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I know that this type of question has been asked on here before but I am still having a hard understanding what is going on. The text that I am learning from is "Linear Algebra Done Right by Sheldon Axler" and I don't think it covers this topic very well. Does anyone have any suggestions? I am studying for my linear comp at the end of January.

Given the matrix

$$A = \left(\begin{matrix}0&-1&-1\\-3&-1&-2\\7&5&6\end{matrix}\right).$$

Find the Jordan form $J$ and an invertible matrix $Q$ such that $A = QJQ^{-1}$.

I know that I want to start by finding the eigenvalues of this matrix. I ended up getting det$(A - \lambda I) = (\lambda + 2)(\lambda62 + 3\lambda - 2) \implies \lambda = -2, \frac{-3 -\sqrt{17}}{2}, \frac{\sqrt{17} - 3}{2}$.

Now, from what I have read on this website, I know now that I want to find the null space of $A - \lambda I$ for each eigenvalue. Then my invertible matrix $Q$ will have entries whose columns are these vectors. Why does one do this? Is there a "nicer" way to go about doing this other than direct computation?

This problem is from a previous linear comp at my university, so I can't imagine that they would have wanted me to the direct calculation.

Hopefully, my questions and explanation of the problem were clear. Thanks for any help in advance.

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The characteristic polynomial is $$ p(\lambda)=\lambda^3-5\lambda^2+8\lambda-4=(\lambda-1)(\lambda-2)^2. $$ The vectors involved in the Jordan blocks associated with eigenvalue $2$ is the column space of $$ A-I = \begin{pmatrix}-1 & -1 & -1 \\ -3 & -2 & -2 \\ 7 & 5 & 5\end{pmatrix}. $$

The column space of $A-I$ is two-dimensional, which gives $\mbox{dim}(\mbox{ker}(A-I))=1$, and it is obvious that $\mbox{ker}(A-I)$ is spanned by $$ \begin{pmatrix}0 \\ 1 \\ -1\end{pmatrix}. $$ (It is obvious because the last two columns of $A-I$ are identical.) Then \begin{align} (A-2I)(A-I)&=\begin{pmatrix}-2 & -1 & -1 \\ -3 & -3 & -2 \\ 7 & 5 & 4\end{pmatrix}. \begin{pmatrix}-1 & -1 & -1 \\ -3 & -2 & -2 \\ 7 & 5 & 5\end{pmatrix} \\ &= \begin{pmatrix}-2 & -1 & -1 \\ -2 & -1 & -1 \\ 6 & 3 & 3\end{pmatrix} \end{align} Because $(A-2I)^2(A-I)=0$, you have \begin{align} (A-2I)\begin{pmatrix} -1 \\ -2 \\ 5\end{pmatrix}&=\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} \\ (A-2I)\begin{pmatrix}-1 \\ -1 \\ 3\end{pmatrix} &= 0. \end{align} The Jordan form is $$ J = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{pmatrix} $$ and the transition matrix $Q$ is $$ Q = \begin{pmatrix} 0 & -1 & -1 \\ 1 & -1 & -2 \\ -1 & 3 & 5 \end{pmatrix}. $$

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Eigenvalues are $2,2,1$, which simplifies the calculation a lot.

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  • $\begingroup$ Well I better double check my work, thank you. $\endgroup$ – Taylor McMillan Jan 4 at 23:03
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    $\begingroup$ A simple way to check if you Eigenvalues $\lambda_j ,j\in \{1...n\}$ for a $n\times n$ Matrix are correct is to check if $\sum_{j=1}^n\lambda_j=trace(A)$ and if $\prod_{j=1}^n\lambda_j=det(A)$ $\endgroup$ – A. P Jan 4 at 23:13
  • $\begingroup$ Thanks @A.P. Also, how do I handle finding the null space for the eigenvalue 2 that has multiplicity 2? $\endgroup$ – Taylor McMillan Jan 4 at 23:24
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    $\begingroup$ simply use the same algorithm as for calculating the Nullspace for the value 1. In this case you will notice that you only get one Vektor, therefore the dimension of the Nullspace for $\lamda=2$ is 1. $\endgroup$ – A. P Jan 4 at 23:37

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