# Selected Exercises

## 3.D

7. Suppose $V$ & $W$ are finite-dimensional. Let $v \in V$. Let $$E = \{ T \in \mathcal L(V, W): T(v) = 0 \}.$$

(a) Show that $E$ is a subspace of $\mathcal L(V, W)$.
(b) Suppose $v \neq 0$. What is $\dim E$?

For (a), to show $E$ is a subspace of $\mathcal L(V, W)$, we need to show that $E$ contains zero and is closed under both addition and multiplication.

If $T = 0$, then $T(v) = 0$ for all $v \in V$, so $0 \in E$.

Now, suppose $T_0, T_1 \in E$. Then,

$$(T_0 + T_1)(v) = T_0(v) + T_1(v) = 0 + 0 = 0,$$

where the first equality comes from the definition of linear maps and the second comes from the fact that both maps are assumed to be in $E$. Thus, $E$ is closed under addition.

Last, suppose $T \in E$ and $\lambda \in F$. Then, $$(\lambda T)(v) = \lambda \circ T(v) = \lambda \circ 0 = 0,$$

where the first equality comes from the definition of linear maps and the rest should be obvious. Thus, $E$ is closed under multiplication.

For (b), we want to find the dimension of $E$. One way to do this is construct an isomorphism between $E$ and a vector space whose dimension we already know, as Axler does as part of showing that $\dim \mathcal L(V, W) = (\dim V)(\dim W)$.

We do this in a kind of roundabout way. First, observe that, given an arbitrary $T \in E$, $v \neq 0 \in V$, we can extend $v$ to a basis of $V$, $v, u_1, u_2, …, u_n$. Since $T \in E$, we know that $\mathcal M(T)$'s first column will consist of all $0$s. Since $\mathcal M: \mathcal L(V, W) \rightarrow F^{m, n}$ is an isomorphism, $\dim E$ is the number of matrices in $F^{m, n}$ where the first column vector is $0$. Thus, $\dim E = (n-1) * m = n * m - m = (\dim V)(\dim W) - \dim W$.

9. Suppose $V$ is finite-dimensional and $S, T \in \mathcal L(V)$. Prove that $ST$ is invertible iff both $S$ and $T$ are invertible.
We start with the forward direction, for which we assume that $ST$ is invertible. By 3.56 and the definitions of injectivity and surjectivity, $\operatorname{null} ST = \{0\}$ and $\operatorname{range} ST = V$.

Since $\operatorname{range} ST \subset \operatorname{range} S$, $V \subset \operatorname{range} S$ but also, by definition, $\operatorname{range} S \subset V$, so $\operatorname{range} S = V$, meaning $S$ is surjective. Because $S$ is an operator, surjectivity implies invertibility.

Now, assume $\operatorname{null} T \neq \{0\}$. Then there exists $v \neq 0 \in V$ such that $T(v) = 0$, meaning $(ST)(v) = 0$. However, we also know that $\operatorname{null} ST = 0$, so we have a contradiction. Hence, $T$ is injective and therefore invertible.

For the backwards direction, we assume that $S$ and $T$ are invertible. To show $ST$ is invertible, we can show that $\operatorname{null} ST = \{0\}$.

Let $v \in V$ where $ST(v) = 0$ and $T(v) = w$. First, $S(w) = 0$ iff $w \in \operatorname{null} S$ iff $w = 0$. Then, $T(v) = 0$ iff $v \in \operatorname{null} T$ iff $v = 0$. Hence, $ST(v) = 0$ iff $v = 0$, meaning $\operatorname{null} ST = \{0\}$.

Therefore, $ST$ is injective and invertible.

10. Suppose $V$ is finite-dimensional and $S,T \in \mathcal L(V)$. Prove that $ST = I$ iff $TS = I$.
Assume $ST = I$. Let $v_0, v_1 \in V$ where $T(v_0) = v_1$. By assumption, it must be true that $S(v_1) = v_0$. Thus, $TS(v_1) = v_1$, and $TS = I$. We can apply analogous logic in the reverse direction.

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

For the forward direction, we assume that $T$ is a scalar multiple of the identity, i.e. $T = \lambda I$ for some $\lambda \in F$.

Thus, $(TS)(v) = T(S(v)) = \lambda * S(v) = S(\lambda v) = S(T(v)) = (ST)(v)$. where the third equality comes from the homogeneity property of linear maps.

For the reverse direction, we take a more arduous path. Assume that $ST = TS$ for every $S \in \mathcal L(V)$.

First, we show that $(v, T(v))$ is a linearly dependent list. Assume that $(v, T(v))$ is linearly independent. Then we can extend $(v, T(v))$ to a basis, $(v, T(v), u_1, …, u_n)$. Let $S$ be the linear map defined as $S(av + bT(v) + c_1 u_1 + \dots + c_n u_n) = bv$. Thus, $S(v + T(v)) = v$ but $S(v + T(v)) = S(v) + S(T(v)) = 0 + T(S(v)) = 0 + T(0) = 0$. This gives us a contradiction since $T$ is assumed to be linearly independent but supposedly $v = 0$. Thus, $(v, T(v))$ must be linearly dependent for all $v \in V$.

While we now know that $T(v) = a_v * v$ for some $a_v \in F$, we still have to show that $a_v$ isn’t dependent on the value of $v$.

Assume that we have $v, w \in V$ and $a_v, a_w \in F$ where $T(v) = a_v v$ and $T(w) = a_w w$. We want to show that $a_v = a_w$ regardless of the values of $v$ and $w$.

First, we do this for the case where $v$ and $w$ are linearly dependent. In this case, we know that $bw = v$ for some $b \in F$. Thus, we have

$$\begin{eqnarray} a_v v & = & T(v) \\\\ & = & T(bw) \\\\ & = & bT(w) \\\\ & = & b a_w w \\\\ & = & a_w b w \\\\ & = & a_w v. \end{eqnarray}$$

Second, we do this for the case where $v$ and $w$ are linearly independent. Assume that $T(v+w) = a_{v+w} (v+w)$ and note that we know that $a_0 v + a_1 w = 0$ implies $a_0 = a_w = 0$. So, we have

$$T(v + w) = a_{v+w} (v+w)$$

but also

$$T(v + w) = T(v) + T(w) = a_v v + a_w w,$$

which implies

$$a_v v + a_w w = a_{v+w} v + a_{v+w} w.$$

Hence,

$$(a_v - a_{v+w}) v + (a_w - a_{v+w}) w = 0.$$

And since $a_v, a_w$ is linearly independent, $a_v = a_{v+w} = a_w$.

Thus, $a_v = a_w$ for all $v, w \in V$ and therefore $T$ is a scalar multiple of the identity map.