This is the matrix given:
\begin{bmatrix}
a & a & a \\ b & b & b \\ c& c & c \\
\end{bmatrix}
I'm not very well-versed in linear algebra, so I'll explain what I do understand, feel free to correct me on any of it.
If this matrix is shown to have 3 distinct eigenvalues (and thus 3 distinct eigenvectors?), then it is diagonalizable. This is because with 3 distinct eigenvalues, there are 3 linearly independent vectors, and they form a basis.
A diagonalized matrix would be a matrix with only values in the diagonal.
How can it be shown that there are three distinct eigenvalues (or not) for this matrix?
Best Answer
The characteristic polynomial is $$det \begin{pmatrix} a-\lambda&a&a\\ b&b-\lambda&b\\ c&c&c-\lambda\\ \end{pmatrix}=-\lambda^3+(a+b+c)\lambda^2$$ This has solutions $\lambda = 0$ and $\lambda = a+b+c$. Thus the eigenvalues are not all distinct. It also has eigenvectors $(-1,0,1)$, $(-1,1,0)$, and $(a/c, b/c, 1)$. If you pick $a$, $b$, and $c$ so that these eigenvectors are linearly independent then you can diagonalize the matrix. As DietrichBurde says we can't have $a+b+c=0$.