det proof in fact

Algebra Level 2

$A,B\in M_n(\mathbb{R}), AB=0,$ find the minimum of $\det(A^2+B^2).$

The answer is 0.

1 solution

Chris Lewis
Feb 14, 2020

It's trivial to achieve $\det \left(A^2+B^2\right)=0$ by taking $A=B=\bold0$ . The question is, can we make the determinant negative? There are two cases to explore:

Case 1: $\det A \neq 0$

The matrix $A$ is invertible; so $B=A^{-1} \bold0=\bold0$ . Then $\det \left(A^2+B^2\right)=\det \left(A^2\right)=\left( \det A\right)^2 \ge 0$ .

This case also covers $\det B \neq 0$ .

Case 2: $\det A = \det B = 0$ :

Work in progress!! (I had an error in a previous post)

An analogous question :

If $\vec A$ and $\vec B$ be mutually perpendicular, what is the minimum value of $(|\vec A+\vec B|)^2$ ? The answer is simple. :)

A Former Brilliant Member - 1 year, 3 months ago

It's wrong totally. A, B are matrix.

Ma Jack - 1 year, 3 months ago

I see your point - I'm trying a different approach but are you going to post a solution?

Chris Lewis - 1 year, 3 months ago

det proof in fact

The answer is 0.

1 solution

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