Let $A$ and $B$ be square $n\times n$ matrices over $\Bbb F$. Show that if $\det(A) = 0$ then $\det(AB) = 0$.

Question

So this question is part (c) to the problem:
Let $T: V \to W$ and $S: W \to U$ be linear transformations of finite-dimensional vectors spaces over the field $\Bbb F$.

Show that $\ker(T)\subseteq\ker(S\circ T)$ and $\operatorname{im}(S\circ T)\subseteq\operatorname{im}(S)$.
Show that $\operatorname{rank}(S\circ T)≤\operatorname{rank}(T)$ and $\operatorname{rank}(S\circ T)≤\operatorname{rank}(S)$.

Now I showed part a and b and I am assuming that we should use these two conclusions to answer part c. So in that case it makes sense to try and define $\ker(A)\subseteq{\ker(BA)}$ because of the definition of linear maps.

We know the definiton of $\ker(T) := \{v\in V\; |\; T(v) = 0\}$ and hence if $\det(A) = 0$ then $\ker(\det) :=\{A\in M_{n\times n}(F)\;|\; \det(A) = 0\}$. And when $\det(BA) = 0$, we have from the result in part 1, i.e. $\ker(\det)\subseteq \ker(\det\circ \det)$. But I cannot makes sense of part 1 and 2 any further. I do know that $BA=AB$ when $A$ is invertible, but since $\det(A) = 0$ then we can't have that equality.

Yes but we proved that only when det(A) does not equal 0. It would a trivial proof because if det(AB) = det(A)det(B) then det(AB) = (0)det(B) = 0 — Sir lethian, May 29 '19 at 02:32
I defined ker(det) as {$A\in M_{n\times n}(F) | det(A) = 0$} — Sir lethian, May 29 '19 at 02:34
Maybe this is helpful: How to show that $\det(AB)=\det(A)\det(B)$? — Bach, May 29 '19 at 02:43

score 1 · Accepted Answer · answered May 29 '19 at 02:42

$\det A=0\Longleftrightarrow\text{rank}A=\dim(\text{Im}(S))<n\Longleftrightarrow\text{Im}(S)\subset U$. You know from part $(b)$ that $\text{Im}(S\circ T)\subseteq\text{Im}(S)\subset U$, which implies $\text{Im}(S\circ T)\subset U$ and $\text{rank}AB<n$. What does this tell you about $\det AB$?

score 0 · Answer 2 · answered May 29 '19 at 03:07

How was the determinant introduced to you? Usually one gets proved quite easily the following property:

Let $A = (a_1 \ldots a_n)$ where $a_1, \ldots , a_n$ are the columns of $A$. Then

$\det A = 0 \Leftrightarrow a_1, \ldots , a_n$ are linearly dependent $\Leftrightarrow \exists x \in \mathbb{F}^n, x\neq 0: Ax = 0$

Using this you can prove your claim very quickly considering two cases:

Case 1: $\det B = \det (b_1 \ldots b_n) = 0$

$$\exists x \in \mathbb{F}^n, x \neq 0: Bx = 0 \Rightarrow ABx= 0 \Rightarrow \det (AB) = 0$$

Case 2: $\det B \neq 0 \Rightarrow \text{im}(B) = \mathbb{F}^n$ $$\exists y \in \mathbb{F}^n, y \neq 0: Ay = 0 \stackrel{\text{im}(B) = \mathbb{F}^n}{\Longrightarrow} \exists x \in \mathbb{F}^n, x \neq 0: y=Bx \Rightarrow Ay=ABx = 0 \Rightarrow \det(AB) = 0$$

Let $A$ and $B$ be square $n\times n$ matrices over $\Bbb F$. Show that if $\det(A) = 0$ then $\det(AB) = 0$.

2 Answers2