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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

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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 $\{[2/\sqrt{6}, -1/\sqrt{6}, -1/\sqrt{6}]^T, [0, 1/\sqrt{2}, -1/\sqrt{2}]^T\}$.

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    $\begingroup$ It's worth noting that the orthogonality of the eigenspaces is not unique to projection matrices, but will occur for any normal matrix, i.e. a matrix that commutes with its own adjoint. This includes any Hermitian matrices, orthogonal matrices, and many others. $\endgroup$ Commented Sep 27, 2023 at 17:32

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