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#1 h Feb 26, 2016, 4:55 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

It's true: $A,B \in M_n(\mathbb{R})$ such that $\det{(A^2+B^2)}=0$ then $\det A=\det B.$

This post has been edited 1 time. Last edited by BEHZOD_UZ, Feb 26, 2016, 4:56 PM

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Tolaso

#2 h Feb 28, 2016, 10:38 PM • 2 Y

Y by Adventure10, Mango247

BEHZOD_UZ wrote:

It's true: $A,B \in M_n(\mathbb{R})$ such that $\det{(A^2+B^2)}=0$ then $\det A=\det B.$

Probably a condition is missing? If $AB=BA$ also holds then the result holds.

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#4 h Mar 2, 2016, 4:06 PM • 1 Y

Y by Adventure10

Of course, it's not true. Take $A=\begin{pmatrix}0&1\\-1&0\end{pmatrix},B=diag(0,1)$ .

Tolaso, you said that you have a proof when $AB=BA$ . Can you give it ?

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#5 h Mar 2, 2016, 5:32 PM • 2 Y

Y by Adventure10, Mango247

In fact, " $\det(A^2+B^2)=0,AB=BA$ implies $\det(A)=\det(B)$ " is true when $n=2$ and false when $n=3$ . For $n=2$ , I used a formal computation with a computer.

For $n=3$ , one has the following counterexample: $A=I_3,B=diag(\begin{pmatrix}0&1\\-1&0\end{pmatrix},2)$ . Note that $\det(A)\not=\pm\det(B)$ .

This post has been edited 1 time. Last edited by loup blanc, Mar 2, 2016, 5:46 PM

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#6 h Mar 3, 2016, 10:28 AM • 2 Y

Y by Adventure10, Mango247

Well, the exercise must be missing another condition. We also need that $A, B$ are diagonizable. I thought that was given. My mistake.

What I have a solution to is this:

Quote:

Let $A, B \in \mathbb{R}^{n \times n}$ that are diagonizable in $\mathbb{R}$ . If $\det (A^2+B^2)=0$ and $AB=BA$ , then prove that $\det A = \det B =0$ .

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#7 h Mar 3, 2016, 3:58 PM • 5 Y

Y by Tolaso, adityaguharoy, Adventure10, Mango247, MS_asdfgzxcvb

We can do better. If (1) $\det(A^2+B^2)=0$ , (2) $AB=BA$ and (3) the spectra of $A,B$ are real, then $\det(A)=\det(B)=0$ .

Proof. According to (2),(3), $A,B$ are simultaneously triangularizable over $\mathbb{R}$ with diagonals (respectively) $(\lambda_i),(\mu_i)$ ; then $\det(A^2+B^2)=\Pi_i(\lambda_i^2+\mu_i^2)=0$ . Consequently, there is $i$ s.t. $\lambda_i=\mu_i=0$ ...

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Tolaso

#8 h Mar 3, 2016, 11:01 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

loup blanc wrote:

We can do better. If (1) $\det(A^2+B^2)=0$ , (2) $AB=BA$ and (3) the spectra of $A,B$ are real, then $\det(A)=\det(B)=0$ .

Proof. According to (2),(3), $A,B$ are simultaneously triangularizable over $\mathbb{R}$ with diagonals (respectively) $(\lambda_i),(\mu_i)$ ; then $\det(A^2+B^2)=\Pi_i(\lambda_i^2+\mu_i^2)=0$ . Consequently, there is $i$ s.t. $\lambda_i=\mu_i=0$ ...

Yep, that would be the main idea , that $A,B$ are simultaneously diagonizable or triangulazible on your part of you.

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#9 h Apr 5, 2025, 6:31 AM

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loup blanc wrote:

In fact, " $\det(A^2+B^2)=0,AB=BA$ implies $\det(A)=\det(B)$ " is true when $n=2$ and false when $n=3$ . For $n=2$ , I used a formal computation with a computer.

Do u have a proof for it (instead of a formal computation)? If so, kindly share it..

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#10 h Apr 5, 2025, 9:46 AM • 1 Y

Y by teomihai

@ Saucepan_man02. Here $n=2$ .
Since $A,B$ are real and $AB=BA$ , $A^2+B^2=(A+iB)(A-iB)$ , $\det(A^2+B^2)=|\det(A+iB)|^2$ and
$\det(A+iB)=\det(A-iB)=0$ ; let $f(x)=\det(A+xB)=\det(B) x^2+ux+\det(A)$ ; then $f(i)=f(-i)=0$
implies that $\det(A)=\det(B)$ $\square$

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