# Diagonalisable real matrix. Conclusion

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. ?

• 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) – nicomezi Sep 7 '16 at 10:25
• c is saying that every real diagonalizable matrix is symmetric. Do you really believe this? – RKD Sep 7 '16 at 10:27
• So, only b. is correct. No, I don't believe it. Nevertheless, Can you help me solve a. and c. ? – Happy man Sep 7 '16 at 10:29
• 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 $D$). – Crostul Sep 7 '16 at 10:32
• From what you know that $\lambda\in \mathbb{R}$ – Happy man Sep 7 '16 at 10:36

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