Zero matrix and Unit matrix

Definition. The zero matrix $O$ is a matrix in which every entry is $0$. There is one of each shape — a $2\times 2$ zero matrix, a $2\times 3$ zero matrix, and so on:

$O = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \qquad O = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

Why it matters. In ordinary number arithmetic, $0$ is the number you can add without changing anything: $5 + 0 = 5$. The zero matrix does exactly the same job for matrices. For any matrix $A$ of the same order:

$A + O = O + A = A$

and subtracting a matrix from itself always lands you on $O$:

$A - A = O$

Because of this, $O$ is called the additive identity. It is the matrix analogue of the number $0$.

A warning that pays off later. Although $O$ acts like $0$, it does not share all of zero's powers. With ordinary numbers, $ab = 0$ forces $a = 0$ or $b = 0$. For matrices this fails: you can multiply two non-zero matrices and still get $O$ (see the worked example). That single fact is the source of half the slips in matrix algebra — you can never "cancel" a matrix the way you cancel a number.

Cambridge tip. Always write the zero matrix as $O$ (a matrix), never as the scalar $0$, in a matrix equation. Examiners want $A - A = O$, not $A - A = 0$.

Definition. The identity matrix (also called the unit matrix) $I$ is a square matrix with $1$s on the leading diagonal — the diagonal running from the top-left corner to the bottom-right — and $0$s everywhere else:

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \;(2\times 2), \qquad I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \;(3\times 3)$

Why it matters. Just as $1$ is the number you can multiply by without changing anything ($7 \times 1 = 7$), the identity matrix is the matrix you can multiply by without changing anything. For any conformable matrix $A$:

$AI = IA = A$

So $I$ is the multiplicative identity. Notice the order does not matter for $I$ — it works on the left and on the right — which is special, because matrix multiplication is usually not commutative.

Why does it work? Each column of $I$ is "select this one column, leave it alone." When you multiply $A$ by $I$, the row-by-column sums pick out each entry of $A$ unchanged (the $0$s kill every other term). Verifying this once, by hand, fixes the idea permanently — see the worked example $AI = A$.

Cambridge tip. The identity must be square to make sense — there is no rectangular identity. When a question writes "$kI$", that means $k$ on the diagonal and $0$ off it, e.g. $3I = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$, not a matrix of all $3$s.

Multiplying by the zero matrix. Just as $7 \times 0 = 0$, multiplying any matrix by $O$ gives $O$ (whenever the multiplication is defined):

$AO = OA = O$

Every row-by-column sum is a sum of products in which one factor is always $0$, so every entry of the result is $0$.

The dangerous difference from numbers. With real numbers we are used to: $ab = 0 \;\Rightarrow\; a = 0 \ \text{or}\ b = 0.$

This is FALSE for matrices. Two perfectly non-zero matrices can multiply to give the zero matrix. For example:

$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$

Matrices like these are called zero divisors. The practical consequence is the rule you must never forget:

You cannot cancel matrices. From $AB = AC$ you may not deduce $B = C$ (that would need $A$ to be invertible, a later topic). From $AB = O$ you may not deduce $A = O$ or $B = O$.

Cambridge tip. If a question says "deduce" or "explain why", and a matrix product comes out as $O$, the marks are usually for stating that neither factor is $O$ — i.e. recognising the zero-divisor phenomenon — not just for the arithmetic.

What a power means. Exactly as $x^3 = x \cdot x \cdot x$ for numbers, a power of a matrix is repeated multiplication:

$A^2 = A\,A, \qquad A^3 = A^2 A = A\,A\,A, \qquad A^{n} = A^{n-1} A.$

The square requirement. For $A^2 = A\,A$ to even be defined, the number of columns of $A$ must equal the number of rows of $A$ — that is, $A$ must be a square matrix ($n \times n$). A $2 \times 3$ matrix has no powers at all; you simply cannot multiply it by itself.

How to compute a power. Build up one step at a time — find $A^2$, then $A^3 = A^2 A$, and so on. Because matrix multiplication is associative, the grouping does not matter: $A^3 = (A\,A)A = A(A\,A)$, so $A^2 A = A A^2$.

Two traps to avoid.

• Do not square entry by entry. $A^2$ is the product $A\,A$, not "each number squared". For $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$, the correct $A^2 = \begin{pmatrix} 1 & 8 \\ 0 & 9 \end{pmatrix}$ — not $\begin{pmatrix} 1 & 4 \\ 0 & 9 \end{pmatrix}$.
• The convention $A^0 = I$. Just as $x^0 = 1$, the zeroth power of a square matrix is defined to be the identity: $A^0 = I$. This makes formulae like $A^n = \begin{pmatrix} 1 & 0 \\ 4n & 1 \end{pmatrix}$ work cleanly at $n = 0$.

Cambridge tip. When a question asks you to "find $A^3$" or "conjecture a formula for $A^n$", show the intermediate $A^2$ explicitly. The method marks are awarded for the working, and a clearly shown $A^2$ protects you if a later arithmetic slip creeps in.