Is the following matrix diagonalizable?

Question

Determine if this matrix is diagonalizable. $$ C= \begin{bmatrix} \frac{\sqrt2}2 & 0 & -\frac{\sqrt2}2 \\ 0 & 1 & 0 \\ \frac{\sqrt2}2 & 0 & \frac{\sqrt2}2 \end{bmatrix} $$ I worked out the determinant of its characteristic equation and got 1 real eigenvalue and 2 other imaginary eigenvalues. I concluded to say that the matrix was NOT diagonalizable because by definition $\dim(C)$ needs to equal # of distinct eigenvalues for the matrix to be diagonalizable. in this case there is only one real distinct eigenvalues and $\dim(C)=3$. Are imaginary eigenvalues distinct? I don't think so. What do you think?

Answer 1

If you have eigenvalues real $\lambda$ and complex $\gamma$, you will have $\overline{\gamma}$ since if a complex number is a root of a polynomial (characteristic equation), so is its conjugate. These are three distinct eigenvalues.

In short, an imaginary eigenvalue still counts.

Note: I did not check your computation because it is hard to read your post with the current formatting.

Answer 2

This matrix represents a rotation around the axis $y$ (see here). So it is not diagonalizable in $M(3;\mathbb{R})$. The only real eigenspace is spanned by the eigenvector $(0,1,0)^T$ of the eigenvalue $\lambda_1=1$. The other two eigenvalues are $\lambda_{2,3}=\frac{1\pm i} {\sqrt{2}}$ and have no real eigenvectors.