# Linear Algebra Done Right - Chapter 3

# 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.