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I'm trying to find the eigenvectors for the matrix $$\begin{bmatrix} A & B & B \\ B & A & B \\ B & B & A \end{bmatrix} $$ .

I determined the eigenvalues to be $\lambda_1=\lambda_2=A-B$ and $\lambda_3=A+2B$. I also detrrmined one eigenvector to be $v_3=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\1\\1 \end{bmatrix}$ .

But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$\begin{bmatrix} 1 & 1 & 1 \\ 0& 0 & 0\\ 0& 0 & 0 \end{bmatrix} $$.

From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:

$$v_1+v_2+v_3=0$$ $$v_1\neq v_2\neq v_3\neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$

Any help very appreciated.

fazan
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  • See https://math.stackexchange.com/a/1521354/265466 for how to read a kernel basis from the rref. Otherwise, this is practically a duplicate of https://math.stackexchange.com/q/2177457/265466 and many other questions. – amd Jan 16 '19 at 23:06

1 Answers1

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We can arbitrarily pick values for $v_1,v_2,v_3$ that satisfy $v_1+v_2+v_3=0$.

In particular we do not need them all to be different from $0$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $1$ either, although we can normalize afterwards if we want to.

We can pick for instance $v_1=1$ and $v_3=0$, which means that $v_2=-1$. Just now we've found the eigenvector $(1,-1,0)$.

Klaas van Aarsen
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