# 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!!