Using diagonalization, prove the matrix equals it's square

[PirateFan308]

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]

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.

 

[PirateFan308]

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]

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]

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??

 

[PirateFan308]

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]

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.

What do you mean, 0 and 1 are independent?? 0 and 1 are numbers, not vectors. Saying that they are independent makes no sense.