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Problem: Let $A = \begin{pmatrix} 6 & 0 \\ -2 & 2 \end{pmatrix}$. Is this matrix diagonalizable over $\mathbb{R}$? If not, is it diagonalizable over $\mathbb{C}$? Compute the eigenvalues $\lambda$ and the corresponding eigenspaces $E_{\lambda}$.

Attempt at solution: The characteristic polynomial of this matrix is $\det(A - x \mathbb{I}_2)$: \begin{align*} \det \begin{pmatrix} 6 - x & 0 \\ -2 & 2-x \end{pmatrix} = x^2 - 6x +10 \end{align*} Since this polynomial has no roots over $\mathbb{R}$, it is not diagonalizable over $\mathbb{R}$. The complex roots of this polynomial are $\lambda_1 = 3+i$ and $\lambda_2 = 3-i$. This are the eigenvalues of $A$. The eigenspace corresponding to $\lambda_1$ is the nullspace of the matrix $(A - \lambda_1 \mathbb{I}_2)$: \begin{align*} E_{\lambda_1} = N(A - \lambda_1 \mathbb{I}_2) = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \in \mathbb{C}^2 : \begin{pmatrix} -i +3 & 0 \\ -2 & -i -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\right\} \end{align*} Now I was wondering what the geometric multiplicity is of $\lambda_1$, i.e. the dimension of the corresponding eigenspace. If this is not equal to the algebraic multiplicity (which is $1$), then the matrix is not diagonalizable over $\mathbb{C}$. How many free variables are there in the matrix \begin{align*} \begin{pmatrix} -i +3 & 0 \\ -2 & -i -1 \end{pmatrix} ? \end{align*} I thought I could row reduce this to $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. But then I have $\dim(E_{\lambda_1}) = 0$, which is impossible since it must always be greater than or equal to $1$? Some help would be appreciated.

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  • $\begingroup$ your determinant calculation was incorrect. $\endgroup$ – Sisi Jul 13 '15 at 18:15
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    $\begingroup$ You got the characteristic polynomial wrong. Work out the determinant again, carefully... $\endgroup$ – David C. Ullrich Jul 13 '15 at 18:15
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Nothing to calculate: Note that Given matrix is Lower triangular, hence Diagonal entries of matrix are eigenvalues. So, eigenvalues are $2,6$ which are distinct and real , Hence matrix is diagonalizable over $\Bbb R$

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  • $\begingroup$ If I compute the eigenspace of $\lambda = 2$, I get: $\begin{pmatrix} 4 & 0 \\ -2 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$. But this system has only the trivial zero solution, so $\dim(E_{\lambda}) = 0 $? But then the geometric and algebraic multiplicity is not equal and so not diagonizable? $\endgroup$ – Kamil Jul 13 '15 at 19:11
  • $\begingroup$ you get that $4x=0$ and $-2x=0$ that means $x=0$ but $y$ is arbitrary so you can take $(0,1)^t$ as eigenvector $\endgroup$ – Chiranjeev_Kumar Jul 13 '15 at 19:17
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You miscalculated the determinant. You should have ended up with the characteristic polynomial $$ p(x) = (x-6)(x-2) $$ check your work, note that $0 \times (-2) = 0$.

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