Show a Matrix is Symmetric Knowing its Eigenvectors I was looking at some past exams for a linear algebra course and came upon this question:
Let $A$ be a $3 \times 3$ real matrix with the following eigenvectors:
$$ \begin{pmatrix}
1 \\
1 \\
0 
\end{pmatrix}, \begin{pmatrix}
1 \\
2 \\
0 
\end{pmatrix}, \begin{pmatrix}
2 \\
1 \\
0 
\end{pmatrix}, \begin{pmatrix}
0 \\
0 \\
1 
\end{pmatrix}$$
Show that $A$ is symmetric.
My attempt: Since eigenvectors 2, 3, and 4 are linearly independent, they form a basis for $\mathbb{R}^3$. We can then diagonalize $A$. This gives that
$$A = \frac13 \begin{pmatrix}
-1 \lambda_2 + 4 \lambda_3& 2\lambda_2 - 2 \lambda_3& 0 \\
-2 \lambda_2 + 2 \lambda_3& 4\lambda_2 - 1 \lambda_3& 0 \\
0& 0& \frac13 \lambda_4 
\end{pmatrix}$$
From this I conclude $A$ is symmetric iff $\lambda_2 = \lambda_3$. But, I'm not sure where to go from here.

It is easy to see that ${\bf v}_2$ and ${\bf v}_3$ are independent; also,
$$3{\bf v}_1={\bf v}_2+{\bf v}_3$$
as noted in comments.  Multiplying by $A$ gives
$$3\lambda_1{\bf v}_1=\lambda_2{\bf v}_2+\lambda_3{\bf v}_3$$
where $\lambda_k$ are the respective eigenvalues.  Therefore
$$\lambda_2{\bf v}_2+\lambda_3{\bf v}_3=\lambda_1{\bf v}_2+\lambda_1{\bf v}_3\ .$$
By independence, $\lambda_1=\lambda_2=\lambda_3$.
