Given is the matrix,


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?


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

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

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