Now, I'm wondering how to prove this one. It seems simple but I might be looking in the wrong direction.

Properties which are used:

$\text{adj}(AB) = \text{adj}(B)\text{adj}(A)\:\ (1)$

$\text{adj}(I) = I\:\ (2)$

$X \implies Y$

Assuming $A$ is regular.

$AX = XA = I$, where $X$ is its inverse.

$\text{adj}(AX) = \text{adj}(XA) = \text{adj}(I)$

$\text{adj}(X)\text{adj}(A) = \text{adj}(A)\text{adj}(X) = I$, here the implication ends.

$\text{adj}(A)$ is regular and $\text{adj}(X)$ is its inverse.

$Y \implies X$

Assuming $\text{adj}(A)$ is regular.

$\text{adj}(A)X = X\text{adj}(A) = I$, where $X$ is the inverse ($X = \text{adj}(B)$).

According to $(1)$ and $(2)$, I can write this as...

$\text{adj}(BA) = \text{adj}(AB) = \text{adj}(I)$...

Now, here I'm stuck. I do not know if adjugate matrix is unique for each matrix. Don't know if I can just "inverse" the process to get the $BA = AB = I$, and show that $A$ is regular.

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    $\begingroup$ If $n$ is the size of the (square) matrix, the statement to be proved is false (in the direction $\Leftarrow$) for $n=1$ (namely the adjugate will be $(1)$, hence regular, irrespective of$~A$; in particular this is so for $A=(0)$). The case $n=1$ is of course not very interesting, but complete proof should have to use $n\neq1$ at some point. $\endgroup$ – Marc van Leeuwen Aug 12 '13 at 4:15

Use this property $$\text{adj}(A)A=A\text{adj}(A)=\det(A)I.$$

  • $\begingroup$ I'm quite confused. I knew about this property but it is, I guess, somewhat useful in the first part. How does this help me to prove that A is invertible/regular? If A is not invertible det(A) = 0? Wouldn't using this property imply that I know that A is regular? $\endgroup$ – user48724 Dec 16 '12 at 23:15
  • $\begingroup$ What I wanted to say is that if this property holds, the X in the second part is $det(A)^{-1}\cdot A$, what if det(A) is 0. $\endgroup$ – user48724 Dec 16 '12 at 23:23
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    $\begingroup$ @Leolinus: Then use this property:$$\text{If } AB=0, \ A\neq0 \text{ then } B \text{ is not invertible }.$$For proof use proof by contradiction. $\endgroup$ – P.. Dec 16 '12 at 23:32
  • $\begingroup$ So, what you are saying is that $adj(A)\cdot A=det(A)\cdot I$, then I should by contradiction derive that $det(A) != 0$ and prove the invertibility. But, this doesn't work. $adj(A)$ is a regular matrix, so, in your property it's the A matrix, which will then imply that B(my A) is not invertible. I do understand that the first equality in your answer holds, which then implies that my adj(A)X = Xadj(A) = I will hold only if X is A/det(A). I'm wondering can I prove it that easily? $\endgroup$ – user48724 Dec 16 '12 at 23:56
  • $\begingroup$ @Leolinus: You know how to prove that $A$ invertible $\Longrightarrow \text{adj}(A)$ invertible. For the converse prove $A$ not invertible $\Longrightarrow \text{adj}(A)$ not invertible (contrapositive). $\endgroup$ – P.. Dec 17 '12 at 0:03

Here's a solution that relies on the notion of rank. Let $A$ be a matrix of size $n\times n$, and let $0\leq r\leq n$ be its rank. Recall that $$A\times \text{Adj}(A)=\text{Adj}(A)\times A=\text{det}(A)I$$

The rank of $\text{Adj}(A)$ equals

  • $0$ if $r\leq n-2$,
  • $1$ if $r=n-1$,
  • $n$ if $r=n$.

Indeed, if $r\leq n-2$, then each and every coefficitent of $\text{Adj}(A)$ equals $0$, since any $n-1$ columns in $A$ are linearly dependent, and obviously remain so if a line of coefficients is erased from them, so all cofactors are $=0$, and $\text{Adj}(A)=0$.

If $r=n-1$, then there A has $n-1$ linearly independent columns, say the first $n-1$ columns are linearly independent. Over a field, the column rank and the row rank coincide, so there are $n-1$ linearly independent rows among those $n-1$ linearly independent lines, and the determinant these lines and columns define is non-zero, so at least one cofactor is $\neq 0$, and $\text{Adj}(A)\neq 0$. On the other hand, because $0=\text{det}(A)I=A\times \text{Adj}(A)$, the columns of $\text{Adj}(A)$ all lie in the nullspace of $A$, which is one dimensional by the rank theorem, so $0<\text{rank}(\text{Adj}(A))\leq 1$.

if $r=n$, then the above equation (and the fact that $\text{det}(A)\neq 0$) tells you that $\text{Adj}(A)$ is invertible, and has inverse $\frac{A}{\text{det(A)}}$. This analysis shows that $Adj(A)$ is invertible iff $A$ is invertible.

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    $\begingroup$ Note that for $n=1$ the second case ($r=n-1$) still gives that rank of the adjugate matrix is $1$, but this now means that the adjugate is invertible (even though $A$ is not). $\endgroup$ – Marc van Leeuwen Aug 12 '13 at 4:29
  • $\begingroup$ @MarcvanLeeuwen you're right. $\endgroup$ – Olivier Bégassat Aug 12 '13 at 14:30

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