Rules That Fail for Matrices

The hook

Matrices can be added and multiplied, so it is tempting to treat them like numbers. This question tests that temptation with four “rules” that are true for numbers, and asks which survive for matrices. Most of them fail, and the way they fail is instructive. Matrix multiplication is not commutative; there are non-zero matrices whose product is zero; and a quadratic equation for matrices can have infinitely many solutions.

These failures are not defects. They are the reason matrices are useful: they describe transformations of space, and doing two transformations in different orders genuinely gives different results. In 1925 Werner Heisenberg discovered that the quantities of quantum mechanics behave like matrices rather than numbers, and that the failure of \(\mathbf{AB} = \mathbf{BA}\) is the mathematical form of the uncertainty principle.

Before you start

If \(ab = 0\) and \(a \neq 0\), divide by \(a\) to get \(b = 0\). The argument needs every non-zero number to have an inverse. Non-zero matrices need not have inverses (those with determinant zero do not), so the argument breaks down. Keep the geometric picture in mind too: a matrix is a transformation of the plane, and a product \(\mathbf{AB}\) means “do \(\mathbf{B}\), then \(\mathbf{A}\)”.

Working it through

Stage 1 Can \(\mathbf{AB} = \mathbf{0}\) but \(\mathbf{BA} \neq \mathbf{0}\)?

Let \(\mathbf{A} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\), which projects onto the \(x\)-axis (killing the \(y\)-axis), and \(\mathbf{B} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\), which sends \((x, y)\) to \((0, x)\), into the \(y\)-axis. Then \[\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \qquad \mathbf{BA} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \neq \mathbf{0}.\] The assertion is false (Figure 7.1.1).■

Stage 2 The difference of two squares

Expanding carefully, keeping the order of every product: \[(\mathbf{A} - \mathbf{B})(\mathbf{A} + \mathbf{B}) = \mathbf{A}^2 + \mathbf{AB} - \mathbf{BA} - \mathbf{B}^2.\] This equals \(\mathbf{A}^2 - \mathbf{B}^2\) exactly when \(\mathbf{AB} = \mathbf{BA}\). With the matrices of Stage 1, \(\mathbf{AB} - \mathbf{BA} = -\mathbf{BA} \neq \mathbf{0}\), so the assertion is false.■

Stage 3 Non-zero solutions of \(\mathbf{AX} = \mathbf{0}\)

If \(\det\mathbf{A} \neq 0\), then \(\mathbf{A}^{-1}\) exists and \(\mathbf{X} = \mathbf{A}^{-1}\mathbf{AX} = \mathbf{0}\): no non-zero solution.

If \(\det\mathbf{A} = 0\), write \(\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) with \(ad = bc\). The vectors \(\begin{pmatrix} b \\ -a \end{pmatrix}\) and \(\begin{pmatrix} d \\ -c \end{pmatrix}\) satisfy \(\mathbf{A}\begin{pmatrix} b \\ -a \end{pmatrix} = \begin{pmatrix} ab - ba \\ cb - da \end{pmatrix} = \mathbf{0}\), and similarly for the other. Since \(\mathbf{A} \neq \mathbf{0}\), at least one of them is non-zero; call it \(\mathbf{v}\). Then the matrix \(\mathbf{X}\) whose two columns are both \(\mathbf{v}\) is non-zero and \(\mathbf{AX} = \mathbf{0}\). So the assertion is true.■

Stage 4 How many square roots?

Take \(\mathbf{A} = -2\mathbf{I}\) and \(\mathbf{B} = \mathbf{I}\), both non-zero. The equation becomes \((\mathbf{X} - \mathbf{I})^2 = \mathbf{0}\). Any matrix \(\mathbf{N}\) with \(\mathbf{N}^2 = \mathbf{0}\) gives a solution \(\mathbf{X} = \mathbf{I} + \mathbf{N}\), and there are infinitely many such \(\mathbf{N}\): \[\mathbf{N} = \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix}, \qquad \mathbf{N}^2 = \mathbf{0} \text{ for every real } t.\] So \(\mathbf{X} = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\) is a solution for every \(t\), and the assertion is false.■

Looking back

• Order matters. Every expansion must keep products in order; \(\mathbf{AB}\) and \(\mathbf{BA}\) are usually different.

• Zero divisors. Non-zero matrices can multiply to zero, which breaks the argument behind “a quadratic has at most two roots”.

• Think of transformations. Each counterexample was easy to find by picturing what the matrices do to the plane.

What’s really going on

Infinitely many square roots

Stage 4 can be made even more striking. The equation \(\mathbf{X}^2 = \mathbf{I}\) has the obvious solutions \(\pm\mathbf{I}\), but also every reflection of the plane in a line through the origin, \(\mathbf{X} = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}\), since reflecting twice does nothing. That is a whole circle of square roots of \(\mathbf{I}\). A polynomial of degree \(n\) with number coefficients has at most \(n\) roots because numbers have no zero divisors; matrices do, and the count fails.

Non-commutativity in physics

In classical mechanics, position \(x\) and momentum \(p\) are numbers and \(xp = px\). In quantum mechanics they become operators (infinite matrices) with \(xp - px = \mathrm{i}\hbar\), where \(\hbar\) is Planck’s constant divided by \(2\pi\). This single non-commutation relation implies Heisenberg’s uncertainty principle: position and momentum cannot both be known exactly. When Heisenberg discovered the rules of quantum mechanics in 1925 he did not know what matrices were; Max Born recognised that the strange multiplication rule Heisenberg had invented was matrix multiplication.

The thinking skill

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Further reading

• Gilbert Strang, Introduction to Linear Algebra (Wellesley-Cambridge). The most popular first linear algebra text, with an emphasis on understanding.

• Sheldon Axler, Linear Algebra Done Right (Springer; free online). A more abstract approach, centred on linear maps rather than matrices.

• Jim Baggott, The Quantum Story (Oxford University Press). The history of quantum mechanics, including Heisenberg’s matrix mechanics.