![]() ![]() It turns out that the two solutions correspond to orthogonal eigenvectors, so we don’t need to do anything else to find a third. This leads to the equation $$ad^2-2ac^2 bcd = 0.$$ (The cross product actually generates two nontrivial equations, but they aren’t independent.) If $a\ne0$, the quadratic formula gives us $c$ in terms of $d$ or vice-versa. If this is an eigenvector of $A$, then $A\mathbf v$ is a scalar multiple of $\mathbf v$, which holds iff $\mathbf v\times A\mathbf v=0$. The vectors $(1,0,1)^T$ and $(0,1,0)^T$ are an obvious basis for the orthogonal complement of $\mathbf v_1$, i.e., we want vectors of the form $\mathbf v = (c,d,c)^T$. We therefore look for other eigenvectors orthogonal to this one. Subtracting the last column from the first produces $(1 \sigma^2-b,0,b-(1 \sigma^2))^T$, so we know that $\mathbf v_1 = (1,0,-1)^T$ is an eigenvector with eigenvalue $1 \sigma^2-b$. The matrix $A$ is real symmetric, so we know that it is orthogonally diagonalizable. ![]() Many times it can be easier to look for eigenvectors directly instead of following the rote process of computing roots of the characteristic polynomial and then computing null spaces for each eigenvalue. ![]() The Eigenvalues I found are $1 \sigma^2-b, 1 \sigma^2-\dfrac)$. ![]() I calculated the three eigenvalues from a matrix and now I want to calculate the associated eigenvectors. ![]()
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