# If (Q^-1)AQ=D, then columns of Q are Eigenvectors of A

1. Oct 29, 2011

### blashmet

1. The problem statement, all variables and given/known data

Prove: if (Q^-1)AQ=D, then each column of Q is an eigenvector of A.

2. Relevant equations

A vector v is an eigenvector of A iff there exists a scalar λ such that:

Av=λv

3. The attempt at a solution

Suppose (Q^-1)AQ=D. We need to show each column of Q is an eigenvector of A.

At this point, should I actually write out (Q^-1)AQ=D in general element form? That is, should I write out (Q^-1),A,Q,D in the form of nxn matrices?

I'm honestly not even sure how to begin the proof, but any help would be appreciated.

Thanks!

2. Oct 29, 2011

### kru_

Well, since Q is invertible, we have that the columns of Q are linearly independent.

Since Q is nxn, we have a linearly independent list of n vectors in an n-dimension subspace, therefore we have a basis for the subspace.

Does that help to get you started?

3. Oct 29, 2011

### deluks917

AQ = QD. The other (maybe better) way to see it is to think about change of basis.

4. Oct 29, 2011

### I like Serena

Welcome to PF, blashmet!

As deluks917 said: AQ=QD

Now write Q as (q1 q2 ...) where q1, q2 are the column vectors.
And write D out as a diagonal matrix with lambda1, lambda2, etcetera.
Leave A as it is.

Now write out the matrix multiplication...

5. Oct 29, 2011

### blashmet

Thanks for the help guys! This is what I have so far:

Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn]

AQ=A[q1,...,qn]=[Aq1,...,Aqn]

QD=[d1q1,d2q2,...,dnqn]

so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn]

I'm stuck at this point. How do I proceed from here?

Thanks! :)

6. Oct 29, 2011

### I like Serena

Good!

You have to prove that each column of Q is an eigenvector of A.

q1 is the first column of Q.
Does it satisfy the criteria for an eigenvector of A?

7. Oct 29, 2011

### blashmet

Hi Serena!

In a formal proof of this, do I need to have all of this first?:

Q(Q^-1)AQ=QD iff AQ=QD=diag[d1,...,dn]

AQ=A[q1,...,qn]=[Aq1,...,Aqn]

QD=[d1q1,d2q2,...,dnqn]

so AQ=QD iff [Aq1,...,Aqn]=[d1q1,...,dnqn]

Now, is there where I show q1 is an eigenvector for A? If so, I'm confused, because the last line of the proof has [d1q1,...,dnqn]. Don't I need to show d1q1 is an eigenvector?

8. Oct 29, 2011

### deluks917

What is the first column of AQ? What is the first coulumn of QD?

9. Oct 29, 2011

### blashmet

Hi deluks917,

i'm not sure how you'd write it, but i know that AQ and QD are nxn square, so it's probably just a shorthand notation for the general element form of the multiplication of 2 nxn square matrices.

can you answer my other questions? (it's ok if you don't know, i just wasn't sure if you saw them

10. Oct 29, 2011

### deluks917

I'm not sure why you are confused. You wrote AQ = QD implies:

[Aq1,...,Aqn]=[d1q1,...,dnqn]

or
Aq1 = d1q1
...
Aqn = dqn

Do you see the eigenvectors?

11. Oct 30, 2011

### blashmet

It looks like the eigenvalues are dn (the diagonal components of the matrix D).

So I suppose the eigenvectors would be qn (column vectors).

I guess my problem is this:

I need to write out a formal proof from beginning to end, and I'm not sure what to add to what I've already written (or if I need to do more than this).

Can you help me with that? :)

12. Oct 30, 2011

### I like Serena

You are done.

d1...dn are the eigenvalues
and q1...qn are the eigenvectors,
since they satisfy Av=λv.
So each column of Q is an eigenvector of A.

13. Oct 30, 2011

### blashmet

Does the following suffice as a complete, formal proof?

"d1...dn are the eigenvalues
and q1...qn are the eigenvectors,
since they satisfy Av=λv.
So each column of Q is an eigenvector of A."

Thanks Serena! :)

14. Oct 30, 2011

### I like Serena

Yes.

15. Oct 30, 2011

### blashmet

Ok great! Thanks Serena!