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- #1

- Feb 5, 2012

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Here's a question that I encountered recently. I would appreciate if you could go through my solution and let me know if you see any mistakes or have any comments.

**Question:**

Given a linear transformation \(f:\,\mathbb{C}\rightarrow \mathbb{C}\) with matrix \(\begin{pmatrix}3&0&8\\3&-1&6\\-2&0&5\end{pmatrix}\). Find the Jordan normal form of \(f\). Indicate \(X_{f}(t),\,\mu_{f}(t)\). What are the root spaces and the eigenspaces for \(f\)?

**My Answer:**

I found the characteristic polynomial to be, \(X_{f}(t)=-(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))\). Therefore the minimal polynomial is,

\(\mu_{f}(t)=\mu_{f}(t)=(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))\)

Since this matrix has three distinct eigenvalues it is diagonalizable. Therefore the Jordan normal form is,

\[J=\begin{pmatrix}-1&0&0\\0&4+i\sqrt{15}&0\\0&0&4-i\sqrt{15}\end{pmatrix}\]

There is only one eigenvector corresponding to \(\lambda = -1\) and one eigenvector for both of the complex eigenvalues. The eigenspace for \(\lambda =-1\) therefore consist of the single vector, \(\begin{pmatrix}0\\1\\0\end{pmatrix}\) and the eigenspaces for both the complex eigenvalues consist of \(\begin{pmatrix}1\\1\\1\end{pmatrix}\).

Now the geometric multiplicity of each of the eigenvalues is one. So the rootspace corresponding to each eigenvalue consist of all the eigenvectors for that particular eigenvalue plus the zero vector. That is the rootspace of \(\lambda =-1\) is \(\left\{ \begin{pmatrix}0\\1\\0\end{pmatrix},\, \begin{pmatrix}0\\0\\0\end{pmatrix} \right\}\), the rootspaces corresponding to both of the complex eigenvalues is \(\left\{ \begin{pmatrix}1\\1\\1\end{pmatrix},\, \begin{pmatrix}0\\0\\0\end{pmatrix} \right\}\).