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I am having trouble finding the eigenvalues and eigenspaces of symmetric symbolic matrices. The matrices given are as follows:

$$ A = \begin{pmatrix} α & -β & -β \\ -β & α & -β \\ -β & -β & α \end{pmatrix} ,\qquad B = \begin{pmatrix} α & -β & -β & -β \\ -β & α & -β & -β \\ -β & -β & α & -β \\ -β & -β & -β & α \end{pmatrix} $$

(1) Find all eigenvalues of both.

(2) Find the eigenspaces of both.

(3) Find the algebraic multiplicity and the geometric multiplicity of every eigenvalue of both.

(4) Justify if the matrices are diagonalizable.

For previous problems I have calculated the determinant of the matrix subtraced by lambda multiplied by the identity matrix and set the determinant equal to zero to solve for lambda (obtaining the eigenvalues). However, I feel like some properties of symmetric matrices might make these problems easier than completing such lengthy and complex calculations. Any guidance would be greatly appreciated.

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upsilon
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2 Answers2

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If $\beta=0$, then the $\alpha$ is the only eigenvalue and any non-zero vector is an eigenvaector.

Now, we can focus on $\beta \ne 0$.

The row sums is the same, hence $\alpha - (n-1)\beta$ is an eigenvalue. The all one vector is an eigenvector.

Also, when $\alpha-\lambda = -\beta$, the matrix is singular and the rank is $1$, $\alpha + \beta$ is an eigenvalue. This eigenspace has dimension $n-1$. You can consider eigenvector of the form of $e_1-e_j, j=2, \ldots, n.$

Symmetric matrices are diagonalizable.

Siong Thye Goh
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A couple of shortcuts.

Looking at A.

All of the rows sum to $\alpha - 2\beta$

$\begin{bmatrix} 1\\ 1\\1 \end {bmatrix}$ is an eigenvector, with $\alpha - 2\beta$ as the corresponding eigenvalue.

And consider $\lambda = \alpha + \beta$

Then $A - \lambda I = \begin{bmatrix} -\beta &-\beta &-\beta\\-\beta &-\beta &-\beta\\-\beta &-\beta &-\beta\end{bmatrix}$

$\alpha + \beta$ is a doubled eigenvalue.

A similar analysis can be done to matrix $B.$

$\alpha - 3\beta, \alpha + \beta$ appear to be the eigenvalues.

Doug M
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