Linear Proof, don't know how to start it

[Dustinsfl]

Let A be nxn and let $$B=I-2A+A^2$$.
Show that if $$\mathbf{x}$$ is an eigenvector of A belonging to an eigenvalue $$\lambda$$ of A, then $$\mathbf{x}$$ is also an eigenvector of B belonging to an eigenvalue $$\mu$$ of B. How are $$\lambda$$ and $$\mu$$ related?

 

[nicksauce]

Well... what happens if you act 1 - 2A + A^2 on x?

 

[Dustinsfl]

Well... what happens if you act 1 - 2A + A^2 on x?

I don't know what you mean by that.

 

[nicksauce]

You want to show that x is an eigenvector B, meaning that for some constant u, Bx = ux.
Thus you need to calculate Bx = (1 -2A + A^2)x = x - 2Ax + A^2x. Can you simplify that further now?

 

[Dustinsfl]

$$B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mu\mathbf{x}$$

?

 

[Dick]

Evaluate A(lambda*x), please?

 

[Dustinsfl]

Evaluate A(lambda*x), please?

$$B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(\lambda)^2\mathbf{x}=\mu$$

 

[Dustinsfl]

Then do the quadratic equation or factor which ever is appropriate?

 

[Dick]

Then do the quadratic equation or factor which ever is appropriate?

You don't have to do much of anything. Just factor x out and say how Bx is related to mu*x.

 

[Dustinsfl]

Should I just solve for mu then by multiplying by 1/x?

 

[Dick]

Should I just solve for mu then by multiplying by 1/x?

No, there's no such thing as 1/x if x is a vector. That's exactly the wrong answer. Just look at x-2*lambda*x+lambda^2*x=mu*x=Bx and tell me what mu is.

 

[Dustinsfl]

So $$\mu=I-2\lambda+A\lambda$$.

 

[Dick]

So $$\mu=I-2\lambda+A\lambda$$.

mu is a scalar. It's a number. It's not a matrix. It's an EIGENVALUE. Try that again. You were almost there. Now you are going backwards.

 

[Dustinsfl]

mu is a scalar. It's a number. It's not a matrix. It's an EIGENVALUE. Try that again. You were almost there. Now you are going backwards.

$$\mu=1-2\lambda+A\lambda$$

But there is still A

 

[Dick]

The A is there because once upon a time I asked you to evaluate A(lambda*x). And you did it right. And then you forgot. Why did you do that? That's why I feel things are going backwards.

 

[Dustinsfl]

$$B\mathbf{x}=\mathbf{x}-2A\mathbf{x}+A^2\mathbf{x}=\mathbf{x}-2\lambda\mathbf{x}+A(\lambda\mathbf{x})=\mathbf{x}-2\lambda\mathbf{x}+(\lambda)^2=\mu\mathbf{x}$$

I did. I just decide not use it apparently.

 

[Dick]

I did. I just decide not use it apparently.

Ok, so 1-2*lambda+lambda^2=mu. Right? I so very hope you agree with this.

 

[Dustinsfl]

I concur.