# Diagonlizing a 3 x 3 matrix with complex eigens

Okay so I've never learned how to diagonlize a 3 x 3 matrix with complex eigens. I have done 2 x2s.

$\begin{bmatrix} -1/4 &1/4 + 1/\sqrt{2} &-1/2+1/(2\sqrt{2}) \\ 1/4 - 1/\sqrt{2}&-1/4 &-1/2 - 1/(2\sqrt{2}) \\ -1/2-1/(2\sqrt{2}) & -1/2+1/(2\sqrt{2}) &1/2 \end{bmatrix}$

On Mathematica, I got the following (to save you all some time)

$\lambda = 1 \sim (-1/2,-1/2,1)$

$\lambda = -1/2 - i\sqrt{3}/2 \sim (1+i\sqrt{3/2}, 1-i\sqrt{3/2},1)$

$\lambda = -1/2 + i\sqrt{3}/2 \sim (1-i\sqrt{3/2},1+i\sqrt{3/2},1)$

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My guess is you think that because your starting matrix is real, the diagonalization should be real as well. (Just a guess! If this isn't your issue with using the complex eigenvalues/vectors, do explain!)

This is not true, as your example already shows. Note that, as your Mathematica output gives, you'll want to diagonalize (up to order...) with $$D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -\frac{1}{2} - \frac{i\sqrt{3}}{2} & 0 \\ 0 & 0 & -\frac{1}{2} + \frac{i\sqrt{3}}{2} \end{pmatrix}$$ and $$P = \begin{pmatrix} -\frac{1}{2} & 1 + i\sqrt{\frac{3}{2}} & 1 - i\sqrt{\frac{3}{2}} \\ -\frac{1}{2} & 1 - i\sqrt{\frac{3}{2}} & 1 + i\sqrt{\frac{3}{2}} \\ 1 & 1 & 1 \end{pmatrix}$$ so that $$P^{-1} = \frac{1}{12}\begin{pmatrix} -4 & -4 & 8 \\ -i(2i + \sqrt{6}) & i(-2i + \sqrt{6}) & 2 \\ i(-2i + \sqrt{6}) & -i(2i + \sqrt{6}) & 2 \end{pmatrix}$$ and $$P^{-1}AP = D$$ or $$PDP^{-1} = A,$$ whichever you prefer.

The reason for this is because the eigenvectors are complex conjugates, so even though the eigenvectors are complex, all of the complex parts will cancel out once you multiply everything through (recall, $(a+bi)(a-bi) = a^2 + b^2$!). Thus, as the other answer already suggests, the method for diagonalizing a real matrix is the same, regardless of what field (real or complex) the eigenvalues/vectors lie in. It just turns out that if you start with a real matrix, your eigenvectors, if complex, will always cancel out when you multiply everything through!

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There is no difference in the procedure depending on the eigenvalues being real or complex. You can repeat the same procedure for diagonalizing matrices with real eigenvalues.

Step 1. Find the eigenvalues of $A$.

Step 2. Find three linearly independent eigenvectors of $A$.

Step 3: Construct $P$ from the vectors in step 2.

Step 4: Construct diagonal matrix $D$ from the corresponding eigenvalues.

Step 5: $A= PDP^{-1}$.

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my $D$ will contain non real eigens, surely you don't expect me to put them in there? –  Hawk Nov 16 '12 at 8:17
@sizz: If you haven't got the courage to write down a matrix with complex entries, no wonder you have difficulty in diagonalizing matrices with complex (non real) eigenvalues. –  Marc van Leeuwen Nov 16 '12 at 8:41