Characteristic roots of symmetric matrices

Question

Is there a formula or method to find the characteristic roots of a symmetric matrix. The problem is: Given $$A= \begin{bmatrix}123&231&312\\231&312&123\\312&123&231\\\end{bmatrix} .$$ I have to prove that 666 is a characteristic root of the matrix, without solving the characteristic polynomial.

Answer 1

Your matrix $A$ has the property that all rows (and columns, but you don't need that here) add to the same number $r$. When that happens, there is an obvious eigenvector for $r$.

Answer 2

We observe that every row of the matrix

$A = \begin{bmatrix} 123&231&312\\231&312&123\\312&123&231 \end{bmatrix} \tag 1$

sums to the same value, which is $666$. We then conclude that

$A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{bmatrix} 123&231&312\\231&312&123\\312&123&231 \end{bmatrix}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 666 \\ 666 \\ 666 \end{pmatrix} = 666\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}; \tag 2$

thus $666$ is an eigenvalue of $A$ with associated eigenvector $(1, 1, 1)^T$.

Answer 3

Without expanding to a polynomial: $$\begin{vmatrix}123-\lambda&231&312\\231&312-\lambda&123\\312&123&231-\lambda\\\end{vmatrix}=0 \overbrace{\Rightarrow}^{R3+R2+R1\to R1} \begin{vmatrix}666-\lambda&666-\lambda&666-\lambda\\231&312&123\\312&123&231\\\end{vmatrix}=0 \Rightarrow \\ (666-\lambda)\begin{vmatrix}1&1&1\\231&312&123\\312&123&231\\\end{vmatrix}=0 \Rightarrow \lambda =666 .$$

Answer 4

$$A-666I= \begin{bmatrix}-543&231&312\\231&-354 &123\\312&123&-435 \\\end{bmatrix} $$

Note that the rows are linearly dependent, so $$\det \begin{bmatrix}-543&231&312\\231&-354 &123\\312&123&-435 \\\end{bmatrix}=0 $$ That is $666$ is an eigenvalue. The eigenvector in this case is $\begin{pmatrix} 1\\1\\1\end{pmatrix}$