Find the eigenvalues of a symmetric matrix

Question

Find the eigenvalues of a $3 \times 3$ symmetric matrix with $1$ on the main diagonal and $\frac{1}{\sqrt 3}$ off the main diagonal.

Since each row on addition give the same value, one of the three eigenvalue is $1+\frac{2}{\sqrt 3}$.

Is there an easy way to find the other two values without using the formula $\det(A-\lambda I_3) = 0$.

Answer 1

If you know one eigenvalue already, you could use $$\sum\lambda=\operatorname{trace}(M)$$ and $$\prod\lambda=\det(M)$$

Answer 2

You might want to take advantage that your $3 \times 3 $ matrix is a circulant matrix of the form: $$ \begin{bmatrix} c_0 & c_{n-1} & c_{n-2} & \cdots & c_1 \\ c_1 & c_0 & c_{n-1} & \cdots & c_{2}\\ \vdots & c_1& c_0& \ddots & \vdots \\ c_{n-2} & &\ddots & \ddots& c_{n-1}\\ c_{n-1} & c_{n-2} &\cdots & c_1 & c_{0} \end{bmatrix}$$ and the eigenvalues can be computed through the formula:

$$ \lambda_j = c_0 + c_1 \omega_j +c_2 \omega_j^2 + \cdots+ c_{n-1}\omega_j^{n-1}, \quad j = 0,1,\ldots,n-1,\tag 1$$

where $\omega_j = \exp\left(\frac{2 \pi i j}{n}\right)$ are the $n-$th roots of unity.

Concretely, $n = 3$ and we have $\omega_0 = 1, \omega_1 = -\frac{1}{2} + \frac{i\sqrt 3}{2}, \omega_2 = -\frac{1}{2} - \frac{i\sqrt 3}{2}.$

According to $(1)$: $$\lambda_0 = 1 + \frac{1}{\sqrt 3} \cdot 1^1 + \frac{1}{\sqrt 3}\cdot 1^2 = 1 +\frac{2}{\sqrt 3}. $$

$$\lambda_1 = 1 + \frac{1}{\sqrt 3}\cdot \left(-\frac 12 +\frac{i\sqrt 3}{2}\right) + \frac{1}{\sqrt 3}\cdot \left(-\frac 12 +\frac{i\sqrt 3}{2}\right)^2 = 1-\frac{1}{\sqrt 3}.$$

Similarly, $\lambda_2 = 1 - \frac{1}{\sqrt 3}.$

Answer 3

This is a matrix of the form $A=aI_n+bee^T$, where $e=(1,\ldots,1)^T$. Hence any orthogonal basis containing the vector $e$ are $n$ eigenvectors, and the eigenvalues of $A$ are $\lambda_1=a+nb$ (obtained from $Ae=\lambda_1e$) and $\lambda_2=\cdots=\lambda_n=a$ (obtained from $Ax=\lambda_kx$ with $x\perp e$).