How to find eigenvector of large matrix without substitution directly. \begin{bmatrix}a&b&b&0&0&0\\b&a&b&0&0&0\\b&b&a&0&0&0\\0&0&0&a&b&b\\0&0&0&b&a&b\\0&0&0&b&b&a\end{bmatrix}
I have a large matrix as above which I already know that 
\begin{bmatrix}a&b&b\\b&a&b\\b&b&a\end{bmatrix} have three eigenvectors equal to \begin{bmatrix}-1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\0\end{bmatrix} and \begin{bmatrix}1\\1\\1\end{bmatrix} 
Is there any way I can find eigenvector for a large matrix above by using the submatrix that I already solved
Thank you 
ps. If you found this question is a duplicate of another, kindly put the link for me please.
 A: Sure: $$\begin{bmatrix}-1\\0\\1\\-1\\0\\1\end{bmatrix},$$ $$\begin{bmatrix}-1\\1\\0\\-1\\1\\0\end{bmatrix},$$ and $$\begin{bmatrix}1\\1\\1\\1\\1\\1\end{bmatrix}.$$
A: Hint: What happens when you apply your matrix to this vector?
$$\begin{bmatrix}-1\\0\\1\\0\\0\\0 \end{bmatrix}$$
Can you generalise this idea to find 6 eigenvectors? Are they linearly independent?
A: In general, if $A$ and $B$ are diagonalizable as $A=MD_AM^{-1}$ and $B=ND_BN^{-1}$ ( where the columns of $M$ are the eigenvectors of $A$ and the columns of $N$ are the eigenvectors of $B$) than we have:
$$
C=\begin{pmatrix}A&0\\0&B
\end{pmatrix}=\begin{pmatrix}MD_AM^{-1}&0\\0&ND_BN^{-1}
\end{pmatrix}=
$$
$$
=\begin{pmatrix}M&0\\0&N
\end{pmatrix}
\begin{pmatrix}D_A&0\\0&D_B
\end{pmatrix}
\begin{pmatrix}M^{-1}&0\\0&N^{-1}
\end{pmatrix}=\begin{pmatrix}M&0\\0&N
\end{pmatrix}
\begin{pmatrix}D_A&0\\0&D_B
\end{pmatrix}
\begin{pmatrix}M&0\\0&N
\end{pmatrix}^{-1}
$$
So $C$ is diagonalizable as 
$$
C=PDP^{-1}
$$
and the columns of $P=\begin{pmatrix}M&0\\0&N
\end{pmatrix}$ are the eigenvectors of $C$.
