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.

**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.

**Attribution***Source : Link , Question Author : Cloud JR K , Answer Author : Nigel Overmars*