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Let $$A = \begin{pmatrix}3 & 1 &-2\\-1& 0 & 5\\ -1&-1&4\end{pmatrix}$$

To find the Jordan canonical form for $A$, need to find a Jordan canonical basis for $T = L_A$

And the Jordan canonical basis for $A$ is $\beta=$$\left\{(-1,2,1), (1,-3,-1), (-1,2,0)\right\}$

In the answer, the Jordan canonical form for $A$ is

$$J = \lbrack T\rbrack_{\beta} = \begin{pmatrix}3 & 0 &0\\0& 2 & 1\\ 0&0&2\end{pmatrix}$$

But I'm having trouble computing $J = \lbrack T\rbrack_{\beta}$, can anyone show me the calculation process? thanks for the help.

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$$J = S^{-1} A S$$

where:

$$S = \begin{pmatrix}-1 & 1 & -1\\2 & -3 & 2\\ 1 & -1 & 0\end{pmatrix}$$

Note what the columns of $S$ are made from.

What happens to $J$ if you swap the positions of your canonical basis $\beta$? Recommend playing with that.

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  • $\begingroup$ The order of columns of $J$ corresponds to the order of the canonical basis $\beta$? $\endgroup$ – user59036 Mar 25 '14 at 21:46

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