$2\times2$ matrices are not big enough

Question

Olga Tausky-Todd had once said that

"If an assertion about matrices is false, there is usually a 2x2 matrix that reveals this."

There are, however, assertions about matrices that are true for $2\times2$ matrices but not for the larger ones. I came across one nice little example yesterday. Actually, every student who has studied first-year linear algebra should know that there are even assertions that are true for $3\times3$ matrices, but false for larger ones --- the rule of Sarrus is one obvious example; a question I answered last year provides another.

So, here is my question. What is your favourite assertion that is true for small matrices but not for larger ones? Here, $1\times1$ matrices are ignored because they form special cases too easily (otherwise, Tausky-Todd would have not made the above comment). The assertions are preferrably simple enough to understand, but their disproofs for larger matrices can be advanced or difficult.

Answer 1

Any two rotation matrices commute.

Answer 2

I like this one: two matrices are similar (conjugate) if and only if they have the same minimal and characteristic polynomials and the same dimensions of eigenspaces corresponding to the same eigenvalue. This statement is true for all $n\times n$ matrices with $n\leq6$, but is false for $n\geq7$.

Answer 3

A matrix is called doubly stochastic if all its entries are nonnegative and each row and column sums to 1. A matrix is called orthostochastic if it is the entry-wise square of some orthogonal matrix. In general, it is easy to see that every orthostochastic matrix is doubly stochastic.

For $2\times 2$ matrices, every doubly stochastic matrix is orthostochastic, but for $3\times 3$ matrices (and hence anything larger) we have the counterexample $$ \frac{1}{2}\left(\begin{matrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix}\right)$$

Answer 4

The irreducible representations of $GL_n(\mathbb F_q)$ are well understood for $n=1, 2$, but there is still a lot that isn't known for $n \geq 3$.

Answer 5

Edit: I found more examples from questions/answers on this site. Unless otherwise specified, the matrix we concern is $n\times n$.

1. Let $M=\pmatrix{A&B\\ C&D}$, where the subblocks $A,B,C,D$ are real square matrices. Then $\det(M)=\det(AD-BC)$. This is true when $M$ is $2\times2$, but in general false for larger matrices. But one may argue that this is not a valid example, because the statement is true only when the subblocks are $1\times1$ matrices.
2. If $A$ is a singular matrix over a field, then every positive integer power of $A$ is a scalar multiple of $A$. This is true for $n=2$ but not for larger $n$s.
3. If $A$ is a real symmetric matrix with all diagonal entries equal to $1$ and all off-diagonal entries lying inside $[-1,1]$, then $\det(A)\le1$. This is true for $n=2,3$ and false for $n=5$. It's probably true when $n=4$ but I haven't investigated the problem any further.
4. Let ${\cal B}$ denotes the set of all 0-1 matrices and let $M=\max_{A\in \cal B}\det A$. For every $d\in\{0,1,\ldots,M\}$, there exists $A\in{\cal B}$ such that $\det(A)=d$. This is true up to $n=6$, false for $n=7$ and there is no known result for $n\ge22$.
5. (See Omran Kouba's answer to a related question.) If $X,Y\in M_n(\mathbb R)$, then $\det(X^2+Y^2)\ge\det(XY-YX)$. This is true for $n=2$ but not for larger $n$s. By the way, the analogous inequality $\det(X^TX+Y^TY)\ge\det(X^TY-Y^TX)$ does hold for all $n$.

Answer 6

Here's a false statement I ran into recently that requires a $3 \times 3$ counterexample: "if $A$ is normal, then so is every principal submatrix of $A$". It is in a sense "tempting" to believe this because the analogous statement applies to Hermitian and skew-Hermitian matrices.

Simplest counterexample: $$ A = \pmatrix{0&1&0\\0&0&1\\1&0&0} $$

Answer 7

Referring to the question Can commuting matrices $X,Y$ always be written as polynomials of some matrix $A$? it can be said that even the stronger property "for two commuting matrices one of them can always be written as a polynomial in the other" holds for $2\times2$ matrices (in fact, unless one is a scalar multiple of the identity both can be written as a polynomial in the other one), but not for $3\times3~$matrices. The weaker property of the question referred to (which allows a third matrix to be used to express both commuting matrices) also turns out to have counterexamples from size$~3\times3$ on, but they are sufficiently rare that I initially thought that weaker property would be generally valid.

Answer 8

Only for $2\times2$ matrices $A$ it's true that $\mathrm{adj}(A)=\mathrm{trace}(A)I-A$.

Answer 9

Recently, I come up with the following conjecture, which is proved to be true for small matrices and false for large matrices.

Conjecture. Let $E \in \mathbb{R}^{d\times d}$ be the matrix of ones and $M \in \mathbb{R}^{d\times d}$ be a bounded matrix representing a metric. That is, $M$ satisfies the following conditions:

• $1 \geq M_{ij} \geq 0\ $ for $\ 1 \leq i, j \leq d$
• $M_{ij} = 0\ \Leftrightarrow\ i = j$
• $M_{ij} = M_{ji}$
• $M_{ij} + M_{jk} \geq M_{ik}$ for $1 \leq i, j, k \leq d$

Then the matrix $A = E - M$ is positive semi-definite.

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When $d = 2$, $M$ and $A$ respectively have the form $$ M = \begin{bmatrix} 0 & a \\ a & 0 \end{bmatrix},\quad A = \begin{bmatrix} 1 & 1 - a \\ 1 - a & 1 \end{bmatrix} $$ with $0 < a \leq 1$. For any $x \in \mathbb{R}^2$, $$ x^TAx = x_1^2 + x_2^2 + 2(1-a)x_1x_2 = (1-a)(x_1 + x_2)^2 + ax_1^2 + ax_2^2 \geq 0 $$ Thus the conjecture is true for $d = 2$.

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When $d = 3$, $M$ and $A$ respectively have the form $$ M = \begin{bmatrix} 0 & a & b \\ a & 0 & c \\ b & c & 0 \end{bmatrix},\quad A = \begin{bmatrix} 1 & 1 - a & 1 - b \\ 1 - a & 1 & 1 - c \\ 1 - b & 1-c & 1 \end{bmatrix} $$ with $0 < a, b, c \leq 1$. Without loss of generality, we assume $c \geq a$ and $c \geq b$. For $\forall x \in \mathbb{R}^3$, \begin{align} x^TAx\ &=\ x_1^2 + x_2^2 + x_3^2 + 2(1-a)x_1x_2 + 2(1-b)x_1x_3 + 2(1-c)x_2x_3 \\ &=\ (1-c)(x_1 + x_2 + x_3)^2 + cx_1^2 + cx_2^2 + cx_3^2 + 2(c-a)x_1x_2 + 2(c-b)x_1x_3 \\ &=\ (1 - c)(x_1 + x_2 + x_3)^2 + (c-a)(x_1 + x_2)^2 + ax_1^2 + ax_2^2 + cx_3^2 + 2(c-b)x_1x_3 \\ &=\ (1 - c)(x_1 + x_2 + x_3)^2 + (c-a)(x_1 + x_2)^2 + (c-b)(x_1 + x_3)^2 \\ &\ \ \quad \quad + (\color{red}{a + b - c})x_1^2 + ax_2^2 + bx_3^2 \\ &\geq\ 0 \end{align} Note that $a + b - c \geq 0$ by triangle inequality. Therefore, the conjecture holds at $d = 3$.

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When $d = 4$, the proof becomes harder. Using a computer, I enumerate all metric matrices whose non-diagonal elements have values from $\{\frac{1}{1000}, \frac{2}{1000}, \cdots, \frac{1000}{1000}\}$. The result shows that all these matrices are positive semi-definite. So the conjecture seems to be true for $d = 4$.

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For $d = 5$, I find the following counterexample whose eigenvalues are $2.428$, $-0.089$, $0.5$, $1.161$ and $1$. $$ M = \begin{bmatrix} 0 & 0.5 & 0.5 & 0.5 & 1 \\ 0.5 & 0 & 0.5 & 0.5 & 1 \\ 0.5 & 0.5 & 0 & 1 & 0.5 \\ 0.5 & 0.5 & 1 & 0 & 0.5 \\ 1 & 1 & 0.5 & 0.5 & 0 \end{bmatrix},\quad A = \begin{bmatrix} 1 & 0.5 & 0.5 & 0.5 & 0 \\ 0.5 & 1 & 0.5 & 0.5 & 0 \\ 0.5 & 0.5 & 1 & 0 & 0.5 \\ 0.5 & 0.5 & 0 & 1 & 0.5 \\ 0 & 0 & 0.5 & 0.5 & 1 \end{bmatrix} $$ So the conjecture does hold in general for $d \geq 5$.

Answer 10

Interesting although quite elementary is property for matrices made from consecutive integers numbers (or more generally from values of arithmetic progression) where rows make an arithmetic progression.

Only $2 \times 2$ matrices made from consecutive integers numbers are non-singular, matrices of higher dimension are singular.

For example matrix $\begin{bmatrix} 1 & 2\\ 3 & 4 \\ \end{bmatrix}$ is non-singular,
but matrices like $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, $\begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{bmatrix}$, $\dots$ are singular.

This can be applied for easy generating of singular matrices if needed.

Answer 11

The statement:

Let $M\in\Bbb R^{n\times n}$ be symmetric. If $\det(M)>0$ and $\operatorname{Tr}(M)>0$, then $M$ is positive definite.

Is true for $n=2$ (note that $\det(M)=\lambda_1\lambda_2$ and $\operatorname{Tr}(M)>0=\lambda_1+\lambda_2$, where $\lambda_i$ are the eigenvalues of $M$) but wrong for $n>2$ (e.g. if $M$ is a diagonal matrix where $2$ diagonal elements are small and strictly negative).

Note that a similar statement can be maid about positive semi-definite matrices.

Answer 12

For matrices of order 2, we have a simple formula for the determinant of the sum of 2 matrices:

If $A,B$ are matrices of order 2 then $$\det(A+B)=\det(A)+\det(B)+\operatorname{tr}(A\operatorname{adj}(B)),$$ where $\operatorname{adj}(B)$ is the classical adjoint of $B$.

Answer 13

Matrices of size $2 \times 2$ are fully understood in terms of the number of elementary multiplications required to multiply two matrices. This is not the case for even slightly bigger matrices. There is a gap between lower and upper bound for the number of multiplications required.

The product of two $2 \times 2$ matrices (denoted as case $<2,2,2>$) can be computed in $7$ elementary multiplications. Winograd showed that seven is actually the minimum number or multiplications required.

Bläser obtained a lower bound of $19$ multiplications for the product of $3 \times 3$ matrices. However, the best known algorithm for the non-commutative product needs $23$ multiplications. The solution of Laderman has not been improved since 1976. Recent work confirmed, that there are many other algorithms with $23$ multiplications. The case $<2,2,3>$ (equivalent to $<2,3,2>$ and $<3,2,2>$) currently has a lower bound of $10$ and a known solution with $11$ multiplications. Hopcroft and Kerr published an algorithm for $<2,3,3>$ which requires $15$ multiplications, while the lower bound is $14$.

Answer 14

Preliminary: let $A$ be real square matrix.

Definition: $A$ is called (positive) stable if $\sigma(A)\subset\mathbb{C}_+$.

Theorem (Lyapunov): $A$ is stable iff $\exists$ pos.-def. $H\colon A^TH+HA$ is pos.-def.

Definition: $A$ is called diagonally stable if $\exists$ pos.-def. diagonal $D\colon A^TD+DA$ is pos.-def.

Definition: $A$ is called $P$-matrix if all principal minors are positive.

The main contribution: the statement

$A$ is diagonally stable $\Leftrightarrow$ $A$ is a $P$-matrix

is true only for $1\times 1$ and $2\times 2$ matrices. For $3\times 3$ and higher, $\Leftarrow$, in general, is false, however there are some special cases (acyclic matrices, $Z$-matrices) when it is true.

P.S. It is still an open problem to characterize diagonal stability in general. The case $3\times 3$ is known (not sure about $4\times 4$).

Answer 15

Claim: There is an algebraic procedure that terminates in finite steps to obtain the eigenvalues of a general $n\times n$ matrix.

This is true for $n \leq 4$ but fails for $n \geq 5$, thanks to Abel-Ruffini.

Answer 16

Let $n\geq 2$ and $A,B\in M_n(\mathbb{C})$ be s.t. $A^2+B^2+2AB=0_n$.

Then $AB=BA$ $\Leftrightarrow n=2$.

For $\Leftarrow$: see my post in What can be said if $A^2+B^2+2AB=0$ for some real $2\times2$ matrices $A$ and $B$?

For $\Rightarrow$: when $n\geq 3$, there are $A,B$ s.t. $A^2=0,B^2=0,AB=0,BA\not= 0$.

Answer 17

Computing the determinant (over an arbitrary ring, not just fields) is easy.

Sure, trivial for $2\times 2$ and $3\times 3$, but many young students don't realize that the naive approach will be unbearably slow (by hand) already for $5\times 5$.

In case anyone objects to this being a rigorous mathematical statement, you can phrase it as ".. Is not polynomial-time".

Answer 18

Let $A \in M_2(\mathbb R)$, $A$ is invertible or $A$ is nilpotent or $A$ is diagonalizable

Proof: Let $\chi_A$ be its characteristic polynomial, then $\chi_A = X^2-$ tr$(A)X + \det A$.

If $\det A \neq 0$ then $A$ is invertible.

Else if tr$(A) = 0$, then by Cayley-Hamilton theorem, $A^2 = 0$ i.e. $A$ is nilpotent.

Else $X^2-$ tr$(A)X = X(X- $tr $(A))$ has distinct simple roots and vanishes $A$, hence $A$ is diagonalizable.