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For given $n\times n$ matrix $A$ singular matrix, prove that $\operatorname{rank}(\operatorname{adj}A) \leq 1$

So from the properties of the adjugate matrix we know that $$ A \cdot \operatorname{adj}(A) = \operatorname{det}(A)\cdot I$$

Since $A$ is singular we know that $\operatorname{det}(A) = 0$, thus $$ A \cdot \operatorname{adj}(A) = 0$$

This is where I'm getting lost, I think I should say that for the above to happen one of the two, $A$ or $\operatorname{adj}(A)$ would have to be the $0$ matrix, but if $A = 0$ then $\operatorname{adj}(A) = 0$ for sure, which means I said nothing.

A leading hint is needed.

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Hmm, when $A$ is singular, isn't one of its eigenvalues zero, therefore $\det(A)=0$... –  draks ... Jun 3 '13 at 18:52
    
we haven't learned eigenvalues yet so I believe we're not supposed to use it... –  Georgey Jun 3 '13 at 18:55
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up vote 4 down vote accepted

Since $A$ is singular, $\mbox{rank}A\leq n-1$.

Case 1: $\mbox{rank}A\leq n-2$. Then $A$ contains no invertible submatrix of order $n-1$. So every minor of order $n-1$ is zero. What can you conclude about $\mbox{adj}(A)$?

Case 2: $\mbox{rank}A= n-1$. By rank-nullity, we get $\dim\ker A=1$. Now $A\cdot \mbox{adj}(A)=0$ means that the range of $\mbox{adj}(A)$ is contained in $\ker A$. So...

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Why can you tell that because $A$ is singular $\rightarrow \space\space rankA \leq n-1$? –  Georgey Jun 3 '13 at 19:18
    
@Georgey rank $A=n$ iff $A$ is surjective iff $A$ is invertible. –  1015 Jun 3 '13 at 19:19
    
I don't even know what a surjective matrix embarrassed –  Georgey Jun 3 '13 at 19:27
    
@Georgey Don't you know, at least, that a square $n\times n$ matrix is invertible iff it has full rank $n$? So if it is not invertible (i.e. singular) it has rank $\leq n-1$. –  1015 Jun 3 '13 at 19:30
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@Georgey Great. It is helpful to understand the rank of an $n\times m$ matrix as the dimension of the range of the linear map induced by $A:K^m\rightarrow K^n$ on the column vectors. Then you see that the rank of $A$ is $n$ iff the range of $A$ is the whole of $K^n$, that is $A$ is surjective. In the square $m=n$ case, this is equivalent to $A$ being invertible by the rank-nullity theorem. –  1015 Jun 3 '13 at 20:00
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