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Compute the Jordan Canonical form of A = $\begin{bmatrix}i & 1\\1 & -1\end{bmatrix}$.


My (feeble) attempt: After I compute the characteristic polynomial, which gives me $x^2=0$, the eigenvalue is x=0 with multiplicity of 2. Any thoughts/comments on where to go from here? Anything is greatly appreciated!

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1 Answer 1

It's characteristic polynomial is : $$t^2+(1-i) t+(-1-i)=\left(t-\frac{1}{2} \left((-1+i)-\sqrt{4+2 i}\right)\right)\left(t-\frac{1}{2} \left((-1+i)+\sqrt{4+2 i}\right)\right)$$

so the matrix $\begin{bmatrix}i & 1\\1 & -1\end{bmatrix}$ can be diagonalizable to $$\begin{bmatrix}\frac{1}{2} \left((-1+i)-\sqrt{4+2 i}\right) &0 \\0 &\frac{1}{2} \left((-1+i)+\sqrt{4+2 i}\right) \end{bmatrix}$$

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Amazing, but what an ugly number. –  Don Larynx Dec 6 '13 at 4:38

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