The eigenvalues of this symmetric Matrix

Question

I am preparing for my linear algebra exam and while studying with old exams I came accross this problem.

1. Find the Eigenvalues of the following symmetric matrix: \begin{pmatrix} 0 & 0& 0& 4\\ 0& 1& 1& 0\\ 0& 1& 1& 0\\ 4& 0& 0& 0 \end{pmatrix}

I first used Laplace expansion as the standard method and came to a 4th-degree polynomial. If I simplify it the right way I am able to see some of the eigenvalues but with so many ways to factor it, it's just a matter of luck and not suitable for an exam. I am obviously overlooking something since the questions on this test are quite simple and intended to prove just basic knowledge.

Note: I have also realized that the eigenvalues are the eigenvalues of the 2X2 matrix with just ones in the middle and the matrix that remains if we remove the ones. Is this an actual property? Is there any symmetric matrix property I am overlooking?

Answer 1

Hint: permute the rows and columns of the matrix to make it equal to $$ \pmatrix{0&4\\ 4&0\\ &&1&1\\ &&1&1}. $$

Answer 2

Consider a $2\times 2$ symmetric characteristic determinant: $$\begin{vmatrix}a-x&b\\b&a-x\end{vmatrix}$$ We can see that there are at most two roots. It is clear from inspection that the two rows may be made identical by using an $x$ such that $a-x = b$, i.e. $x = a-b$. There is at most one more root. It's instructive to attempt to get the $x$ on multiple rows so that we may find roots in one sweep fell. We may perform elementary operations (on the characteristic determinant, not on the original) to achieve this. Adding the second col to the first we have $$\begin{vmatrix}a+b-x&b\\a+b-x&a-x\end{vmatrix}$$ It is clear we may make the first column vanish (and hence the whole determinant) by using an $x$ such that $a+b - x = 0$, i.e. $x = a+b$. We have found both our roots.

Let's now consider a $3\times 3$ symmetric characteristic determinant:

$$\begin{vmatrix}a-x&b&c\\b&d-x&e\\c&e&f-x\end{vmatrix}$$

It is not as instructive to find the eigenvalues by inspection. However, if we have a "doubly symmetric" matrix (I don't know what the actual name is, but where $[a_{ij}] = [a_{ji}] = [a_{(n-i+1)(n-j+1)}]$), which your matrix satisfies, such as: $$\begin{vmatrix}a-x&b&c\\b&d-x&b\\c&b&a-x\end{vmatrix}$$

Now, it is just a similar process. We have at most 3 roots. From inspection one root is $a-c$. Another is $d-b$. The third may be found similarly (by adding the first and third columns) as $a-2b+c$. (Of course there can be no more roots and it is clear since if we add the third column to the second we get an inconsistent system).

This process may be done for any degree system, including yours, where it returns $(0,2,4,4)$. In fact, there probably is an inductive formula to get the roots for such a characteristic polynomial.

$$\begin{vmatrix} -x&0&0&4\\0&1-x&1&0\\0&1&1-x&0\\4&0&0&-x \end{vmatrix}$$

Inspection gives us $x = 4$ and $x = 0$. Adding the first and fourth columns again gives us $ x = 4$. Adding the second and third columns gives us $x = 2$. For a sanity check on the double root it could be factorized knowing $x = 0, x = 2, x = 4$ or the derivative may be used from where it is clear that $x = 4$ is a double root.

It seems to me to be a matter of clever manipulation rather than some knowledge. Perhaps the knowledge of the row/column operations and their effect on a determinant.