InsideDarkWeb.com

Let $x=begin{bmatrix}3cr4end{bmatrix}$ and $A=begin{bmatrix}0&x^Tcr x&0end{bmatrix}$ is A diagonizable?

I had a problem: let $x=begin{bmatrix}3cr4end{bmatrix}$ and $A=begin{bmatrix}0&x^Tcr x&0end{bmatrix}$ is A diagonizable?

But when I plug in the matrix x and its transpose into A the dimensions don’t work out and we have empty slots where there should be elements and I was wondering if I’m just tripping or if this problem is unsolvable?

Mathematics Asked on November 14, 2021

1 Answers

One Answer

Presumably the bottom right $0$ is actually a $2 times 2$ matrix of $0$, i.e. $$A = left[matrix{0 & 3 & 4cr 3 & 0 & 0cr 4 & 0 & 0cr} right]$$

Answered by Robert Israel on November 14, 2021

Add your own answers!

Related Questions

Abelian subgroup of prime power index

1  Asked on February 9, 2021 by devendra-singh-rana

       

Manifolds Definition

1  Asked on February 8, 2021

   

How many passwords possible?

3  Asked on February 8, 2021 by sagigever

 

Sets; is (A∩B)∪C the same as A∩(B∪C)?

1  Asked on February 8, 2021 by user839131

 

Graph theory: strong regular graph

1  Asked on February 7, 2021 by spencer-ireland

 

Find solution of this ODE

1  Asked on February 7, 2021 by ongky-denny-wijaya

 

Ask a Question

Get help from others!

© 2021 InsideDarkWeb.com. All rights reserved.