# 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

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

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