My quantum mech teacher told me to find the eigenvalues and eigenvectors of some matrices without resorting to boring computations when possible. I was trying to find them for the following matrix:
$$ M = \begin{bmatrix}
1 & 1& 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{bmatrix} $$
Thanks to what I learned in the previous course I followed, I immediately noticed that this matrix is just a projector multiplied by a scalar, namely: I can rewrite it as:
$$M = k \cdot P_{\mathbf{\hat{u}_1}}$$
where $\mathbf{\hat{u}}$ is the unit vector on which I'm projecting, in our case it's easy to see that:
$$k = 3$$
$$ P_\mathbf{\hat{u}_1} = \mathbf{\hat{u}_1}\mathbf{\hat{u}^\ast_1}, \mathbf{\hat{u}_1} = \left[\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}}\right]^T$$
Now, noticing that our original matrix is hermitian (symmetric with only real coefficients, and it is also associated with a self-adjoint operator), I can write the spectral representation of $M$ as follows:
$$ M = \sum_{k=1}^{3} \lambda_k P_k \Rightarrow \{\lambda_1 = 3, \lambda_2 = \lambda_3 = 0 \}$$
Furthermore, I already know that the eigenvector corresponding to $\lambda_1$ is exactly $\mathbf{u}_1$.
This method turned out to be alright, except for one big flaw: I can't derive the eigenvectors associated with $\lambda_{2,3}$. Is there a way to reconstruct them from all the pieces of information I gathered? If this weren't the case, then I'd have to discard my approach. In the latter scenario, is there a quick way or some sort of clever trick to notice for computing the eigenvalues and eigenvectors for the matrix I was given: what I know is that if a matrix is hermitian, then it only admits real-valued eigenvectors. This doesn't seem to be of much help though.
Any help is much appreciated
Best Answer
The eigenspace associated with the zero eigenvalues consists of the kernel of the projection $P_{\bf \hat u_1}$, which is just all the vectors perpendicular to $\bf \hat u_1$, or the vectors $[x,y,z]^T$ with $x+y+z=0$. Since this eigenspace has dimension 2, you have a choice of what basis you use to span it. One choice would be to pick any two vectors out of the set $\{[2/\sqrt{6}, -1/\sqrt{6}, -1/\sqrt{6}]^T$, $[-1/\sqrt{6}, 2/\sqrt{6}, -1/\sqrt{6}]^T$, $[-1/\sqrt{6}, -1/\sqrt{6}, 2/\sqrt{6}]^T$}, although this is not an orthogonal basis; an example of an orthogonal (and orthonormal) basis is $\{[2/\sqrt{6}, -1/\sqrt{6}, -1/\sqrt{6}]^T, [0, 1/\sqrt{2}, -1/\sqrt{2}]^T\}$.