1
$\begingroup$

The title says it all. This polynomial splits over $\mathbb{C}$ pretty obviously, $(t - \sqrt{i})(t + \sqrt{i})(t - i\sqrt{i})(t + i\sqrt{i}),$ but the matrix needs real entries, so I can't just make a diagonal matrix.

So my next idea is that $A^4 = -I,$ write $A^4 = B^2,$ where $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$ Then all I need to do is find $A$ such that $A^2 = B.$ Not possible (I think), probably because I started working with $2 \times 2$ matrices whereas the question does not give a size to he matrix, it just says "find a matrix with minimal polynomial ..."

This leaves me with questioning my approach, and that is why I've posted this question, how does one go about constructing a matrix given that it needs to satisfy a minimal polynomial, restricted to having elements in some field (preferably $\mathbb{R}$ or $\mathbb{C}$)? Solutions would be appriciated, but I am mostly interested in the procedure for constructing such matrices given a general minimal polynomial.


For fun, the subquestion associated with the original question is, "Show that the usual real linear map $v \to Av$ has no non-trivial subspace."

$\endgroup$
  • 5
    $\begingroup$ The degree of the minimal polynomial cannot be larger than the size of the matrix. $\endgroup$ – Quang Hoang Jun 2 '16 at 15:42
  • 4
    $\begingroup$ Look up "companion matrix". $\endgroup$ – Robert Israel Jun 2 '16 at 15:43
  • $\begingroup$ Very cool, hadn't come across the companion matrix before, thanks for the help! $\endgroup$ – Merkh Jun 2 '16 at 15:50
1
$\begingroup$

Hint to find such a matrix:

  1. decompose $t^4+1$ into quadratic/linear factors. Here, we have two quadratic factors $p_1(t)$ and $p_2(t)$.

  2. for each factor $p_i(t)$, find a $2\times 2$ matrix $A_i$ with minimal polynomial $p_i$.

  3. $A=\left(\begin{array}{ll}A_1&0\\0&A_2\end{array}\right)$.

Note:

  1. Since each $p_i$ is irreducible, $A_i$ can be taken in the form $$\left(\begin{array}{rr} a& -b\\b&a\end{array}\right),$$ i.e, a rotation in $\mathbb{R}^2$. Like in Jack's answer, you should be able to find them both.
  2. Clearly, $A$ has two stable subspaces :-).
  3. $A$ should not have an invariant space, since each would yield an eigenvector hence an eigenvalue. But the minimal polynomial of $A$ has no real roots.
$\endgroup$
1
$\begingroup$

Put $$A = \left(\begin{matrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0 \end{matrix} \right).$$ Then $$A^2 = \left(\begin{matrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0 \end{matrix} \right), \,\,\,\,\, A^3 = \left(\begin{matrix} 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0 \end{matrix} \right), \,\,\,\,\, A^4 = \left(\begin{matrix} -1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{matrix} \right). $$ Thus $A^4 + I = 0$. Further, $A^3, A^2, A, I$ are linearly independent, so there is no third degree polynomial which annihilates $A$. Thus $t^4 + 1$ is the minimal polynomial of $A$.

I'm led to believe that if $A$ is $4\times 4$, the second part isn't necessarily true and we can only prove that $A$ has no $1$ or $3$ dimensional invariant subspaces while there are cases where $A$ has a $2$ dimensional invariant subspace. That being said, I'm struggling to find an example.

As an aside, did you find this on the UCLA Basic Exam? I remember this question came up on one of those.

EDIT: I had posted a counterexample to the fact that $A$ has no non-trivial invariant subspaces, but Quang Hoang pointed out in the comments that my counterexample was incorrect.

$\endgroup$
  • $\begingroup$ You counter example is not correct. That matrix has $(t^4+1)(t-1)$ as its minimal polynomial. $\endgroup$ – Quang Hoang Jun 2 '16 at 16:04
  • $\begingroup$ Oh you're right. I will edit. Thanks! $\endgroup$ – User8128 Jun 2 '16 at 16:09
  • $\begingroup$ Ha yes, maybe 2014 spring exam? Been studying for September's qualifiers. $\endgroup$ – Merkh Jun 2 '16 at 16:10
  • 2
    $\begingroup$ If you're coming in this fall as a first year, the boot camp is incredibly helpful in preparing for the Basic so I'd definitely advise you to attend. Good luck! $\endgroup$ – User8128 Jun 2 '16 at 16:15
0
$\begingroup$

Figured it out, thanks to @Robert Israel's comment to look for companion matrices:

The answer to this specific question is $A = \begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0 \end{pmatrix}$.
Thanks for the responses everybody.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.