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Consider the matrix $A= \begin{pmatrix}1 & 1& 1\\ 0 & 1& 1\\ 0& 0& 1\end{pmatrix}$, a $3\times 3$ matrix.

A generalised eigenvector of $A$ is a non-zero vector $v$, an element of $\mathbb{R}^3$, such that $[(A-bI)^k] v = 0$, for some scalar $b$ and some integer $k\geqslant 1$

Give an example of a vector $v$, element of $\mathbb{R}^3$, such that $v$ is a generalised eigenvector of $A$ but not an eigenvector of $A$.

I am not too sure how to find the generalised eigenvectors.

I started by $\det(A-bI)^k = k^3 \det(A-bI) = 0$

$\implies \det(A-bI) = 0$ to find eigenvalues for the generalised eigenvectors, but I realized won't this just be the same as finding eigenvectors?? Can someone please tell me what I did wrong or how I should do it

Thanks

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  • $\begingroup$ is 1 1 1 the column entries or row entries? $\endgroup$ – Bijesh K.S Oct 30 '16 at 7:51
  • $\begingroup$ 1 1 1 is the first row entries $\endgroup$ – Howard Oct 30 '16 at 7:52
  • $\begingroup$ Please check my edit to see how to typeset a matrix. $\endgroup$ – Andreas Caranti Oct 30 '16 at 7:58
  • $\begingroup$ Ok thanks, I understand now $\endgroup$ – Howard Oct 30 '16 at 8:00
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The vector $$ \begin{bmatrix}0\\0\\1\end{bmatrix} $$ is clearly not an eigenvector (please do check it).

However, since $(A - I)^{3} = 0$ (please do check it!), all nonzero vectors are generalised eigenvectors with respect to the eigenvalue $1$.

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  • $\begingroup$ Why isn't it an eigenvector? I multiplied it by A and got itself again. $\endgroup$ – Howard Oct 30 '16 at 8:08
  • $\begingroup$ @Howard, sorry, you're perfectly right, it's just that I am used to row vectors... it's fixed now. Again, my apologies. $\endgroup$ – Andreas Caranti Oct 30 '16 at 8:24
  • $\begingroup$ Oh right, makes sense to me now, thanks a lot $\endgroup$ – Howard Oct 30 '16 at 8:32
  • $\begingroup$ @Howard, please consider upvoting and/or accepting my answer if you found it useful. $\endgroup$ – Andreas Caranti Oct 30 '16 at 12:29

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