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Consider the following matrix: $$ \left[ \begin{array}{ccc} 2 & 0 & 2 \\ 0 & -2 & 0 \\ 2 & 0 & -1 \end{array} \right] $$ What would be the standard way to find the eigenvectors for this matrix? since it's Hermitian they must be orthogonal.

By inspection it can be found that the eigenvalue is $-2$ for $[0, 1, 0]^T$ and then $\left[ \begin{array}{ccc} 2 & 2 \\ 2 & -1 \end{array} \right]$ can be analyzed further to find that of the other two eigenvalues, of which one is $-2$, one results in a degeneracy. Since we already know an eigenvector we can use the scalar product to find another orthogonal to it and thus create are basis.

But how could you do this if you don't see the eigenvector by inspection and just calculate the eigenvalues blindly. I just get $2x=-z$ as the plane of solutions, how do I proceed?

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  • $\begingroup$ Did you mean $2z=x$? $\endgroup$
    – BoZenKhaa
    Jun 7 '15 at 14:27
  • $\begingroup$ $\left[ \begin{array}{ccc} 2-\lambda & 0 & 2 \\ 0 & -2-\lambda & 0 \\ 2 & 0 & -1-\lambda \end{array} \right] $, for $\lambda=-2$, $\left[ \begin{array}{ccc} 4 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & 1 \end{array} \right] $ thus $2x+z=0$ $\endgroup$ Jun 7 '15 at 14:29
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Solving directly for $\lambda = 3$, I get: $\left[ \begin{array}{ccc} 2&0&2\\ 0&-2&0\\ 2&0&-1\\ \end{array} \right] \left[ \begin{array}{ccc} x\\ y\\ z\\ \end{array} \right] = 3 \left[ \begin{array}{ccc} x\\ y\\ z\\ \end{array} \right]$.

By solving for $x, y, z$, we get $y=0$ and $2z=x$. These equations define a line, not plane, in $\mathbb{R}^3$. As such, the eigenvector for $3$ is given by any choice of $x$, such as $x=2$ which gives $\left[ \begin{array}{ccc} 2\\ 0\\ 1\\ \end{array} \right]$. You can check that it is indeed an eigenvector for $\lambda =3$.

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Hint:

You're supposed to compute the characteristic polynomial of the matrix (let's call it $A$): $$\chi_A(\lambda)=\det(A-\lambda I)$$ The eigenvalues are the roots of this polynomial, and the eigenvectors for a given root are just the vectors in the kernel of $A-\lambda I$.

You should find a factorisation: $\chi_A(\lambda)=(\lambda+2)^2(3-\lambda)$, whence the third eigenvalue: $\,\lambda=3$.

The subspace $E_{-2}$ has dimension $2$, so the matrix is diagonalisable.

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