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I have two examples of matrices where you would compute JCF's, I'm confused about what you would do in the second example to compute the canonical basis.

Example 1

Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & -4 & 2 \\ \end{bmatrix}$, $A^{2} = \begin{bmatrix} 0 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$ and $A^{3} = 0$.

In this case, $\lambda = 0 $ so:

  • dim (null $(A - 0I)$) = dim (span$\{(1, 0, 0)\}$) = 1

  • dim (null $(A - 0I)^{2}$) = dim (span$\{(1, 0, 0), (0, 0, 1)\}$) = 2

  • dim (null $(A - 0I)^{3}$) = dim (span$\{(1, 0, 0), (0,1,0),(0, 0, 1)\}$) = 3

Thus we need a cycle of length 3: $\{A^{2}x, Ax, x\}$. We need a $v$ such that $N^{2}x = v$ and $v \in$ null $(A - 0I) \cap$ Im$(A - 0I)^{2}$. The only such $v$ is $(1,0,0)$ thus following some calculations we get the cycle $\{(1,0,0),(0,1,2),(0,0,1)\}$

So this is fine.

Example 2

This is where i'm confused. Let $A = \begin{bmatrix} 2 & 0 & 2 & 0 \\ 0 & 2 & 0 & 6 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}$. Our $\lambda = 2$ here with the multiplicity 4.

  • dim (null $(A - 2I)$) = dim (span$\{(1, 0, 0,0), (0,1,0,0)\}$) = 2

  • dim (null $(A - 2I)^{2}$) = dim (span$\{(1, 0, 0,0),(0,1,0,0),(0, 0, 1,0),(0,0,0,1)\}$) = 4

This time we need two cycles (both of length 2): $\{Ax,x,Ay,y\}$ but I can't seem to find a $v \in$ null $(A - 2I) \cap$ Im$(A - 2I)^{2}$ because from my understanding (which could be wrong), there isn't really anything in Im$(A - 2I)^{2}$, is there? So how do we find the canonical basis?

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Of course there's something in $(A - 2I)^2$; as you've said, $\operatorname{null}[(A - 2I)^2]$ is all of $\Bbb R^4$. So, all you really need are two vectors in $\operatorname{null}(A - 2I)$.

More specifically, we begin by finding a basis of $\operatorname{null}(A - 2I)$. In particular, take $v_1 = (1,0,0,0)$ and $v_2 = (0,1,0,0)$. Each of these vectors will be the beginning of a cycle.

For our first chain, we need a vector $v_1^{(2)}$ such that $$ (A - 2I)v_1^{(2)} = v_1 $$ For instance, take $v_1^{(2)} = (0,0,1/2,0)$. For our second chain, we need a vector $v_2^{(2)}$ such that $$ (A - 2I)v_2^{(2)} = v_2 $$ For instance, take $v_2^{(2)} = (0,0,0,1/6)$. With that done, our Jordan basis will be $$ \{v_1,v_1^{(2)},v_2,v_2^{(2)}\} $$

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  • $\begingroup$ Notably, if you know anything about tensor/Kronecker products, we can make the process shorter by noting that the matrix can be written as $$ A = 2I + \pmatrix{0&1\\0&0} \otimes \pmatrix{2&0\\0&6} $$ $\endgroup$ – Ben Grossmann Jun 24 '17 at 22:08
  • $\begingroup$ THANK YOU! This was super helpful! :) $\endgroup$ – Saneea Mustafa Jun 26 '17 at 18:43
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You know there will be two Jordan blocks of size $2$. Hence you to find two linearly independent vectors $u_1$ and $v_1$ in $\ker (A-2I)^2\setminus\ker(A-2I)$ and set $u_0=(A-2I)u_1$, $\;v_0=(A-2I)v_1$. Then $(u_0, u_1, v_0,v_1)$ is your Jordan basis.

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  • $\begingroup$ Your method is "backwards" to how this is usually taught, but on consideration it seems a lot more logical. $\endgroup$ – Ben Grossmann Jun 24 '17 at 22:17
  • $\begingroup$ It's the way I teach it. If you proceed upwards, it's often more complicated , because you have to make adjustments to ensure linear independence. $\endgroup$ – Bernard Jun 24 '17 at 22:25
  • $\begingroup$ Well, I like it! I'll try to remember it. $\endgroup$ – Ben Grossmann Jun 24 '17 at 22:40

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