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I had a linear algebra test today, and there were 6 questions which were of the type "say true with proof, or false with counterexamples". The sad part was (as far as I figured out) only one of them was true, and the other five needed counterexamples. These 5 were (as far as I can recall)

  1. Two $3\times 3$ matrices on $\mathbb C$ with the same minimal polynomial must be similar.

  2. Two $3\times 3$ nilpotent matrices must be similar.

  3. Two $4\times 4$ nilpotent matrices with same minimal and characteristic polynomial must be similar.

  4. Product of two nilpotent matrices is nilpotent.

  5. For any operator $T$ on $V$, we have $V=\text{range }T\oplus \text{ker }T$.

(Calm down, I'm not asking these five questions together)

Now, although (I believe) I could correctly arrive at these counterexamples, it made me completely exhausted. In other math courses, when we were constructing counterexamples, life didn't seem to be this difficult as I was mostly familiar with the properties of the real numbers or functions $f:\mathbb R\to \mathbb R$, or sequences of real numbers. I had very concrete idea of how they worked, and doing some brute force calculations weren't that difficult. In here however, things seemed really hard- I had no idea how to quickly guess a matrix with a desired minimal or characteristic polynomial, I didn't know how to construct "the simplest" $4\times 4$ nilpotent matrix (except that upper triangular matrices with $0$ diagonal is nilpotent), and matrices cannot be quickly multiplied with brute force like real numbers.

This got me thinking. I want to know what are some of the tricks and properties, or some of the more famous candidate operators and matrices that we should remember to construct these quick counterexamples.

I understand that my question sounds a little vague. So, here's some more explanation. In real analysis, when we needed to find pathological examples of continuous (/discontinuous) functions, the Dirichlet function was more often than not called for aid. Similarly, in sequences, $\{(-1)^n\}_{n=1}^\infty$ was often our first guess; in cases concerning metric spaces, the Discrete Metric Space was often the first choice for a pathological example. So, my question is whether there are any such "first guess" matrices, or any candidate matrices designed specially to bend the rules that we may use. In other words, what are the properties that I should remember (considering the fact that there are too many nitty-gritties to remember all of them) to have sufficient hints to construct counterexamples?

Although I've accepted an answer, please feel free to still continue to pour interesting answers into this post, and recieve upvotes (at least) from me.

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  • $\begingroup$ For 4., consider $\pmatrix{0&1\\0&0}$ and $\pmatrix{0&0\\1&0}$ $\endgroup$ Commented Sep 13, 2021 at 19:30
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    $\begingroup$ @J.W.Tanner that's exactly what I used. But, that's not my real question. My question is about whether there are some general properties that we should keep in mind while trying to find out matrix counterexamples. $\endgroup$ Commented Sep 13, 2021 at 19:52
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    $\begingroup$ @user8675309 your answer was quite helpful. I would like to request you to add that answer to my post as well for future references. I'm sure I'll need it quite a few times in future. $\endgroup$ Commented Sep 14, 2021 at 19:07
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    $\begingroup$ Try to find examples where all the entries are ones and zeros, as those are pretty easy to multiply (and the more zeros, the better!). $\endgroup$ Commented Sep 15, 2021 at 4:20
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    $\begingroup$ Some things, you have to learn by experience, as you did. $\endgroup$ Commented Sep 15, 2021 at 13:10

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There's an approach to these problems that I largely attribute to Artin's Algebra which is: when evaluating whether a claim is true or not, try to evaluate with some very simple representatives. Frequently you'll find a counterexample. (Alternatively, if true, you can frequently bootstrap a general proof from a simple one.) Diagonal matrices are of course easiest, but beyond that

(i.) upper triangular matrices
(ii.) Hermitian matrices are very easy to work with and have properties you can easily exploit.
(iii.) block diagonal matrices are also quite nice.
This motivates a very useful technique: work out some $2\times 2$ and perhaps $3\times 3$ examples then use them in simple block structures to create more complicated examples. In general block diagonal matrices of the form
$A= \left[\begin{matrix}Y & \mathbf 0\\\mathbf 0 & Z\end{matrix}\right]$
where ($Y$ and $Z$ are square) are extremely easy to work with. There are underlying interpretations in terms of direct sums (and direct products) that can be helpful. I trust you know that the characteristic polynomial splits along the blocks, i.e.

$\det\big(\lambda I -A\big)=\det\big(\lambda I -Y\big)\cdot \det\big(\lambda I -Z\big)$

(iv.) involutions and idempotence also have very nice properties, useful elsewhere


For this post, every nilpotent matrix is similar to a strictly upper triangular matrix, so use them as a representatives in $2-4$ (and even 5).

The $2\times 2$ case is simplest so start with that. Here we only have representatives $\mathbf 0$ and $N = \left[\begin{matrix}0 & \alpha\\0 & 0\end{matrix}\right]$ for some $\alpha \neq 0$. These are obviously not similar, justification: (a) the conjugacy class of the zero matrix only contains the zero matrix or (b) they have different ranks so can't be similar.

(1) requires some tinkering though I'd attack this in terms of either diagonal matrices or use (iii).

(2) Consider re-using the the $2\times 2$ case (and trivial $1\times 1$ case) in blocks $\left[\begin{matrix}\mathbf 0 & \mathbf 0\\\mathbf 0 & 0\end{matrix}\right]$ and $\left[\begin{matrix}N & \mathbf 0\\\mathbf 0 & 0\end{matrix}\right]$. Same rationale for why they can't be similar.
(3) since $4=2\cdot 2$, consider the $2\times 2$ case, twice, via a block matrix $A:= \left[\begin{matrix}N & \mathbf 0\\\mathbf 0 & N\end{matrix}\right]$ and $A':= \left[\begin{matrix}N & \mathbf 0\\\mathbf 0 & \mathbf 0\end{matrix}\right]$
$\text{rank } A =2 \gt 1 =\text{rank } A'$ so they aren't similar but they have the same minimal and characteristic polynomials.

note: you said

I had no idea how to quickly guess a matrix with a desired minimal or characteristic polynomial

This is one of the advantages of the block diagonal structure. You learn the minimal (and characteristic) polynomials for simple (e.g. $2\times 2$) cases then infer the result for some more complicated matrix living in a higher dimensional vector space. i.e.

$p\big(A\big):= \left[\begin{matrix}p\big(N\big) & \mathbf 0\\\mathbf 0 & p\big(N\big)\end{matrix}\right]$
so it's immediate that $A$ has the same minimal polynomial as $N$. And you can check that $A'$ has same min polynomial -- anything of lower degree wouldn't annihilate $N$ on its leading block.

(4) Make using of Hermicity so $Z: =N^*N$ is a product of two nilpotent matrices but e.g. Hermitian matrices are diagonalizable and the only diagonalizable nilpotent matrix is the zero matrix, but $\text{rank}\big(N^*N\big)=\text{rank}\big(N\big)=1\gt 0$, etc. There are many ways to finish, but the big idea is you can convert an awkward matrix into a diagonalizable one via Hermicity.

(5) a simple approach is to recognize that you have been getting hit with nilpotent matrices so use them. Again with $2\times 2$ case we have $N(N) =N^2 = \mathbf 0$ but $N\neq \mathbf 0$ so $N$ kills $N\cdot V \neq \mathbf 0$ which tells you something about $\ker N \cap \text{im } N$.

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    $\begingroup$ This is a great answer (+1). This is the type of answer I was looking for. Thanks! $\endgroup$ Commented Sep 15, 2021 at 13:01
  • $\begingroup$ This is a nice answer! $\endgroup$
    – Pole_Star
    Commented Sep 17, 2022 at 10:40
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Some of these problems can be nicely approached with the help of rational canonical form. Note that two matrices are similar if and only if their invariant factors $p_1 \mid \cdots \mid p_k$ are the same. The minimal polynomial is the polynomial $p_k$ (the invariant factor of highest degree), and the characteristic polynomial is the product of all invariant factors, i.e. the polynomial $p(x) = p_1(x) \cdots p_k(x)$.


Problem 1: We want to find two matrices with the same minimal polynomial (i.e. the same largest factor) that are not similar. Clearly, these matrices must have more than one factor for this to hold. One approach is to take the matrices with invariant factors $$ p_1(x) = x, \quad p_2(x) = x(x-1) = x^2 - x\\ q_1(x) = x-1, \quad p_2(x) = x(x-1) = x^2 - x. $$ This leads to the rational form matrices $$ \left[ \begin{array}{c|cc} 0 & 0 & 0\\ \hline 0 & 0 & 0\\0 & 1&1\end{array} \right], \quad \left[ \begin{array}{c|cc} 1 & 0 & 0\\ \hline 0 & 0 & 0\\0 & 1&1\end{array} \right]. $$ Up to similarity, this is the same example as Brauer's.

Problem 2: A matrix is nilpotent if and only if its minimal polynomial is of the form $p(x) = x^k$.

Problem 3: We're looking for matrices with the same minimal and characteristic polynomials, but distinct invariant factos. One approach is to take $$ p_1(x) = x, \quad p_2(x) = x, \quad p_3(x) = x^2\\ q_1(x) = x^2, \quad q_2(x) = x^2. $$

Problem 4: This requires some degree of trial and error. One often fruitful approach is to try matrices whose entries are $0$ and $1$. JW's example takes advantage of the fact that a matrix and its transpose will be similar, which means that if one is nilpotent then so is the other.

Problem 5: It helps to consider nilpotent operators. In fact, it turns out that $V = \operatorname{range}(T) \oplus \ker(T)$ holds if and only if $T$ and $T^2$ are of the same rank.

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  • $\begingroup$ Rational canonical form is something I wasn't aware of. A nice and useful answer anyways though (+1). Thanks. $\endgroup$ Commented Sep 15, 2021 at 13:04
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Constructing counterexamples becomes easier, when you bring matrices to their Jordan canonical form (over $\mathbb{C}$). Let $J_n(\lambda)$ be the Jordan block of size $n\times n$ with eigenvalue $\lambda$ (lets say lower triangular).

  1. False, consider $\mathrm{diag}(0,0,1)$ and $\mathrm{diag}(0,1,1)$.
  2. False, consider $0_{3\times 3}$ and $J_3(0)$.
  3. False, consider $\mathrm{diag}(J_2(0),J_2(0))$ and $\mathrm{diag}(J_2(0),0,0)$.
  4. False, example by J.W.Tanner in comments: $J_2(0)$ and $J_2(0)^t$.
  5. False, consider $T=J_2(0)$.

Note that if matrices are not similar over $\mathbb{C}$, then they are not similar over smaller fields. All my examples make sense even over finite fields.

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  • $\begingroup$ This is something that I also felt when I was doing some google searches last night. But, the thing is we haven't learnt Jordan forms as yet (I guess, that's the trick our professor pulled out). So, this maybe a good answer (+1), but I need a month or so to appreciate it :( $\endgroup$ Commented Sep 14, 2021 at 6:36
  • $\begingroup$ Wow, this is quite surprising to hear. Without Jordan I would have faced similar difficulties as you did. It almost sounds like the professor forgot that Jordan was not covered yet. $\endgroup$ Commented Sep 14, 2021 at 7:44
  • $\begingroup$ Based on previous questions, the asker seems to be familiar with rational canonical form, which is covered in Hoffmann and Kunze before Jordan form. Similar tricks can be applied using this canonical form as a guide. $\endgroup$ Commented Sep 14, 2021 at 19:24
  • $\begingroup$ @BenGrossmann unfortunately no :( Our syllabus was kind of from chapter 6.3 to 6.8, that is from Annihilating polynomials to Primary Decomposition Theorem. $\endgroup$ Commented Sep 15, 2021 at 13:19

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