# Using diagonalization, prove the matrix equals it's square

## Homework Statement

Suppose that A is a 2x2 matrix with eigenvalues 0 and 1. Using diagonalization, show that A2 = A

## The Attempt at a Solution

Let $$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

Av=λv where $$v=\begin{pmatrix}x\\y\end{pmatrix}$$ and x,y≠0
If λ=0 then $$ax+by=0$$ and $$cx+dy=0$$
If λ=1 then $$ax+by=1$$ and $$cx+dy=1$$

so Av-λv=0, then Av-λIv=0, then (A-λI)v=0. Since v≠0, then (A-λI)=0
so for λ=0 $$\begin{pmatrix}a&b\\c&d\end{pmatrix}$$ and $$ax+by=0$$ and $$cx+dy=0$$
For λ=1 $$\begin{pmatrix}a-1&b\\c&d-1\end{pmatrix}$$ and $$ax-x+by=1$$ and $$cx+dy-y=1$$

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.

$$X^{-1}AX= \begin{pmatrix}0&0\\0&1\end{pmatrix}$$

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

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

micromass
Staff Emeritus
Homework Helper
You don't need to specifically find your matrix X. Only knowing that the matrix X exists would suffice. So the only thing you need is that there exists a matrix X such that

$$XAX^{-1}$$

is diagonal. You don't need the specific form of X.

You don't need to specifically find your matrix X. Only knowing that the matrix X exists would suffice. So the only thing you need is that there exists a matrix X such that

$$XAX^{-1}$$

is diagonal. You don't need the specific form of X.

In order for the matrix X to exist, it must be invertible (it must be nxn) and since it must be able to multiply by A (2x2), X must also be 2x2.

Since there are two distinct eigenvalues, there must be two linearly independant eigenvectors. These two distinct eigenvectors can for X.

I'm still a bit confused on how to prove exactly why there must be two linearly independent eigenvectors.

micromass
Staff Emeritus
Homework Helper
In order for the matrix X to exist, it must be invertible (it must be nxn) and since it must be able to multiply by A (2x2), X must also be 2x2.

Since there are two distinct eigenvalues, there must be two linearly independant eigenvectors. These two distinct eigenvectors can for X.

I'm still a bit confused on how to prove exactly why there must be two linearly independent eigenvectors.

You can prove this directly. If v and w are eigenvectors belonging to distinct eigenvalues, then v and w are independent. Thus if $Av=\lambda v$ and if $Aw=\mu w$ and if $\lambda =\mu$, then v and w are independent.

Try to prove this. Prove it by contradiction. Assume that v and w are dependent. What do you know then??

HallsofIvy
Homework Helper
You can prove this directly. If v and w are eigenvectors belonging to distinct eigenvalues, then v and w are independent. Thus if $Av=\lambda v$ and if $Aw=\mu w$ and if $\lambda =\mu$, then v and w are independent.
You mean "if $\lambda\ne \mu$".

Try to prove this. Prove it by contradiction. Assume that v and w are dependent. What do you know then??

Try to prove this. Prove it by contradiction. Assume that v and w are dependent. What do you know then??

If v and w are dependent, then w=cv which could be put into $Av=\lambda v$ and $Aw=\mu cv$ so $\lambda =\mu c$ and they are dependent. But 0 and 1 are independent, so this is a contradiction.

micromass
Staff Emeritus
If v and w are dependent, then w=cv which could be put into $Av=\lambda v$ and $Aw=\mu cv$ so $\lambda =\mu c$ and they are dependent. But 0 and 1 are independent, so this is a contradiction.