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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.

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  • $\begingroup$ You pretty much already done it, you just need to understand that you've done it to finalize. $\endgroup$
    – Git Gud
    Commented May 10, 2015 at 22:11

1 Answer 1

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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.

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  • $\begingroup$ Thank you for the answer. You mean that I don't have enough eigenvectors for creating columns of the $\textit{change of basis matrix}$? $\endgroup$
    – marco11
    Commented May 10, 2015 at 22:23
  • $\begingroup$ @marco11 Yes this is exactly what I'm saying! $\endgroup$ Commented May 10, 2015 at 22:28
  • $\begingroup$ Is this enough to say if they ask (for example during exam) for the proof of this? $\endgroup$
    – marco11
    Commented May 10, 2015 at 22:33
  • $\begingroup$ @marco11 I have no idea what the standards are for whatever course you are taking. This is, however, a very standard fact in Linear Algebra. This is a better question for your instructor/professor than for me. $\endgroup$ Commented May 10, 2015 at 22:34
  • $\begingroup$ Alright. Thank you very much for your answer, Overall, I found it very helpful and instructive. $\endgroup$
    – marco11
    Commented May 10, 2015 at 22:45

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