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I am looking for a Jordan basis for $\phi: \mathbb{R}^3 \to \mathbb{R}^3, \phi((x_1,x_2,x_3))=(4x_1-x_2,4x_1,8x_1-4x_2+2x_3)$.

The matrix of the transformation in the standard basis is

$$A=M(\phi)_{\mathcal{st}}= \begin{bmatrix} 4 & -1 & 0 \\ 4 & 0 & 0 \\ 8 & -4 & 2 \\ \end{bmatrix}$$

The characteristic polynomial is $p_\phi(\lambda)=(2-\lambda)^3$, $r((A-2I)^0)-r((A-2I)^1)=3-1=2$ so the Jordan matrix must have two blocks of size $\ge 1$. I choose one of the possible Jordan forms $$M(\phi)_{\mathcal{A}}= \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix}$$ From this, I know that for vectors from $\mathcal{A}$ we must have $$\phi(\alpha_1)=2\alpha_1$$ $$\phi(\alpha_2)=\alpha_1+2\alpha_2$$ $$\phi(\alpha_3)=2\alpha_3$$ This means that $\alpha_1$ and $\alpha_3$ are the eigenvectors, for instance $\alpha_1=(1,2,0),\alpha_3=(0,0,1)$. Now $$\phi(\alpha_2)=\alpha_1+2\alpha_2$$ $$\phi(\alpha_2)-2\alpha_2=\alpha_1$$ $$\phi(\alpha_2)-2\text{id}(\alpha_2)=\alpha_1$$ $$(\phi-2\text{id})(\alpha_2)=\alpha_1$$ so $$(A-2I)\alpha_2=\alpha_1$$ $$\begin{bmatrix} 2 & -1 & 0 \\ 4 & -2 & 0 \\ 8 & -4 & 0 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ 0 \\ \end{bmatrix}$$ This system doesn't have a solution and we are one vector short. What am I missing?

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    $\begingroup$ Why must $(A-2I)\alpha_2$ equal $\alpha_1$? Why not $\alpha_3$? Or some linear combination of them, like $\begin{pmatrix}1\\ 2\\ 4\end{pmatrix}$? $\endgroup$
    – Bob Jones
    Mar 21, 2017 at 7:43
  • $\begingroup$ @BobJones We are looking for $\mathcal{A}$. $M(\phi)_{\mathcal{A}}$ gives us the condition $\phi(\alpha_2)=\alpha_1+2\alpha_2$ (from the definition of $M(\phi)_{\mathcal{B}}=M(\phi)_{\mathcal{B}}^{{\mathcal{B}}}$, where $\mathcal{B}$ is a basis). Then we use this to get to $(\phi-2\text{id})(\alpha_2)=\alpha_1$ which, in matrix form, can be written as the equation you're asking about. I was able to work through several examples using this approach and get to a system I was able to solve. This time, I feel I'm missing something basic. $\endgroup$
    – Zelazny
    Mar 21, 2017 at 8:22
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    $\begingroup$ Yes, the thing you're missing is that the eigenvectors space does not have to be generated by those two vectors; you cannot just pick two vectors in the space and declare them to be $\alpha_1$ and $\alpha_3$. Because $\alpha_1$ must be in the image of $A-2I$, you have to take that into account and pick an $\alpha_1$ that works for that. Like my vector I gave. Then you may choose $\alpha_2$ in the preimage, and $\alpha_3$ to complete the eigenspace. $\endgroup$
    – Bob Jones
    Mar 21, 2017 at 14:40

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All we know at this stage is that $\{ (1,2,0), (0,0,1) \}$ is a basis for ${\rm Ker}(A-2I)$.

We cannot just pick any random vector in ${\rm Ker}(A - 2I)$ to be our $\alpha_1$; we need to find one that works.

First, let us add a vector to $\{ (1,2,0), (0,0,1) \}$ to form a complete basis for $\mathbb R^3$. Call this extra basis vector $\alpha_2$. It doesn't matter which vector we choose for $\alpha_2$, as long as it is linearly independent of $(1,2,0)$ and $(0,0,1)$. I'm going to choose $$\alpha_2 = (1,0,0).$$

This means that $\alpha_1$ must be $$ \alpha_1 = (A - 2I) \alpha_2 = (2,4,8). $$ [Note that, once $\alpha_2$ has been chosen, we do not get to choose $\alpha_1$!]

Finally, we can pick $\alpha_3$ to be any vector in the linear span of $\{ (1,2,0),(0,0,1)\}$ that is not a multiple of $\alpha_1$. Again, it doesn't matter which one we choose, so I'm going to go for $$ \alpha_3 = (0,0,1).$$

Thus we have constructed a basis $$\alpha_1 = (2,4,8), \ \ \ \alpha_2 = (1,0,0), \ \ \ \alpha_3 = (0,0,1)$$ that obeys the desired properties, $$ A \alpha_1 = 2 \alpha_1, \ \ \ A \alpha_2 = \alpha_1 + 2\alpha_2, \ \ \ A \alpha_3 = 2 \alpha_3.$$

By the way, it is better to "start at the top and work down" than to "start at the bottom and work up". Said another way, it is better to start with a basis vector $\alpha_2$ in ${\rm Ker}(A - 2I)^2 = \mathbb R^3$ that is not in ${\rm Ker}(A-2I)$, and then complete the basis of $\mathbb R^3$ by choosing a vector $\alpha_3$ in ${\rm Ker}(A- 2I)$ complementary to both $\alpha_2$ and $\alpha_1 = (A - 2I)\alpha_2$. You can easily generalise this approach and turn it into an algorithm that works in all cases.

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