresponds to $v_1,\dots ,v_m$ spanning $V$?
• (b) What property of $T$ corresponds to $v_1,\dots ,v_m$ being linearly indepedent?

(a) If $v_1,\dots ,v_m$ spans $V$, for all $v\in V$, there must exist scalars $z_1,\dots ,z_m$ such that

$$v=z_1{v}_1+\cdots +z_m{v}_m$$

which is the definition of the linear transformation $T(z_1,\dots ,z_m)=z_1{v}_1+\cdots +z_m{v}_m$. Since every $v\in V$ can be written as $T(z_1,\dots ,z_m)$, we can say that $\text{range }T=V$. Thus, $T$ is surjective.

(b) If $v_1,\dots ,v_m$ are linearly independent, if we have $z_1{v}_1+\cdots +z_m{v}_m=0$, we need $z_1,\dots ,z_m=0$.

The key here again is to recognize that the $z_1{v}_1+\cdots +z_m{v}_m$ is the operation applied by our map $T$. This means that if we apply the map and the result is zero as below:

$$T(z_1,\dots ,z_m)=z_1{v}_1+\dots z_m{v}_m=0$$

the only possible case when this happens is when $z_1=\cdots =z_m=0$ for linear independence. This implies that $\text{null }T=0$, and since injectivity implies null space is 0 and vice versa, the map $T$ must be injective.

Problem 4

Show that $\{T\in \mathcal{L}({\mathbb{R}}^5,{\mathbb{R}}^4):\dim \text{null }T>2\}$ is not a subspace of $\mathcal{L}({\mathbb{R}}^5,{\mathbb{R}}^4)$.

We can show this is not closed under addition with a counterexample. Suppose we have $T_1,T_2\in T$, such that

$$\begin{array}{r}{T}_1(x_1,x_2,x_3,x_4,x_5)=(0,0,x_3,x_4)\\ T_2(x_1,x_2,x_3,x_4,x_5)=(x_1,x_2,0,0)\end{array}$$

Then, we have

$$\begin{array}{rl}\text{null }(T_1)& =\{(x_1,x_2,0,0,x_5):x_1,x_2,x_5\in \mathbb{R}\}\\ \text{null }(T_2)& =\{(0,0,x_3,x_4,x_5):x_3,x_4,x_5\in \mathbb{R}\}\end{array}$$

This means that both $\text{null }(T_1)$ and $\text{null }(T_2)$ have dimension $3$.

However, note that:

$$\begin{array}{rl}(T_1+T_2)(x_1,x_2,x_3,x_4,x_5)& =(0,0,x_3,x_4)+(x_1,x_2,0,0)\\ & =(x_1,x_2,x_3,x_4)\end{array}$$

such that

$$\text{null }(T_1+T_2)=\{((0,0,0,0,x_5)):x_5\in \mathbb{R}\}$$

which means that $\dim \text{null }(T_1+T_2)=1$. Thus, $(T_1+T_2)\notin T$, since members of $T$ must have null space dimension higher than 2. This means that $T$ is not closed under addition, and hence it cannot be a subspace of $\mathcal{L}({\mathbb{R}}^5,{\mathbb{R}}^4)$.

Problem 5

Give an example of $T\in \mathcal{L}({\mathbb{R}}^4)$ such that $\text{range }T=\text{null }T$.

An example would be

$$T(x_1,x_2,x_3,x_4)=(0,0,x_1,x_2)$$

as this would give

$$\begin{array}{rl}\text{null }T& =\{(0,0,x_3,x_4):x_3,x_4\in \mathbb{R}\}\\ \text{range }T& =\{(0,0,x_1,x_2):x_1,x_2\in \mathbb{R}\}\end{array}$$

In both the null space and the range, the first two elements are zero and the last two elements are arbitrary scalars.

Problem 6

Prove that there does not exist $T\in \mathcal{L}({\mathbb{R}}^5)$ such that $\text{range }T=\text{null }T$.

The rank-nullity theorem requires $\dim V=\dim \text{null }T+\dim \text{range }T$. In this case $V$ is ${\mathbb{R}}^5$, so $\dim V=5$. If we define $d=\text{range }T=\text{null }T$, we would need $5=2d$ or $d=2.5$. Since dimension needs to be a whole number, this is not possible.

Problem 7

Suppose $V$ and $W$ are finite-dimensional with $2\le \dim V\le \dim W$. Show that $\{T\in \mathcal{L}(V,W):T\text{ is not injective}\}$ is not a subspace of $\mathcal{L}(V,W)$.

Suppose $\dim V=n$ and $\dim W=m$, with $2\le n\le m$. Let $v_1,\dots ,v_n$ be a basis of $V$ and $w_1,\dots ,w_m$ be a basis of $W$.

Define $T_1,T_2\in \mathcal{L}(V,W)$ such that

$$\begin{array}{rl}{T}_1(a_1{v}_1+\cdots +a_n{v}_n)& =a_2{w}_2+\cdots +a_n{w}_n\\ T_2(a_1{v}_1+\cdots +a_n{v}_n)& =a_1{w}_1+\cdots +a_{n-1}{w}_{n-1}\end{array}$$

To find the null space of $T_1$, we consider $a_2{w}_2+\cdots +a_n{w}_n=0$. Since $w_2,\dots ,w_n$ are linearly independent (as they are part of a basis), the only solution is $a_2=a_3=\cdots =a_n=0$. This means that $a_2{w}_2+\cdots +a_n{w}_n=0$ for any $v=a_1{v}_1$, meaning that the null space consists of all scalar multiples of $v_1$, which gives $\dim \text{null }{T}_1=1$. Similarly, $\dim \text{null }{T}_2=1$.

Since $\dim \text{null }{T}_1=\dim \text{null }{T}_2=1$, these maps are not injective (injectivity implies null space is 0 and vice versa).

However:

$$\begin{array}{rl}(T_1+T_2)(a_1{v}_1+\cdots +a_n{v}_n)& =T_1(a_1{v}_1+\cdots +a_n{v}_n)+T_2(a_1{v}_1+\cdots +a_n{v}_n)\\ & =a_1{w}_1+2a_2{w}_2+\cdots +2a_{n-1}{w}_{n-1}+a_n{w}_n\end{array}$$

Imitating the process above and finding the null space of $(T_1+T_2)$ by finding $v$ such that $(T_1+T_2)(v)=0$ would show us that the only solution is $v=0$, as we have the $n$ terms this time instead of $n-1$. This means that $T_1+T_2$ is injective. Hence, we have found a counterexample showing that set is not closed under addition, and cannot be a subspace of $\mathcal{L}(V,W)$. $■$

Problem 8

Suppose $V$ and $W$ are finite-dimensional with $\dim V\ge \dim W\ge 2$. Show that $\{T\in \mathcal{L}(V,W):T\text{ is not surjective}\}$ is not a subspace of $\mathcal{L}(V,W)$.

Let $T_1,T_2\in \mathcal{L}({\mathbb{R}}^2)$ such that $\dim T_1=\dim T_2=2$. Then define

$$\begin{array}{r}{T}_1(x,y)=(x,0)\\ T_2(x,y)=(0,y)\end{array}$$

Neither $T_1$ or $T_2$ are surjective since both of their ranges are one-dimensional subspaces of ${\mathbb{R}}^2$.

However

$$\begin{array}{rl}(T_1+T_2)(x,y)& =T_1(x,y)+T_2(x,y)\\ & =(x,0)+(0,y)\\ & =(x,y)\end{array}$$

So $T_1+T_2$ has range ${\mathbb{R}}^2$, making it surjective.

Thus, since closure under addition does not hold, we have shown that this set cannot be a subspace of $\mathcal{L}(V,W)$.

Problem 9

Suppose $T\in \mathcal{L}(V,W)$ is injective and $v_1,\dots ,v_n$ is linearly independent in $V$. Prove that $T{v}_1,\dots ,T{v}_n$ is linearly independent in $W$.

Let

$$0=a_{1}T{v}_1+\cdots +a_{n}T{v}_n$$

where $a_1,\dots ,a_n\in \mathbb{F}$.

By linearity, this implies that $T(a_1{v}_1+\cdots +a_n{v}_n)=0$.

Since $T$ is injective, the equation $T(a_1{v}_1+\cdots +a_n{v}_n)=0$ implies that $a_1{v}_1+\cdots +a_n{v}_n=0$.

• This is because linear maps take 0 to 0, so in an injective map, $0$ can be the only input that gets mapped to $0$.

Thus, we’ve shown that the original equation is equivalent to solving $a_1{v}_1+\cdots +a_n{v}_n=0$. Because $v_1,\dots ,v_n$ is linearly independent, the only solution is $a_1=\cdots =a_n=0$. This means the only solution to $a_{1}T{v}_1+\cdots +a_{n}T{v}_n=0$ is also $a_1=\cdots =a_n=0$. Thus, $T{v}_1,\dots ,T{v}_n$ is also linearly independent.

Problem 10

Suppose $v_1,\dots ,v_n$ spans $V$ and $T\in \mathcal{L}(V,W)$. Show that $T{v}_1,\dots ,T{v}_n$ spans $\text{range }T$.

• For every $w\in \text{range }T$, there exists some $v$ such that $w=Tv$.
• As $v_1,\dots ,v_n$ spans $V$, there exists $a_1,\dots ,a_n$ for which $v=a_1{v}_1+\cdots +a_n{v}_n$.
• Thus, any $w$ can be written as $w=Tv=T(a_1{v}_1+\cdots +a_n{v}_n)=a_{1}T{v}_1+\cdots +a_{n}T{v}_n$. This means that any $w\in \text{range }T$ is can be expressed as a linear combination of $T{v}_1,\dots ,T{v}_n$.
• Therefore, $T{v}_1,\dots ,T{v}_n$ spans $\text{range }T$.

Problem 11

Suppose that $V$ is finite-dimensional and $T\in \mathcal{L}(V,W)$. Prove that there exists a subspace $U$ of $V$ such that

$$U\cap \text{null }T=\{0\}\text{and}\text{range }T=\{Tu:u\in U\}$$

Problem 12

Suppose $T$ is a linear map from ${\mathbb{F}}^4$ to ${\mathbb{F}}^2$ such that

$$\text{null }T=\{(x_1,x_2,x_3,x_4)\in {\mathbb{F}}^4:x_1=5x_2\text{ and }{x}_3=7x_4\}.$$

Prove that $T$ is surjective.

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32