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$\begin{pmatrix}8&k\\ 0&4\end{pmatrix}$

So I know that for a matrix to be diagonalizable, it needs to have 2 distinct real eigenvalues.

So I calculated the characteristic polynomial to be:

$x^2-12x+32-k$

Therefore, for the discriminant to be greater than $0$, I got the inequality:

$k>-4$

Therefore, I thought that it would diagonalizable for all $k$ except $k=-4$, but this is not correct.

Any help?

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    $\begingroup$ Your computation of the characteristic polynomial is incorrect. $\endgroup$ – vadim123 Feb 19 '18 at 21:00
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    $\begingroup$ The eigenvalues of a diagonalizable matrix need not be distinct. Take for instance the identity matrix. $\endgroup$ – Lorenzo Quarisa Feb 19 '18 at 21:01
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    $\begingroup$ The eigenvalues of an upper triangular matrix are on the diagonal! $\endgroup$ – Fakemistake Feb 19 '18 at 21:01
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    $\begingroup$ A matrix is diagonalizable when, for every eigenvalue, its geometric multiplicity (dimension of the corresponding eigenspace) is equal to its algebraic multiplicity (multiplicity of the root in the characteristic equation). $\endgroup$ – Adrian Keister Feb 19 '18 at 21:03
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    $\begingroup$ It is diagonalizable for any $\;k\;$ ...this already should be clear from the above comments: if an $\;n\times n\;$ matrix has $\;n\;$ different eigenvalues then it is diagonalizable. $\endgroup$ – DonAntonio Feb 19 '18 at 21:04
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Hint:

First it depends on which field you want it to be diagonalisable, e.g. $\mathbf Q$, $\mathbf R$, $\mathbf C$?

Supposing the characteristic polynomial splits on the base field, the necessary and sufficient condition for a matrix to be diagonlisable, is NOT to have distinct eigenvalues (this is only a sufficient condition), but the geometric multiplicity being equal to the algebraic multiplicity, i.e. if $\lambda$ is an eigenvalue with algebraic multiplicity $m_\lambda$ and eigenspace $E_\lambda$: $$\dim E_\lambda=m_\lambda$$ (it is always true that $\;\dim E_\lambda \le m_\lambda$).

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The characteristic polynomial is $(t-8)(t-4)$, so the eigenvalues are $t_1 = 4$ and $t_2 = 8$. By inspection, when $v_2 = (1,0)^T$, $\begin{pmatrix}8&k\\ 0&4\end{pmatrix} v_2 = 8v_2$.

To find another eigenvector $v_1$.

$$\begin{pmatrix}4&k\\ 0&0\end{pmatrix} \begin{pmatrix}v_{11}\\ v_{12}\end{pmatrix} = 0$$

So choose $v_1 = (k,-4)^T$, which correspond to $t_1 = 4$, and $v_2 = (1,0)^T$ corresponding to $t_2 = 8$.

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Note that to diagonalize the matrix you need a basis of eigenvectors for $\mathbb{R^2}$.

When we find two distinct eigenvalues that is given for free.

If eigenvalues are not distinct we need to check whether or not for the eigenvalues 2 (which has algebraic multiplicity 2) we can find 2 linearly independent eigenvectors, in this case you can diagonalize the matrix.

Otherwise you can only obtain a Jordan form.

In this case since $\lambda_1=8$ and $\lambda_2=4$, we can diagonalize $\forall k$.

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It is more concrete to just go ahead and see what happens. With the eigenvalues $8,4$ we can get a matrix with eigenvectors as columns from $$ R = \left( \begin{array}{rr} 1 & -k \\ 0 & 4 \end{array} \right). $$ We can confirm this with $$ \left( \begin{array}{rr} 8 & k \\ 0 & 4 \end{array} \right) \left( \begin{array}{rr} 1 & -k \\ 0 & 4 \end{array} \right) = \left( \begin{array}{rr} 8 & -4k \\ 0 & 16 \end{array} \right) $$ Then $$ R^{-1} = \left( \begin{array}{rr} 1 & -\frac{k}{4} \\ 0 & \frac{1}{4} \end{array} \right) $$ and $$ \left( \begin{array}{rr} 1 & -\frac{k}{4} \\ 0 & \frac{1}{4} \end{array} \right) \left( \begin{array}{rr} 8 & k \\ 0 & 4 \end{array} \right) \left( \begin{array}{rr} 1 & -k \\ 0 & 4 \end{array} \right) = \left( \begin{array}{rr} 8 & 0 \\ 0 & 4 \end{array} \right) $$ Whatever the appearances, the value of $k$ does not matter. Since someone mentioned it, we are taking the characteristic of the field to not be two.

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