Are there quicker ways to find the eigenvalues of a matrix?

Question

I need to find the eigenvalues of this matrix: $$ A= \begin{pmatrix} 1&4&16\\ 4&16&1\\ 16&1&4 \end{pmatrix} $$ Of course, I could do this by solving $\text{det}(\lambda I - A)=0$, but is there a shorter way to do this for this particular matrix which doesn't rely on making a lot of calculations by hand?

Edit: Not asking for homework help. I want to know if there is a 'smart' way to do this that exploits the structure of this matrix. This question was asked in a previous question paper of an online standardized exam which allows no calculator (except the basic on-screen calculator) and leaves approximately 3 min/question. So, having a smart way to do it is pretty much necessary.

Answer 1

Using @Arthur 's recognition that one eigenvalue is $\lambda_1 = 21$.

You know the sum of the eigenvalues is equal to the trace of the matrix $= 21$.

Thus the remaining 2 eigenvalues satisfy $\lambda_2+\lambda_3=0$.

Also $\lambda_1 \,\lambda_2 \, \lambda_3 = \det(A)$ combining with the above, leading to, say, $-21\lambda_2^2=\det(A)$, from which you can obviously find $\lambda_2$ and hence $\lambda_3$

Answer 2

All rows sum to $21$, so $(1, 1, 1)^T$ is an eigenvector with eigenvalue $21$. To find the remaining two eigenvalues, you can divide the characteristic polynomial by $(\lambda - 21)$, and get a quadratic equation. The matrix is symmetric, so you know the equation has real solutions (real symmetric matrices always have only real eigenvalues, and orthogonal eigenvectors).