So I have this matrix of a linear map $ \varphi$: $ \begin{pmatrix} 8 &-6 &-1 \\ 5 &-3 &-1 \\ 4 & -2 & -2 \end{pmatrix} $, where $ \varphi \in (\mathbb{Q}^3)$. I have to find all one dimensional $ \varphi$ invariant subspaces of $ \mathbb{Q}^3$ . So my question is I have to find the eigenvalues of the matrix, see which of them belong to $ \mathbb{Q}^3$ and with those values find the eigenvectors and write $ l(v)$ where $ v$ is the eigenvector. Also please tell me if I'm wrong , but my textbook doesn't explain it very well, an invariant subspace of $ \varphi$ is one dimensional if the eigenvalue $ \lambda$ belongs to $ \mathbb{R}$ is real ,and is two dimensional if $ \lambda$ is imaginary for example $ 2\imath$.Could you also explain to me in short why that is so. Also please tell me if I have mistaken somewhere and excuse me if this question is silly it's just that my hole textbook is 80 pages of badly written explanations with no examples, but my program is different from the american or english one so textbooks in english don't really work for me. Sorry for the long post and big thanks to anyone who can help!
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1$\begingroup$ had you heard about the algebra's fundamental theorem? $\endgroup$– janmarqzCommented Jan 8, 2014 at 21:05
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1$\begingroup$ there also polynomials which have rational eigenvalues, and if the polynomial of $\varphi$ should have one, then there would exists son eigen rational eigenvector $\endgroup$– janmarqzCommented Jan 8, 2014 at 21:16
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1$\begingroup$ The eigenspace corresponding to a real eigenvalue needs not, in general, be one dimensional. Easy counter examples are available. What your textbook may be saying is that, if $\lambda$ is a complex eigenvalue of a matrix (non-zero imaginary part), then its complex conjugate is also an eigenvalue of the same matrix. So here you get two eigenspaces. $\endgroup$– AnyADCommented Jan 8, 2014 at 21:22
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1$\begingroup$ for you benefit check: wolframalpha.com/input/… $\endgroup$– janmarqzCommented Jan 8, 2014 at 21:26
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1$\begingroup$ @janmarqz Thank you so that means I have 1 one dimensional invariant subspase and it is $l(v)$ where $ v $ is equal to $ (-1,1,0) $ $\endgroup$– randomnameCommented Jan 8, 2014 at 21:28
1 Answer
An invariant subspace is a vector subspace $W$ such that for each $w\in W$ we have $\varphi(w)\in W$ too. A one-dimensional invariant subspace has a basis $\{w_1\}$, and we must have $\varphi(w_1) = \lambda w_1$ where $\lambda$ is the sole coordinate for the image vector wrt. the basis chosen.
We can also conclude that this $\lambda$ (because it is a coordinate) belongs to the field of scalars.
You seem to be working with $\mathbb{Q}^3$, a vector space over the field $\mathbb{Q}$. We can conclude that each $\varphi$-invariant subspace of dimension one corresponds to some rational eigenvalue.
On the other hand one can easily realize that each rational eigenvalue corresponds to one or more one-dimensional invariant subspaces (just pick an eigenvector).
The only difficulty that can arise is if some rational eigenvalue gives rise to more than one one-dimensional invariant eigenspace (because its entire eigenspace is multidimensional). However one can show that this only happens if the eigenvalue in question is a double (or multiple) root of the characteristc polynomial of $\varphi$. That does not happen with the matrix you have. When it does happen, one simply finds the full eigenspace, and any one-dimensional subspace of that eigenspace will clearly be invariant.
