2
$\begingroup$

Let $A$ be a non-singular square matrix. We know that $A \cdot \operatorname{adj}A = \det A \cdot I$. This implies that $\det\left(\operatorname{adj} A\right) = \left(\det A\right)^{n-1}$. Hence $\operatorname{adj} A$ is non-singular.

But how to understand it in terms of linearly independent rows or columns?

It means that when each entry of a non singular matrix is replaced by its co-factor, linear independence of rows and columns is preserved. I have no idea how to think about it. Any help is highly appreciated. Thanks!

$\endgroup$
  • 1
    $\begingroup$ Normally, adjugates are not understood to be related to linear dependence (though I would be happy to learn otherwise), so the proof you sketched using determinants seems to be the best one to me. $\endgroup$ – darij grinberg Aug 10 '15 at 11:30
0
$\begingroup$

Not what you asked but I want to sat that there is a more general theorem saying that

a. If $A$ is invertible then adj$(A)$ is invertible.

b. If rank$(A)=n-1$, ($A$ is an $n\times n$ matrix), then rank$($adj$(A)=1$.

c. If rank$(A)<n-1$, adj$(A)=0$.

If rank$(A)<n-1$ then the determinant of all the $n-1\times n-1$ submatrices is zero.

If rank$(A)=n-1$, then since adj$(A)\cdot A=0$ it follows that col$(A)\subseteq$ker$(adj(A))$. Since dim(col$(A))=rank(A)$ the result follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.