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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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    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}.$2012-08-22
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    @Jennifer, you've cancelled a factor of $\det A$ from both sides, no? If $\det A=0$, that requires some justification.2012-08-22
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    @GerryMyerson you're right.2012-08-22
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    If $det( A ) = 0$, you can't write $A^{-1}$ either...2012-08-22
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    @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.2012-08-22

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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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    "Continuity"? Why to introduce topology here? Perhaps you meant "inductively"?2012-08-22
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    I assume it's the *continuity* of the determinant as a polynomial.2012-08-22
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    @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.2012-08-22
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    @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.2012-08-22
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    @GerryMyerson how do you use AA′=(detA)I without saying A^(-1)=(1/detA)A′ ?2012-08-22
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    Got it, @GerryMyerson. Thanks2012-08-22
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    @MiguelAlvarez , that very basic equality is true *always*, whether $\,A\,$ is regular or singular.2012-08-22
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    True, thanks @DonAntonio2012-08-22
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    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).2012-08-22
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    @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 $\det A'=(\det A)^{n-1}$.2012-08-22
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    OK - missed that2012-08-22
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    And over an arbitrary field...2012-08-22
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    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 $\det(A(0)')=(\det A(0))^{n-1}$.2012-08-22
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    @Robert, over an arbitrary field, I guess I'd go with EuYu's answer.2012-08-22
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    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.2012-08-22
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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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    Thanks. I've just noticed it & removed my comment.2012-08-22
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    I think it is because $\,AB=\det A\cdot I = 0=\,$ the zero matrix, @JenniferDylan2012-08-22