# Finding the Generalized Eigenspace

Given is the matrix,

\begin{bmatrix}0&0&-2&0&0\\0&0&1&0&0\\1&1&2&0&0\\-1&-1&-2&-1&-2\\1&1&2&1&2\\\end{bmatrix}

Find the generalized eigenspace for the eigenvalue λ = 0.

I have found two solutions for the eigenvector for the eigenvalue 0.

$$λ_1 = \begin{bmatrix}1\\-1\\0\\0\\0\\\end{bmatrix} λ_2 = \begin{bmatrix}0\\0\\0\\-2\\1\\\end{bmatrix}$$

The eigenvalue 0 has an algebraic multiplicity of 2. See the characteristic polynomial $$λ^2(1-λ)^3$$

I find the same vectors $$λ_1, λ_2$$ also for the generalized eigenvector.

Are these steps correct? And what is the generalized eigenspace? Are my vectors the bases of the space?

Thanks.

Edit: Sorry, my bad the question asks for generalized eigenspace not for eigenvector.

• Seems that you have to find the generalized eigenvector for $\lambda=1$. – Peter May 3 '16 at 16:17
• The question asks for λ = 0 – AriNubar May 3 '16 at 16:18
• But for $\lambda=0$, we have no generalized eigenvector, because there are $2$ linear independent vectors solving $Ax=0$ – Peter May 3 '16 at 16:19
• @AriNubar as Peter said, there must be a mistake in the question itself or in the way that you copied it. – Omnomnomnom May 3 '16 at 16:21
• $$S=\pmatrix{-1&0&0&-2&4\\1&0&0&1&-2\\0&0&0&1&-1\\0&-2&-1&0&0\\0&1&1&0&0}$$ – Peter May 3 '16 at 16:47