Polynomial of Jordan Decomposition

[azdang]

Homework Statement

Let A be a real or complex nxn matrix with Jordan decomposition A = $$X \Lambda X^{-1}$$ where $$\Lambda$$ is a diagonal matrix with diagonal elements $$\lambda_1,...,$$ $$\lambda_n$$. Show that for any polynomial p(x):
p(A)=$$Xp(\Lambda)X^{-1}$$

$$p(\Lambda)$$ should really be the matrix with p($$\lambda_j$$) on its diagonal for j=1,...,n but I couldn't figure out how to make that matrix in latex.

The Attempt at a Solution

I'm guessing there should be a way to take p of both sides and somehow extract the X and X inverse, but I can't seem to figure it out. Does anyone see anything? Thank you.

 

[tiny-tim]

Hi azdang!

(for matrices etc in LaTeX, see http://www.physics.udel.edu/~dubois/lshort2e/node56.html#SECTION00850000000000000000 )

Hint: if the polynomial is A², then A²= XΛX^-1XΛX^-1 = XΛ²X^-1 = … ?

 

[azdang]

Hey tiny-tim. Thank you for your response. Quick question: Is it okay to assume p(A) = $$A^2$$? (This would be the polynomial p(x)=$$x^2$$ right?) Can we say that because the problem say 'for any polynomial'? Or is there a more general way to prove this?

Also, I do see then that $$\Lambda^2=p(\Lambda)$$, so that is all clear. Thank you very much. I'm just wondering if we should find a way to represent p(x) more generally?

 

[tiny-tim]

Is it okay to assume p(A) = $$A^2$$?
…
I'm just wondering if we should find a way to represent p(x) more generally?

Yes, we need to deal with a general p(x).

So you do still need to prove it for p(A) = Aⁿ, for any n …

my n = 2 was just an example for you.

 

[azdang]

Alright, cool.

So, we could say p(x) = xⁿ.

Then, p(A)=Aⁿ=$$(X \Lambda X^{-1})^n$$=$$X \Lambda X^{-1}X \Lambda X^{-1}...X \Lambda X^{-1}$$.

All of the X's and X^-1 will cancel out except for the ones on the edges so we are left with: p(A)=$$X \Lambda^n X^{-1}$$.

So, we know p($$\Lambda$$)=$$\Lambda^n$$. Can we say that that is equivalent to the matrix with p($$\lambda_j$$) on its diagonal where j=1,...,n because it is a diagonal matrix?

 

[tiny-tim]

just got up :zzz: …

… So, we know p($$\Lambda$$)=$$\Lambda^n$$. Can we say that that is equivalent to the matrix with p($$\lambda_j$$) on its diagonal where j=1,...,n because it is a diagonal matrix?

Yes, of course (for Λ^m).

(I don't think the examiner would even expect a reason for that )

And then prove that it's "additive", in the sense that if B = XΛ_BX^-1 and C = xΛ_CX^-1 then Λ_B+C = … ?

 

[azdang]

I'm kind of confused by this last part. I don't really understand the subscripts on Lambda or where B and C are coming from. Thank you again, tiny-tim, for helping me get through this :)

 

[tiny-tim]

I'm kind of confused by this last part. I don't really understand the subscripts on Lambda or where B and C are coming from. Thank you again, tiny-tim, for helping me get through this :)

I mean, if B and C are real or complex nxn matrices with Jordan decomposition B = XΛ_BX^-1 and C = XΛ_CX^-1 and B + C = XΛX^-1 where Λ Λ_B and Λ_C are diagonal matrices, what is the equation relating Λ Λ_B and Λ_C ?

 

[azdang]

But we cannot be guaranteed that the X's in the Jordan decompositions for B and C are the same, can we?

 

[tiny-tim]

But we cannot be guaranteed that the X's in the Jordan decompositions for B and C are the same, can we?

??

Assume they are … then prove the proposition … and then use it to answer the original question.

 

[azdang]

Well then, does $$X\Lambda_B X^{-1} + X\Lambda_C X^{-1}=X(\Lambda_B + \Lambda_C)x^{-1}$$?

Then $$p(\Lambda_B + \Lambda_C)=p(\Lambda_B)+p(\Lambda_C)$$, so the matrix would be the diagonal matrix with $$\lambda_B_j + \lambda_C_j$$ on the diagonal, j=1,...n, right? I'm not sure if I'm understanding all this, or why we are showing it is additive. Sorry!

 

[tiny-tim]

Well then, does $$X\Lambda_B X^{-1} + X\Lambda_C X^{-1}=X(\Lambda_B + \Lambda_C)x^{-1}$$?

That's right!

Then $$p(\Lambda_B + \Lambda_C)=p(\Lambda_B)+p(\Lambda_C)$$, so the matrix would be the diagonal matrix with $$\lambda_B_j + \lambda_C_j$$ on the diagonal, j=1,...n, right? I'm not sure if I'm understanding all this, or why we are showing it is additive.

Because it's easy to prove the question for p(A) = A^m, but not for p(A) = A^m + A^l …

so we just prove it for A^m, show that it's the same X for any m, and then use the theorem above to prove that it's true for any sum of powers of A, ie for any polynomial.

 

[azdang]

Okay, I think I'm understanding, though I might have a hard time explaining it. One thing in what you just said that confused me, what do you mean 'show that it's the same X for any m'. I really can't thank you enough for walking me through this btw.

 

[tiny-tim]

One thing in what you just said that confused me, what do you mean 'show that it's the same X for any m'.

I mean, for example, A² = XΛ²X^-1 and A³ = XΛ³X^-1 (and it's the same X, which was worrying you earlier )

 

[azdang]

Oooh, okay. Yes, I can see right away that they are the same X!

So, I should really just show this for:
p(x)=$$\alpha_0+\alpha_1x+\alpha_2x^2+...+\alpha_nx^n$$ for scalars alpha.

P.S. I totally just did this ^ and I TOTALLY get it. Again, thank you very very much!