Worksheet 5

Solution.

We will do both parts at the same time by giving a left and right inverse for $A^t$. Recall that

Note that the order of multiplication is reversed when we take transposes. Since $A$ is invertible, $A^{-1}$ exists and

Since $I_n^t = I_n$, transposing both sides of the equations above yields

In summary,

$A^t$ has a left and right inverse, so it's invertible, and we have shown that the left and right inverse are both $\p{A^{-1}}^t$, i.e., $\p{A^t}^{-1} = \p{A^{-1}}^t$.

Solution.

1. Note that $A$ is surjective: given $w \in \F^n$, we have

   $$\p{AB}w = I_nw = w \implies A\p{Bw} = w$$

   since matrix multiplication is associative. If we set $v = Bw \in \F^n$, then $Av = w$, so because $w$ was arbitrary, this shows that $A$ is surjective.

   Because $A$ is a square matrix, surjective is equivalent to invertible, so $A$ is invertible.

2. Now that we know that $A$ is invertible, we know that $A^{-1}$ exists. Thus, multiplying both sides of $AB = I_n$ by $A^{-1}$,

   $$A^{-1}AB = A^{-1}I_n \implies B = A^{-1}.$$

   Hence, $B$ is equal to the invertible matrix $A^{-1}$, so $B$ itself is invertible with $B^{-1} = \p{A^{-1}}^{-1} = A$.

3. Let $\func{T}{V}{W}$, $\func{S}{W}{V}$ be linear maps between vector spaces, where $\dim{V} = \dim{W} = n < \infty$. If $T \circ S = I_W$, then $T$ and $S$ are both invertible with $T = S^{-1}$ and $S = T^{-1}$.

   To prove this, let's show that $T$ is surjective. The proof is the same as before: let $w \in W$, and notice that

   $$\p{T \circ S}\p{w} = I_W\p{w} \implies T\p{S\p{w}} = w.$$

   As before, setting $v = S\p{w}$ gives $T\p{v} = w$, so $T$ is surjective, hence invertible. I'll reprove this here in case you haven't seen this fact:

   Because $T$ is surjective, it follows that $\dim R\p{T} = \dim W = n$. Since $V$ is finite-dimensional, we may apply rank-nullity to get

   $$\begin{aligned} \dim V = \dim R\p{T} + \dim N\p{T} &\implies n = n + \dim N\p{T} \\ &\implies \dim N\p{T} = 0 \\ &\implies N\p{T} = \set{0}, \end{aligned}$$

   so $T$ is injective. Thus, $T$ is a bijection, and the inverse of a linear map is linear, so $T$ is an isomorphism. (Recall that isomorphism here means that $T$ is a linear bijection and that $T^{-1}$ is a linear map.)

   The same calculations as above still work:

   $$T^{-1} \circ T \circ S = T^{-1} \circ I_W \implies S = T^{-1} \implies S^{-1} = \p{T^{-1}}^{-1} = T.$$