4
$\begingroup$

A is a non singular square matrix of order 2 such that |A + |A|adjA| = 0, where adjA represents adjoint of matrix A, and |A| represents det(A) (determinant of matrix A)

Evaluate |A – |A|adjA|.

I believe this can be solved using Cayley Hamilton Theorem, but I'm not sure exactly how one would proceed to find the value of the required determinant. The motivation behind my thought was appearance of |A - kI| = 0 in the determinant to be evaluated, where k is a constant.

#1 Of course, |A|adjA can be reduced to |A|² A-1, leaving us with |A + |A|² A-1|. I don't know how to take it from here. A detailed solution using Cayley Hamilton theorem would be appreciated.

#2 I guess that for a square matrix of order n, the required determinant would be equal to n². Could someone please help prove or disprove the result, by generalising for a nxn matrix?

Thanks a lot!

$\endgroup$
1
$\begingroup$

We can take your expression $|A + |A|^2A^{-1}| = 0$, noting that we have assumed that $A$ is invertible, and find that $$ |A| \cdot |A + |A|^2A^{-1}| = 0 \implies\\ |(A + |A|^2A^{-1})A| = 0 \implies\\ |A^2 + |A|^2I| = 0 $$ In other words, if $A$ is invertible, then the condition given is equivalent to stating that $-|A|^2$ is an eigenvalue of $A^2$, which is true if and only if either $i|A|$ or $-i|A|$ is an eigenvalue of $A$ (if $A$ is a real matrix, then both must be eigenvalues). Since the product of the eigenvalues of $A$ is $|A|$, we can determine both eigenvalues of $A$, which allows us to evaluate $$ |A - |A|\operatorname{adj}(A)| = \frac{|A^2 - |A|^2I|}{|A|} $$ In the $n \times n$ case, we should not expect that knowing $|A + |A|^2A^{-1}| = 0$ is enough information to determine $|A - |A|\operatorname{adj}(A)|$. The case in which $|A| = 0$ should be handled separately, but perhaps continuity arguments suffice.


So, we have determined that the eigenvalues of $A$ are either $i|A|$ and $-i$ or $-iA$ and $i$. We compute $$ |A^2 - |A|^2 I| = \prod_{i=1}^2(\lambda_i^2 - |A|^2) = (-1 - |A|^2)(-|A|^2 - |A|^2) = 2|A|^2(1 + |A|^2) $$

$\endgroup$
  • $\begingroup$ Hmm, so what exactly is the value of that determinant? I'm afraid I still don't get it. $\endgroup$ – arya_stark Dec 27 '17 at 17:53
  • 1
    $\begingroup$ So we have determined that the eigenvalues of $A$ are either $i|A|$ and $-i$ or $-iA$ and $i$. We compute $$ |A^2 - |A|^2 I| = \prod_{i=1}^2(\lambda_i^2 - |A|^2) = (-1 - |A|^2)(-|A|^2 - |A|^2) = 2|A|^2(1 + |A|^2) $$ $\endgroup$ – Omnomnomnom Dec 27 '17 at 17:57
  • $\begingroup$ Could you just add it to the answer? LaTeX doesn't seem to work in the comments, and I'm not familiar with the code. $\endgroup$ – arya_stark Dec 27 '17 at 17:58
  • $\begingroup$ @user28968 That's strange; LaTeX works in the comments for me, and it is fairly common for people to use code like this in the comments. You may want to post a question about that to this site's meta (if you want the code in comments to compile properly). $\endgroup$ – Omnomnomnom Dec 27 '17 at 18:00
  • $\begingroup$ Another thing, the answer given for this problem is 4, which holds when |A| = ±1, according to the expression you arrived at. How do we show that |A| = ±1? $\endgroup$ – arya_stark Dec 28 '17 at 2:54
1
$\begingroup$

$\text{adj A}=|A|A^{-1}$, at least when $A$ is non-singular, so $$A-|A|\text{adj A}=A-|A|^2A^{-1}=A^{-1}(A-|A|I)(A+|A|I).$$ The determinant of this is $$|A|^{-1}\chi(|A|)\chi(-|A|)$$ where $\chi(x)=\det(A-xI)$ is the characteristic polynomial of $A$. If we write $\chi(x)=x^2-tx+|A|$ (where $t$ is the trace of $A$) then $\chi(\pm |A|)=|A|(1\mp t+|A|)$ etc.

$\endgroup$
  • $\begingroup$ So how can we compute the required determinant? $\endgroup$ – arya_stark Dec 27 '17 at 17:54

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.