The question is:

Suppose there is a $3$-dimension vector space $V$ over the field of real number $R$, and there is a $3\times3$ matrix $A$ whose entries are real numbers. $A$ is the matrix representation of a linear transformation $\mathscr{A}$ under some basis of $V$. What dimension of invariant subspaces must it has? Choices are $1, 2, 3$.

Here is what I think:

Since the characteristic polynomial of $A$ is of degree $3$, it must have a root in $R$, thus it has a characteristic subspace of dimension $1$, which is also its invariant subspace. And obviously $V$ is its invariant subspace, thus $3$ is right.

But I do not know how to check whether it must have a $2$-dimensional invariant subspace. How can I proceed from this? If possible, I'm also very willing to know how to deal with similar problem for $4\times4$ matrix or higher dimension. Thank you!

  • $\begingroup$ Do you know the primary decomposition theorem? $\endgroup$ – user113529 Sep 5 '17 at 5:15
  • $\begingroup$ @user43687 Sorry I'm not familiar with that. But I have some basic understanding of minimal polynomial. Would you like to explain it? I don't quite understand what Wiki says. Thank you! $\endgroup$ – Edward Wang Sep 5 '17 at 5:38
  • $\begingroup$ Your observation on my previous answer was right. Embarassingly, in my haste I gave you the wrong explanation of rational canonical form. For one, the minimal polynomial is not the product, but rather simply dn. I don't have time to fix the answer...so I will delete it. If I have time, I will give a proper answer. Apologies. $\endgroup$ – user113529 Sep 5 '17 at 14:42
  • $\begingroup$ @user43687 You already helped a lot and thank you. If you have time, I'm happy to learn from your revised answer. $\endgroup$ – Edward Wang Sep 5 '17 at 17:54
  • $\begingroup$ I think I know the answer. I will post when I have time... $\endgroup$ – Edward Wang Sep 13 '17 at 16:25

Consider its cyclic decomposition: $$ V=F[\mathscr{A}]\alpha_1 \oplus...\oplus F[\mathscr{A}]\alpha_i, i\in \{1,2,3\} $$ If $i$ is $2$ or $3$, it's done. We only need to consider when $i=1$, then $V$ is a cyclic space, thus $\mathscr{A}$ only has one elementary divisor. Note that irreducible polynomials in $R$ must have degree $1$ or $2$, its Jordan normal form must be: $$ \begin{bmatrix} \lambda & & \\ 1 & \lambda & \\ & 1 & \lambda \end{bmatrix} $$ $\lambda$ being its eigenvalue. Then we see a $2\times2$ submatrix in the down-right. Thus it must have a dimension $2$ subspace.

  • $\begingroup$ @user296602 I have posted my answer to the question. $\endgroup$ – Edward Wang Sep 14 '17 at 14:49

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