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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.

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    $\begingroup$ 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 $\det(\text{adj}(A)) =\det(A)^{n-1}.$ $\endgroup$
    – user2468
    Aug 22, 2012 at 4:52
  • $\begingroup$ @Jennifer, you've cancelled a factor of $\det A$ from both sides, no? If $\det A=0$, that requires some justification. $\endgroup$ Aug 22, 2012 at 4:56
  • $\begingroup$ @GerryMyerson you're right. $\endgroup$
    – user2468
    Aug 22, 2012 at 4:59
  • $\begingroup$ If $det( A ) = 0$, you can't write $A^{-1}$ either... $\endgroup$
    – xavierm02
    Aug 22, 2012 at 11:23
  • $\begingroup$ @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. $\endgroup$ Aug 22, 2012 at 18:35

3 Answers 3

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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.

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  • $\begingroup$ Thanks. I've just noticed it & removed my comment. $\endgroup$
    – user2468
    Aug 22, 2012 at 5:04
  • $\begingroup$ I think it is because $\,AB=\det A\cdot I = 0=\,$ the zero matrix, @JenniferDylan $\endgroup$
    – DonAntonio
    Aug 22, 2012 at 5:04
  • $\begingroup$ Sir, how we get "which implies $A$ is the zero matrix" $\endgroup$ Feb 3, 2022 at 13:23
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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.

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  • $\begingroup$ "Continuity"? Why to introduce topology here? Perhaps you meant "inductively"? $\endgroup$
    – DonAntonio
    Aug 22, 2012 at 4:46
  • $\begingroup$ I assume it's the continuity of the determinant as a polynomial. $\endgroup$
    – user2468
    Aug 22, 2012 at 4:48
  • $\begingroup$ @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. $\endgroup$
    – DonAntonio
    Aug 22, 2012 at 4:50
  • $\begingroup$ @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. $\endgroup$ Aug 22, 2012 at 4:54
  • $\begingroup$ @GerryMyerson how do you use AA′=(detA)I without saying A^(-1)=(1/detA)A′ ? $\endgroup$ Aug 22, 2012 at 5:04
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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.

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