0
$\begingroup$

A matrix $A\in\mathbb{R}^{n,n}$ is diagonalisable on $\mathbb{R}$. Then:

a. characteristic polynomial of $A$ is product of polynomials degree $1$ with real coefficients

b. matrix $A^2$ is also diagonalisable

c. $A=A^T$ .

b. is obviously true, we know that $A^2=(PDP^{-1})^2 = PD^2P^{-1}$ Due to $D^2$ is still diagonal then b. is true.
a. We know that $A$ is similar to some diagonal matrix $D$. Therefore, $A$ has the same characteristic polynomial as $D$, so it is product of $(x-\lambda)$ where $\lambda$ is any eigenvalue. Lets consider:
$A=\left( \begin{array}{ccc} 2 & 0 \\ 0 & 2 \\ \end{array} \right)=\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{ccc} 2& 0 \\ 0 & 2 \\ \end{array} \right)\left( \begin{array}{ccc} 0& 1 \\ 1 & 0 \\ \end{array} \right)$ is diagonalisable, but $P_A(x)=(x-2)^2$. So it is not prodcut of degree 1 polynomials. Hence, a. is not true.

What about these solutions ? Can you help me solve c. ?

$\endgroup$
  • $\begingroup$ For a and your example, $(x-2)^2=(x-2)(x-2)$, then $P_A$ is product of polynomials degree 1 with real coefficients. (they are the same I agree but the assumption does nos require them to be distincts) $\endgroup$ – nicomezi Sep 7 '16 at 10:25
  • 1
    $\begingroup$ c is saying that every real diagonalizable matrix is symmetric. Do you really believe this? $\endgroup$ – RKD Sep 7 '16 at 10:27
  • $\begingroup$ So, only b. is correct. No, I don't believe it. Nevertheless, Can you help me solve a. and c. ? $\endgroup$ – Happy man Sep 7 '16 at 10:29
  • 1
    $\begingroup$ a) is true. Indeed, the characteristic polynomial factorizes as $\prod_{\lambda} (x- \lambda)$, where $\{ \lambda \}$ are the eigenvalues of $A$ (i.e. the diagonal entries of $D$). $\endgroup$ – Crostul Sep 7 '16 at 10:32
  • $\begingroup$ From what you know that $\lambda\in \mathbb{R}$ $\endgroup$ – Happy man Sep 7 '16 at 10:36
0
$\begingroup$

For $c)$ , just take the matrix $\pmatrix{1&2\\3&4}$ which is not symmetric , but real diagonalizable.

$\endgroup$
  • $\begingroup$ What about b. ? $\endgroup$ – Happy man Sep 7 '16 at 10:36
  • $\begingroup$ Sorry, I meant a) $\endgroup$ – Happy man Sep 7 '16 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.