A hub and nine spokes

Idea. The matrix has rank 2, so only two eigenvalues are nonzero. The trace and the sum of squared entries pin them down without any characteristic polynomial.

Step 1: rank. Rows 2 to 10 are all equal to $(1,0,\dots ,0)$, and row 1 is $(0,1,\dots ,1)$. So the rank is 2, and 0 is an eigenvalue 8 times. Call the other two $\lambda$ and $\mu$.

Step 2: trace. The diagonal is zero, so $\lambda +\mu =0$.

Step 3: squares. For a symmetric matrix, $\sum {\lambda }_i^2=\mathrm{tr}(A^2)={\sum }_{i,j}{a}_{ij}^2$. The matrix has 18 entries equal to 1, so ${\lambda }^2+{\mu }^2=18$, and with $\mu =-\lambda$, ${\lambda }^2=9$.

So the eigenvalues are $3$, $-3$ and $0$ (eight times), and the largest is 3.

Check with an eigenvector. Take $x=(3,1,1,\dots ,1)$. Then $(Ax{)}_1=9=3\cdot 3$ and $(Ax{)}_j=3=3\cdot 1$ for $j\ge 2$. In general a star with $n-1$ leaves has largest eigenvalue $\sqrt{n-1}$.