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

  • Thread starter blashmet
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  • #1
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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!
 

Answers and Replies

  • #2
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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?
 
  • #3
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AQ = QD. The other (maybe better) way to see it is to think about change of basis.
 
  • #4
I like Serena
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Welcome to PF, blashmet! :smile:

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
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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
I like Serena
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Good! :wink:

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
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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?

Thanks for your help! :smile:
 
  • #8
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What is the first column of AQ? What is the first coulumn of QD?
 
  • #9
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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
382
4
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
15
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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
I like Serena
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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
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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
I like Serena
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Yes.
 
  • #15
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Ok great! Thanks Serena!
 

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