# Is the following matrix diagonalizable?

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?

• It's almost impossible to understand what you wrote for $\;C\;$ . Please do follow the easy directions of the site to write mathematics. – DonAntonio May 11 '16 at 19:23

## 2 Answers

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.

• A rotation matrix is not diagonalizable over the field of all reals but is diagonalizable over the field of all complex, how did you determine that M(3;all real)? – Peter B. May 11 '16 at 20:40
• I've said that the matrix is not diagonalizable in $M(3,\mathbb{R})$, but obviously it is diagonalizable in $M(3,\mathbb{C})$ since it has three different eigenvalues in $\mathbb{C}$, so we have three linearly independent eigenspaces. – Emilio Novati May 12 '16 at 14:37

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.