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Given a characteristic polynomial $P$ of matrix $A$ I need to show that the characteristic polynomial $O$ of $A^2$ can't have more different real roots than $P$.

I know that the characteristic polynomial for both cases can be calculated like this:
$P = |A - \lambda I| = 0$
$O = |A^2 - \lambda^2 I| = 0$
But in a general case with $n*n$ matrices they become way too complicated.

Can anyone guide me in the right direction?
Thanks!

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That's wrong, isn't it? $A = \begin{pmatrix} 0 & -1 \\\ 1 & 0 \end{pmatrix}$ has charakteristic polynomial $P(t) = t^2 + 1$ and $A^2 = -\mathrm{Id}$, which has $Q(t) =(t+1)^2$? –  martini May 10 '12 at 13:00
    
Shouldn't it be $Q(t) = (t^2 + 1)^2$? In this case both $P(t)$and $Q(t)$ have 0 real roots. –  fikster May 10 '12 at 13:13
    
Because $Q(t) = A^2 - t^2I = \begin{vmatrix} -1 - t^2 & 0\\ 0 & -1 - t^2 \end{vmatrix} = (1+t^2)^2$ –  fikster May 10 '12 at 13:21
    
In your equation for the characteristic polynomial of $A^2$ you shouldn't have $\lambda^2$, you should just have $\lambda$. It is true though that the square of any eigenvalue for $A$ will automatically be an eigenvalue of $A^2$. –  rschwieb May 10 '12 at 13:21
    
It's plain to see that $A^2=\begin{pmatrix} -1 & 0 \\\ 0 & -1 \end{pmatrix}$ has $-1$ as an eigenvalue. You can't go wrong: every vector is an eigenvector for that particular $A^2$! –  rschwieb May 10 '12 at 13:30
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2 Answers

This is false, for somewhat trivial reasons. Let $A$ be a matrix with eigen values $\lambda_1,\dots,\lambda_n$ then note that the eigenvalues of $A^2$ are $\lambda_1^2,\dots,\lambda_n^2$ since if $v_i$ is an eigenvector of $\lambda_i$ then

$$A^2v_i=A(\lambda v_i)=\lambda Av_i=\lambda^2v_i.$$

For instance if the eigenvalues are purely imaginary the number of real eigenvalues increases. To take an example from the comments

$$A=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$$

then $\mathsf{char}(A)=x^2+1$ which has imaginary roots $\pm i$ but $\mathsf{char}(A^2)=(x+1)^2$ has real repeated root $-1$.

Are there some more conditions on your matrix perhaps?

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I only assumed it was true, since I didn't think about using complex roots. What if I changed the question from real roots to complex roots? –  fikster May 10 '12 at 13:34
    
Most plausibly, the question should have been "can't have less" rather than "can't have more". Or, "can't have more complex roots" rather than "can't have more real roots". –  rschwieb May 10 '12 at 13:35
    
@fikster If you change the question to complex roots, there's not much to prove. Since every eigenvalue of a real matrix is going to be complex. I suppose you could consider complex eigenvalues that aren't purely real, but that seems a bit silly. Do you have some sort of context for this question? Are you trying to prove something else and this is a step in that proof? –  JSchlather May 10 '12 at 13:38
    
The question was about different (unique) roots, so if I changed it to complex roots, then: $char(A) = x^2 +1$ has 2 unique complex roots, while $char(A^2) = (x+1)^2$ has one unique (complex) root, which is less. –  fikster May 10 '12 at 13:46
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Since the roots of these polynomials are eigenvalues, let's just think in terms of eigenvalues.

The eigenvalues of $A^2$ are just the eigenvalues of $A$, squared (right? It's tempting to believe, and scribbling on a napkin made it seem so but I could be overlooking something silly).

In that case, the real eigenvalues of $A$ would stay real after squaring, and some complex eigenvalues might become real after squaring.

This would show that $A^2$ has at least as many real eigenvalues as $A$. (This is not the original question, but it seems like this might have been the intended question.)

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