Find the largest eigenvalue of the following matrix
$$\begin{bmatrix}
1 & 4 & 16\\
4 & 16 & 1\\
16 & 1 & 4
\end{bmatrix}$$
This matrix is symmetric and, thus, the eigenvalues are real. I solved for the possible eigenvalues and, fortunately, I found that the answer is $21$.
My approach:
The determinant on simplification leads to the following third degree polynomial.
$$\begin{vmatrix}
1-\lambda & 4 &16\\
4 &16-\lambda&1\\
16&1&4-\lambda
\end{vmatrix}
= \lambda^3-21\lambda^2-189\lambda+3969.$$
At a first glance seen how many people find the roots of this polynomial with pen and paper using elementary algebra. I managed to find the roots and they are $21$, $\sqrt{189}$, and $-\sqrt{189}$ and the largest value is $21$.
Now the problem is that my professor stared at this matrix for a few seconds and said that the largest eigenvalue is $21$. Obviously, he hadn't gone through all these steps to find that answer. So what enabled him answer this in a few seconds? Please don't say that he already knew the answer.
Is there any easy way to find the answer in a few seconds? What property of this matrix makes it easy to compute that answer?
Thanks in advance.
Best Answer
Requested by @Federico Poloni:
Let $A$ be a matrix with positive entries, then from the Perron-Frobenius theorem it follows that the dominant eigenvalue (i.e. the largest one) is bounded between the lowest sum of a row and the biggest sum of a row. Since in this case both are equal to $21$, so must the eigenvalue.
In short: since the matrix has positive entries and all rows sum to $21$, the largest eigenvalue must be $21$ too.