Question

Let A be a diagonalizable matrix with every eigenvalue of A either 0 or 1. Prove that A is idempotent (A^2=A).

Answer 1

This is what I've got so far:

\[\text{By the definition of an eigenvalue, there exists nonzero vector } \vec x \text{ such that} \]\[A\vec x =\lambda \vec x \]\[\text{But }\lambda =\lambda^2 \text{ for all }\lambda=0,1. \text{ Hence, }\]\[A\vec x =\lambda^2 \vec x\]\[A\vec x=\lambda \left( \lambda \vec x \right)\]

Answer 2

\[\text{Multiplying both sides by A on the \left}\]\[AA\vec x = A\lambda^2 \vec x\]\[\text{But }\lambda^2 = \lambda \text{ so we have}\]\[A^2 \vec x=A\lambda \vec x\]

Answer 3

\[A^2 \vec x=\lambda A \vec x\]\[\text{But }A \vec x=\lambda \vec x \text{ so we have}\]\[A^2 \vec x=\lambda (\lambda \vec x)\]\[A^2 \vec x=\lambda^2 \vec x\]\[A^2 \vec x=\lambda \vec x\] Hence A is nilpotent

Answer 4

Does that suffice?

Answer 5

Correction: idempotent

Answer 6

D with eigenvalues on the main diagonal yeah

Answer 7

\[A^2=P^{-1}DP\]

Answer 8

er
D^2

Answer 9

yeah that's much easier... I have no experience doing a lot of proofs

Answer 10

hate going about it the long way when there's a simple solution

Answer 11

thanks for your help

Answer 12

your prof asks you to prove it?

Answer 13

yep

Answer 14

Prove theorem, it means you are junior math major in university?

Answer 15

i'll minor in math