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I am working on the following Linear Algebra problem:

(a) Suppose $T: \mathbb{R}^4 \longrightarrow \mathbb{R}^4$ is a linear transformation with characteristic polynomial $x^2(x-1)^2$. Describe the 3-dimensional invariant subspaces of $T$. (Your answer may have different cases depending on $T$.)

(b) Let $A$ be the matrix \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix} with characteristic and minimal polynomial $x^5-1$ over $\mathbb{Q}$. Show that any $A$-invariant subspace of $\mathbb{Q}^5$ has dimension $0,1,4$ or $5$.

(c) Suppose $T: \mathbb{R}^4 \longrightarrow \mathbb{R}^4$ is a linear transformation with characteristic polynomial $(x^2-x)(x^2+1)$. Describe the $2$-dimensional $T$-invariant subspaces of $\mathbb{R}^4$.

In parts (a) and (c), I suppose I'm not sure what is meant by describing the specified $T$-invariant subspaces.

The Jordan form for a square $n \times n$ matrix $A$ gives a decomposition of the $n$-dimensional Euclidean space into invariant subspaces of $A$, where every Jordan block corresponds to an invariant subspace. Thus, for parts (a) and (c), is it enough to find the Jordan form for each possible minimal polynomial that divides the given characteristic polynomial, and simply say that the $m$-dimensional $T$-invariant subspaces are those corresponding to size $m$ Jordan blocks for a particular eigenvalue? If not, how can I describe the requested $T$-invariant subspaces?

In part (b), the characteristic polynomial and minimal polynomial factor as $(x-1)(x^4 + x^3 + x^2 + x + 1)$ over $\mathbb{Q}$. How can I show there are $A$-invariant subspaces of dimension $0,1,4$ or $5$ if there is only one possible minimal polynomial, and it only has $1$ eigenvalue of $\lambda = 1$ in $\mathbb{Q}$ ? It seems there's only one Jordan normal form to consider, and thus, only one possible dimension of the $A$-invariant subspace.

Thanks!

Mo Behzad Kang
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    A subspace $W$ is $T$-invariant if $T^{-1}W = W$ (I think). An $n$-dimensional $T$-invariant subspace is a subspace that is $n$-dimensional. – mathworker21 Nov 19 '19 at 02:45
  • @mathworker21 the definition of an invariant subspace I had in mind is that a subspace $W$ is $T$-invariant if $T(W) \subseteq W$. – Mo Behzad Kang Nov 19 '19 at 02:46
  • @DWiggles Right. Must have been a typo in the original problem. – Mo Behzad Kang Nov 19 '19 at 02:52
  • For part (b), there is indeed only one possible Jordan normal form. And that should tell you all the possibilities for $A$-invariant subspaces. Each $A$-invariant subspace will be a direct sum of eigenspaces corresponding to different blocks in the JNF. – D Wiggles Nov 19 '19 at 02:55
  • @DWiggles What is meant by "eigenspaces corresponding to different blocks in the JNF" ? Here, there's only one eigenvalue in $\mathbb{Q}$ given by $\lambda = 1$. Perhaps this is what's confusing me a lot. – Mo Behzad Kang Nov 19 '19 at 03:15
  • The eigenvalue need not lie in $\Bbb Q$. For instance, $\pmatrix{0&-1\1&0}$ has characteristic polynomial $\lambda^2 + 1$ with no roots in $\Bbb R$. The roots lie in $\Bbb Q$ if and only the polynomial splits into linear factors over $\Bbb Q$. This is not the case. – Fimpellizzeri Nov 19 '19 at 22:19
  • @Fimpellizieri I see. In that case, there are $5$ distinct eigenvalues here, right ? Corresponding to the 5th roots of unity ? Do I proceed from here by finding the dimensions of the eigenspaces for each of these eigenvalues ? – Mo Behzad Kang Nov 19 '19 at 22:43
  • In $\Bbb Q$ there is only one eigenvalue $\lambda = 1$. We could extend the example above to $\Bbb R^3$ with $M = \pmatrix{0&-1&0\1&0&0\0&0&1}$. Still we would have that $i$ and $-i$ are roots of $M$'s characteristic polynomial not in $\Bbb R$, which of course means there can be no eigenspace in $\Bbb R^3$ associated to either of them. However, by construction we obviously have that $\Bbb R^2 \times {0}$ is an $M$-invariant subspace of $\Bbb R^3$. This is not a coincidence. – Fimpellizzeri Nov 20 '19 at 03:56

1 Answers1

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Taking the Jordan normal form is the way to go. However, in the first case we don't have the minimal polynomial, so we must take all possible cases for the minimal polynomial and work with each of them.

Start with $a$, and consider $x^2(x-1)^2$. What can be the minimal polynomial? Note that the minimal polynomial must be a factor of the char. polynomial, but also must have as roots, all the roots of the char. polynomial. It follows that we have these possibilities : $$ x(x-1), x^2(x-1) , x(x-1)^2 , x^2(x-1)^2 $$

The invariant subspaces of a matrix have dimensions given by the degrees of the co-prime factors of the minimal polynomial, and whatever is left behind is decided by which eigenvalues are left, and how large their blocks can be (i.e. the leftover can be made into blocks, none of which should have size more than that specified by the minimal polynomial). This key statement, which should be mentioned along with the Jordan normal form statement IMO, drives the entire exercise.

So we break into cases :

  • $x^2(x-1)^2$ : Two invariant subspaces of dimension two each.

  • $x^2(x-1)$ : One invariant subspace of dimension two, and one invariant subspace of dimension one. That leaves only one dimension free for any other block, so that gives one more invariant subspace of dimension one.

  • $x(x-1)^2$ : Same analysis as above leads to the conclusion : one invariant subspace of dimension $2$ and two of dimension $1$.

  • $x(x-1)$ : Certainly there are two invariant subspaces of dimension $1$ each. That leaves us two dimensions to work with : but these cannot go under the same block, since they are of different eigenvalues $0 \neq 1$. So they go under different blocks, giving four invariant subspaces of dimension $1$ each (in other words, a diagonalizable matrix!)

So how do we describe the $3$ dimensional invariant subspaces of $T$? We may do it as follows , based on minimal polynomial :

  • $x^2(x-1)^2$ : There's no three-dimensional invariant subspace, because both invariant subspaces are of dimension two each, so any subspace of dimension $3$ must contain a non-zero vector from both subspaces, hence must contain the whole of $\mathbb R^4$, a contradiction.

  • $x^2(x-1)$ : Take the direct sum of the two-dimensional invariant subspace corresponding to $0$ (the kernel, in fact!) and any of the two eigenvectors of $1$.

  • $(x-1)^2x$ : Take the direct sum of the two-dimensional invariant subspace corresponding to $1$ with any one of $2$ eigenspaces of $0$.

  • $x(x-1)$ : Take the direct sum of any three of the eigenspaces.

That characterizes all the $3$ dimensional invariant subspaces, and completes problem $a$.

The point of $b$ is that the minimal polynomial is $x^5 - 1 = (x-1)(x^4+x^3+x^2+x+1)$ as an irreducible break up in $\mathbb Q$. By what I mentioned earlier, the disjoint invariant subspaces must be of dimensions $1$ and $4$. Now, of course you could take just the zero vector or the entire space, so that gives the answer.

Use the same principle for $c$ : the irreducible break up is $x(x-1)(x^2 + 1)$. But what can the minimal polynomial be? It shares all the roots of the characteristic polynomial. But the roots of the characteristic polynomial are all distinct! Therefore the min. polynomial must also be the same i.e. $x(x-1)(x^2+1)$.

From here, we deduce that there are two invariant subspaces of dimension $1$ which are eigenspaces of $0$ and $1$, and one invariant subspace of dimension $2$ corresponding to $x^2+1$. Now it is easy to say what all the two-dimensional invariant subspaces are.

Sarvesh Ravichandran Iyer
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  • Hi. Thank you for the thorough answer. It is a big help. So in part (c), the two-dimensional invariant subspaces are simply the subspace corresponding to the irreducible factor $x^2 + 1$, and the direct sum of the 1-dimensional invariant subspace corresponding to the eigenspace of $\lambda = 0$ with the 1-dimensional invariant subspace corresponding to the eigenspace of $\lambda = 1$, correct ? – Mo Behzad Kang Nov 21 '19 at 22:28
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    Correct , you are right! – Sarvesh Ravichandran Iyer Nov 22 '19 at 00:48