What is the largest eigenvalue of the following matrix?

Solution 1:

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.

Solution 2:

The trick is that $\frac1{21}$ of your matrix is a doubly stochastic matrix with positive entries, hence the bound of 21 for the largest eigenvalue is a straightforward consequence of the Perron-Frobenius theorem.

Solution 3:

If you sum the row, all the rows they to the same number (21).

That indicates that $\begin {bmatrix} 1\\1\\1 \end{bmatrix}$ must be an eigenvector and 21 is the associated eigenvalue.

The trace of the matrix equal 21, and the sum of the eigenvalues equals the trace.

The remaining two eigenvalues are the negative of one another.

And $3969 < 21^3$ so the other absolute value of the other two eigenvalues are each less than $21$

Solution 4:

I believe there is a more elementary argument that has not yet been mentioned.

Claim: If $A$ is an $n$-by-$n$ matrix with nonnegative entries and all row sums equal to $r$, then the largest eigenvalue of $A$ (in absolute value) is $r$.

Proof: As has been noted, the all-$1$s vector is an eigenvector of $A$ with eigenvalue $r$. Conversely, let $x=(x_1,\ldots,x_n)^T$ be an eigenvector of $A$ with eigenvalue $\lambda$. Let $x_i$ be (an) element of largest absolute value, and by scaling by $-1$, we may assume that $x_i>0$ without loss of generality. Now the $i$-th component of $Ax$ is $\lambda x_i$, and its absolute value satisfies $$|\lambda|x_i=|(Ax)_i|=|\sum_{j=1}^nA_{ij}x_j|\leq\sum_{j=1}^n|A_{ij}||x_j|\leq\sum_{j=1}^nA_{ij}x_i=x_i\sum_{j=1}^nA_{ij}=rx_i,$$ so $|\lambda|\leq r$.