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.

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

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...

  • $\begingroup$ Why can you tell that because $A$ is singular $\rightarrow \space\space rankA \leq n-1$? $\endgroup$ – Georgey Jun 3 '13 at 19:18
  • $\begingroup$ @Georgey rank $A=n$ iff $A$ is surjective iff $A$ is invertible. $\endgroup$ – Julien Jun 3 '13 at 19:19
  • $\begingroup$ I don't even know what a surjective matrix embarrassed $\endgroup$ – Georgey Jun 3 '13 at 19:27
  • $\begingroup$ @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 $\leq n-1$. $\endgroup$ – Julien Jun 3 '13 at 19:30
  • 1
    $\begingroup$ @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. $\endgroup$ – Julien Jun 3 '13 at 20:00

This is an old question, but I would like to add a solution that only uses properties of matrices i.e. no abstract algebra.
Sylvester's rank inequality states that for two matrices $X,Y \in \mathcal{M_n(\mathbb{C})}$ we have that $\operatorname{rank}(XY)\ge \operatorname{rank}X+\operatorname{rank} Y-n$.
Since $A$ is singular, we have $\operatorname{rank}A\le n-1$.
We have the equality $A\cdot \operatorname{adj}A=O_n$.
Hence, from Sylvester's rank inequality $0=\operatorname{rank}(A\cdot \operatorname{adj}A)\ge \operatorname{rank}A + \operatorname{rank}(\operatorname{adj}A)-n$.(1)
Case 1. $\operatorname{rank}A\le n-2=>\operatorname{adj}(A)=O_n$ as in Julien's solution.
Case 2. $\operatorname{rank}A=n-1$.
From (1) we have that $\operatorname{rank}(\operatorname{adj}A)\le 1$, so $\operatorname{rank}(\operatorname{adj}A)\in \{0,1\}$.
If $\operatorname{rank}(\operatorname{adj}A)=0$, then $\operatorname{adj}A=O_n$ and this contradicts $\operatorname{rank} A=n-1$.
Hence, $\operatorname{rank}(\operatorname{adj}A)=1$ in this case.
In both cases $\operatorname{rank}(\operatorname{adj}A)\le 1$.

  • 1
    $\begingroup$ Most simple yet elegant solution to the question $\endgroup$ – L-- Mar 31 '20 at 4:37

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