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I am trying to diagonalize the following matrices:

$$A = \begin{pmatrix}0 & 1\\-1 & 2\end{pmatrix}\qquad B = \begin{pmatrix}1 & 2\\-1&-1\end{pmatrix}$$

For matrix $A$, I find an eigenvalue of $1$ with algebraic multiplicity of $2$. I find, though, that the dimension of its eigenspace consists of $1$ vector? Can I still construct a matrix to diagonalize with somehow?

For matrix B, I get the same scenario for eigenvalues (value of $1$ repeated twice), but am unsure if my eigenspace, consisting of $(1,0)$, is correct? Then again, I am left in the same situation.

Are those matrices un-diagonalizable? (spelling?) Can someone help me out?

Thank you.

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    $\begingroup$ But the characteristic polynomial of $B$ is $x^2+1,$ why do you say that $B$ has $1$ as an eigenvalue? $\endgroup$ Commented Apr 24, 2014 at 3:21
  • $\begingroup$ The characteristic polynomial I got for B is (1-x)^2.. $\endgroup$ Commented Apr 24, 2014 at 3:33
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    $\begingroup$ Calculation: $$char(B)=\det\begin{pmatrix}1-x & 2\\-1&-1-x\end{pmatrix}=x^2-1+2=x^2+1.$$ $\endgroup$ Commented Apr 24, 2014 at 3:36
  • $\begingroup$ Yeah you are right, I always have trouble with dealing with factorizing the polynomial to find the eigenvalues. Thank you. $\endgroup$ Commented Apr 24, 2014 at 3:37
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    $\begingroup$ No, the second has eigenvalues $\pm i$ and is diagonalizable as my answer shows. $\endgroup$ Commented Apr 24, 2014 at 4:03

2 Answers 2

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Hints:

The first is not diagonalizable, but we can use the Jordan Normal Form. We get:

$$A = PJP^{-1} =\left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \\ \end{array} \right) \left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & 1 \\ \end{array} \right)$$

The second is diagonalizable and we get:

$$A = PJP^{-1} = \left( \begin{array}{cc} i-1 & -i-1 \\ 1 & 1 \\ \end{array} \right) \left( \begin{array}{cc} -i & 0 \\ 0 & i \\ \end{array} \right) \left( \begin{array}{cc} \frac{1}{2 i} & \frac{i+1}{2 i} \\ -\frac{1}{2 i} & \frac{i-1}{2 i} \\ \end{array} \right)$$

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  • $\begingroup$ When you say "you can use Jordan Normal Form," that means that the matrix is not diagonalizable, right? $\endgroup$ Commented Apr 24, 2014 at 3:22
  • $\begingroup$ Ok. (My LA class did not cover Jordan normal form, so I didn't know if that could be considered a special case of diagonal, or not...) $\endgroup$ Commented Apr 24, 2014 at 3:30
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    $\begingroup$ Here is a nice write-up for the 2x2 case: math.berkeley.edu/~ogus/old/Math_54-05/webfoils/jordan.pdf $\endgroup$ Commented Apr 24, 2014 at 3:32
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If you have an eigenvalue with multiplicity n, then you must have n linearly independent eigenvectors from that eigenvalue in order for the matrix to be diagonalizable.

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