# Finding eigenvalues/vectors of a matrix and proving it is not diagonalisable.

I have got the following matrix. $$\begin{pmatrix} -7 &4 \\ -9 &5 \end{pmatrix}$$ I need to find the eigenvalues, eigenvectors and $\textbf{prove}$ that it is not diagonalisable.

I have managed to show that the only eigenvalue is $\lambda=-1$ (from the equation $det(A-\lambda I)=0$). Then I have calculated the only eigenvector, which is $(2,3)$. Now I need to prove that the matrix is not diagonalisable. I think it would be reasonable to prove that if eigenvalues are repeating, then the matrix is not diagonalisable.

Have anyone got any suggestions how to proceed? Thank you very much.

• You pretty much already done it, you just need to understand that you've done it to finalize. Commented May 10, 2015 at 22:11

An $n\times n$ matrix $A$ is diagonalizable if and only if there exists a basis $\{v_1,\dotsc,v_n\}$ for $\Bbb R^n$ consisting of eigenvectors of $A$. Here you have a $2\times 2$ matrix and you've shown that every eigenvector is a scalar multiple of $\vec v=(2,3)$. What can you conclude?
Note that your suggestion that "if eigenvalues are repeating, then the matrix is not diagonalizable" is false. For example, the identity matrix $I_n$ has only one eigenvalue $\lambda=1$ and this eigenvalue has algebraic multiplicity $n$. However, $I_n$ is clearly diagonalizable.
• Thank you for the answer. You mean that I don't have enough eigenvectors for creating columns of the $\textit{change of basis matrix}$? Commented May 10, 2015 at 22:23