A $\{0,1\}$-matrix with positive spectrum must have all eigenvalues equal to $1$ Here's a cute problem that was frequently given by the late Herbert Wilf during his talks. 

Problem: Let $A$ be an $n \times n$ matrix with entries from $\{0,1\}$ having all positive eigenvalues. Prove that all of the eigenvalues of $A$ are $1$.

Proof:


   Use the AM-GM inequality to relate the trace and determinant.


Is there any other proof?
If one wants to use the AM-GM inequality, you could proceed as follows:
Since $A$ has all $1$'s or $0$'s on the diagonal, it follows that $tr(A)\leq n$.
Now calculating the determinant by expanding along any row/column, one can easily see that the determinant is an integer, since it is a sum of products of matrix entries (up to sign). Since all eigenvalues are positive, this integer must be positive. AM-GM inequality implies
$$det(A)^{\frac{1}{n}}=\left(\prod_{i}\lambda_{i}\right)^{\frac{1}{n}}\leq \frac{1}{n}\sum_{i=1}^{n}\lambda_{i}\leq 1.$$
Since $det(A)\neq 0$, and $m^{\frac{1}{n}}&gt;1$ for $m&gt;1$, the above inequality forces $det(A)=1$.
We therefore have equality which happens precisely when $\lambda_{i}=\lambda_{j}$ for all $i,j$. Combining this with the above equality gives the result.
Since $\det(A) \neq 0$, $A$ is nonsingular and all diagonals of $A$ must equal one.  Next, we know the relationship between the determinant and trace is
\begin{align*}
\det(A) &amp;= \exp(tr(\ln(A))) \\
&amp;= \exp(0) \\
&amp;= \exp(\sum_i \ln(\lambda_i)) \\
\end{align*}.  Step 2 holds because all diagonals of $A$ are one, so their log is zero, and step 3 because natural logs are analytic. This suggests $$\sum_i \ln(\lambda_i) = 0$$, or, $$\prod_i \lambda_i = 1$$.  

Then, because all diagonals of $A$ are one, $tr(A) = \sum_i \lambda_i = n$ must be true.  Hence, the AM-GM inequality states $$ \left[ \prod_{i} \lambda_i  \right]^{\frac{1}{n}} \leq \frac{1}{n} \sum_{i} \lambda_{i}$$, with equality holding only when $\lambda_i = \lambda_j$  $\forall i,j$. Thus, $$ 1^{\frac{1}{n}} = \frac{1}{n} \cdot n$$, and all eigenvalues of $A$ must be equal.  Since $\prod_i \lambda_i = 1$, all eigenvalues must be one.