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I have the linear transformation

$T \in \mathcal{L}(\mathbb{C}^{5})$ defined by $T(x_1,x_2,x_3,x_4,x_5)=(2x_1,x_2+x_4,2x_3,x_2+x_4,-x_1+x_3+2x_5)$

The matrix with respect to the standard basis is $A=\begin{bmatrix} 2&0&0&0&0\\ 0&1&0&1&0\\ 0&0&2&0&0\\ 0&1&0&1&0\\ -1&0&1&0&2\end{bmatrix}$

I know the eigenvalues are $2$ and $0$. The eigenspaces are $E(0,T)=\{(0,x_2,0,-x_2,0,0):x_2 \in \mathbb{C}\}$

$E(2,T)=\{(x_1,x_2,x_1,x_2,x_5):x_1,x_2,x_5 \in \mathbb{C}\}$

I considered the matrix $(A-2I)^5$ which I found on my calculator to be

$(A-2I)^5=\begin{bmatrix} 0&0&0&0&0\\ 0&-16&0&16&0\\ 0&0&0&0&0\\ 0&16&0&-16&0\\ 0&0&0&0&0\end{bmatrix}$

I then calculated the null space of this matrix to yield the generalized eigenspace: $G(2,T)=\{(x_1,x_2,x_3,x_2,x_5):x_1,x_2,x_5 \in \mathbb{C}\}$

In a linear algebra book they give an example with the matrix,

In the example they give the matrix

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

They claim a basis of the null space of $(A-2I)^2$ is

$\begin{bmatrix} 1\\ -3\\ 1 \end{bmatrix}, \begin{bmatrix} -1\\ 2\\ 0 \end{bmatrix} $

However when calculating the way I did with the first matrix I get

$\begin{bmatrix} 1/2\\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -1/2\\ 0\\ 1 \end{bmatrix} $

as a basis for the generalized eigenspace.

I am wondering why in the first example I have given I can just calculate $\text{null}(A-5I)^5$ to give me the generalized eigenvectors. In other examples I have seen they were not able to do this, Which method is correct?

I thought you cannot just calculate the null space of $(A-\lambda I)^{\text{dimV}}$ to get the generalized eigenspace for a matrix of a linear map with $V$ as the domain.

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  • $\begingroup$ You can certainly find a basis for $(A-\lambda I)^{\dim v}$ directly, but not every basis of $(A-\lambda I)^{\dim V}$ is a Jordan basis. Observe that if you use the basis that you computed in $P^{-1}AP$ you don’t get the Jordan normal form of $A$. $\endgroup$ – amd May 8 at 1:54
  • $\begingroup$ @miles If my answer is not clear, please tell me. $\endgroup$ – Jan May 15 at 20:45
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Since you don't have shown your calculations, I can't tell you where the mistake is. Furthermore, the solutions from the book you have quoted are wrong.

The matrix $A$ has two linearly independent eigenvectors, take for instance $(-1, 3, 1)$ and $(-1, 2, 1)$. Furthermore, note that

$$(A - 2I)^2 = \begin{pmatrix} 2 & 1 & -1 \\ -4 & -2 & 2 \\ -2 & -2 & 1 \end{pmatrix}.$$

A row-reduced echelon from of this matrix is

$$\begin{pmatrix} 2 & 1 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$

You now can choose $(1, 0, 2)$ as a generalized eigenvector.

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