Polynomial of Jordan Decomposition

1. Apr 2, 2009

azdang

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

3. 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.

Last edited: Apr 2, 2009
2. Apr 2, 2009

tiny-tim

Hi azdang!

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

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

Last edited by a moderator: May 4, 2017
3. Apr 2, 2009

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?

4. Apr 2, 2009

tiny-tim

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.

5. Apr 2, 2009

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?

6. Apr 3, 2009

tiny-tim

just got up :zzz: …

Yes, of course (for Λm).

(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 = XΛBX-1 and C = xΛCX-1 then ΛB+C = … ?

7. Apr 3, 2009

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 :)

8. Apr 3, 2009

tiny-tim

I mean, if B and C are real or complex nxn matrices with Jordan decomposition B = XΛBX-1 and C = XΛCX-1 and B + C = XΛX-1 where Λ ΛB and ΛC are diagonal matrices, what is the equation relating Λ ΛB and ΛC ?

9. Apr 3, 2009

azdang

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

10. Apr 3, 2009

tiny-tim

??

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

11. Apr 3, 2009

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!!

12. Apr 3, 2009

tiny-tim

That's right!
Because it's easy to prove the question for p(A) = Am, but not for p(A) = Am + Al

so we just prove it for Am, 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.

13. Apr 3, 2009

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.

14. Apr 3, 2009

tiny-tim

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

15. Apr 3, 2009

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!!