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## Homework Statement

Suppose that A is a 2x2 matrix with eigenvalues 0 and 1. Using diagonalization, show that A

^{2}= A

## The Attempt at a Solution

Let [tex]A=\begin{pmatrix}a&b\\c&d\end{pmatrix}[/tex]

Av=λv where [tex]v=\begin{pmatrix}x\\y\end{pmatrix}[/tex] and x,y≠0

If λ=0 then [tex]ax+by=0[/tex] and [tex]cx+dy=0[/tex]

If λ=1 then [tex]ax+by=1[/tex] and [tex]cx+dy=1[/tex]

so Av-λv=0, then Av-λIv=0, then (A-λI)v=0. Since v≠0, then (A-λI)=0

so for λ=0 [tex]\begin{pmatrix}a&b\\c&d\end{pmatrix}[/tex] and [tex]ax+by=0[/tex] and [tex]cx+dy=0[/tex]

For λ=1 [tex]\begin{pmatrix}a-1&b\\c&d-1\end{pmatrix}[/tex] and [tex]ax-x+by=1[/tex] and [tex]cx+dy-y=1[/tex]

We must find two lin. ind. vectors such that we can create X where the first column of X is the first vector, and the second column of X is the second vector.

[tex]X^{-1}AX= \begin{pmatrix}0&0\\0&1\end{pmatrix}[/tex]

If this is true, then [tex]X^{-1}A^{2}X= \begin{pmatrix}0&0\\0&1\end{pmatrix}[/tex]

The problem is, I'm not quite sure how to prove any of this