Answer
False
Work Step by Step
Assume the matrix $A=\begin{bmatrix}
1 & 0\\
0 & 2
\end{bmatrix}\\
\rightarrow \det (A-\lambda I)=0\\
\rightarrow \lambda_1=1,\lambda_2=2$
Then we have:
$v_1=\begin{bmatrix}
1 \\
0
\end{bmatrix}\\
v_2=\begin{bmatrix}
0\\
1
\end{bmatrix}$
But we can notice that a linear combination of a set of eigenvectors of a matrix A:
$v_1+v_2=\begin{bmatrix}
1 \\
1
\end{bmatrix}$ is not an eigenvector of A.
Hence, the statement is not true.
