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How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?

I have been trying to solve this but I can't use $$\det(A^{-1})=\det \Big(\frac{1}{\det(A)} \operatorname{adj}(A) \Big)$$ because $\det(A)=0$ and $\frac{1}{0}$ is not allowed.

Miguel Alvarez
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    The key here is to avoid any formulation which uses $A^{-1}$. Move things to other side, and use Gerry's answer. Hint for clarity though: $\det(\det(A)I) = \det\pmatrix{\det(A) & \dots \ \dots & \dots \ \dots & \det(A)} = \underbrace{\det(A)\det(A)\cdots\det(A)}_{n\text{ times}}.$ That gives you $\det(\text{adj}(A)) =\det(A)^{n-1}.$ –  Aug 22 '12 at 04:52
  • @Jennifer, you've cancelled a factor of $\det A$ from both sides, no? If $\det A=0$, that requires some justification. – Gerry Myerson Aug 22 '12 at 04:56
  • @GerryMyerson you're right. –  Aug 22 '12 at 04:59
  • If $det( A ) = 0$, you can't write $A^{-1}$ either... – xavierm02 Aug 22 '12 at 11:23
  • @miguel Do you know how to turn Gerry's hint into a rigorous proof? If - as for many students - this continuity argument is not clear, then you should ask for further details before accepting an answer. When you are learning about such matters it is crucial that you understand the details of such arguments (esp. since there are many pitfalls in this area). Do not settle for handwaving - rigor is essential. – Bill Dubuque Aug 22 '12 at 18:35

3 Answers3

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Let $B$ denote the adjugate matrix of $A$. Suppose for the sake of contradiction that $\det(B) \neq 0$. Then $B$ is invertible. Since the equation $AB =(\det{A})I = 0$ is true, we have then $$AB\vec{v} = \vec{0}\ \ \forall \vec{v}$$ which implies $A$ is the zero matrix. But then the adjugate of the $0$ matrix is clearly $0$ itself which contradicts the fact that $B$ was invertible.

EuYu
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Let's write $A'$ for the adjoint of $A$. $AA'=(\det A)I$, so $\det A\det A'=(\det A)^n$ (where $A$ is an $n\times n$ matrix). If $\det A\ne0$, this yields $\det A'=(\det A)^{n-1}$. By continuity, this last equation is true even when $\det A=0$, and you're done.

Gerry Myerson
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  • "Continuity"? Why to introduce topology here? Perhaps you meant "inductively"? – DonAntonio Aug 22 '12 at 04:46
  • I assume it's the continuity of the determinant as a polynomial. –  Aug 22 '12 at 04:48
  • @JenniferDylan , so do I, but what for was my question. Perhaps there's something I'm missing, though I believe the equation $,\det A'=\left(\det A\right)^{n-1},$ already proves what the OP wanted. – DonAntonio Aug 22 '12 at 04:50
  • @Don, that equation was derived under the hypothesis $\det A\ne0$, so it can't be used, without some justification, in the case $\det A=0$. Topology supplies the justification. – Gerry Myerson Aug 22 '12 at 04:54
  • @GerryMyerson how do you use AA′=(detA)I without saying A^(-1)=(1/detA)A′ ? – Miguel Alvarez Aug 22 '12 at 05:04
  • Got it, @GerryMyerson. Thanks – DonAntonio Aug 22 '12 at 05:14
  • @MiguelAlvarez , that very basic equality is true always, whether $,A,$ is regular or singular. – DonAntonio Aug 22 '12 at 05:15
  • True, thanks @DonAntonio – Miguel Alvarez Aug 22 '12 at 05:17
  • I think the continuity argument is a good one, and a handy argument to know about, but I do think that $AA'=(\det A)I$ can be proven without assuming $\det A \neq 0$ just by writing down the left hand side and multiplying through (it helps to know that the determinant of a matrix is an alternating form, which makes this straightforward, but even without, the cancellation works). – Mark Bennet Aug 22 '12 at 05:28
  • @Mark, $\det A\ne0$ isn't needed (and wasn't assumed) for $AA'=(\det A)I$; it's needed for going from $\det A\det A'=(\det A)^n$ to $\det A'=(\det A)^{n-1}$. – Gerry Myerson Aug 22 '12 at 06:24
  • OK - missed that – Mark Bennet Aug 22 '12 at 06:40
  • And over an arbitrary field... – Robert Israel Aug 22 '12 at 08:07
  • A basic version of the continuity argument is to consider $A(t)=A+tI$. Then $\det(A(t)')=(\det A(t))^{n-1}$ for every $t$ such that $A(t)$ is invertible, hence at least for every $t\ne0$ such that $|t|\leqslant t_0$ for some $t_0\gt0$. Both $t\mapsto\det(A(t)')$ and $t\mapsto(\det A(t))^{n-1}$ are polynomial functions of $t$ hence continuous functions of $t$, in particular $\det(A(0)')=(\det A(0))^{n-1}$. – Did Aug 22 '12 at 08:24
  • @Robert, over an arbitrary field, I guess I'd go with EuYu's answer. – Gerry Myerson Aug 22 '12 at 13:00
  • In this approach, the proof of continuity is the crux of the OP's problem, so I think if you propose this as an answer (vs. a comment), then you should at least sketch the proof, esp. since it often trips up many students (and even some teachers) in my experience. – Bill Dubuque Aug 22 '12 at 18:34
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Hereafter, $A^*$ stands for adjoint, $|A|$ stands for the determinant of $A$.

Actually we have the conclusion that $$ \operatorname{rank}(A^*) = \begin{cases} n & \operatorname{rank}(A) = n \\ 1 & \operatorname{rank}(A) = n-1 \\ 0 & \operatorname{rank}(A) = 0 \end{cases} $$

To prove the first one, notice that $A A^* = |A|E$ where $E$ is the identity matrix and $\operatorname{rank}(A)=n$ means $|A|\neq 0$. So $A^*$ is invertible and $\operatorname{rank}(A^*)=n$

To prove the second one, notice that $A A^* = |A|E = \mathbf{0}$. This means every column of $A^*$ is in the null space of $A$, which has dimension one. But the alternative definition of rank (see below) tells us $A^* \neq \mathbf{0}$, therefore $\operatorname{rank}(A^*)=1$.

To prove last one, simply use an alternative definition (or calculation) of rank of matrix: $\operatorname{rank}(A)$ is the size of its largest non-vanishing minor. See more at this wikipedia page under the subsection Determinantal rank – size of largest non-vanishing minor.

Neo
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