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.
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.