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

## Homework Statement

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

## Homework Equations

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

Av=λv

## 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!

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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?

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

I like Serena
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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...

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! :)

I like Serena
Homework Helper
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?

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?

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

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

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?

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? :)

I like Serena
Homework Helper
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.

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! :)

I like Serena
Homework Helper
Yes.

Ok great! Thanks Serena!