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Operations on matrices

Three operations turn the set of matrices into an arithmetic system: addition, scalar multiplication, and matrix multiplication. Addition and scalar multiplication are easy: combine entry by entry. Matrix multiplication is more involved , and its rules surprise students at first. This subtopic walks through each operation carefully and lists the algebraic laws they obey.

Addition

If A=(aij)A = (a_{ij}) and B=(bij)B = (b_{ij}) have the same order m×nm \times n, their sum is defined entry-wise:

(A+B)ij=aij+bij.(A + B)_{ij} = a_{ij} + b_{ij}.

So A+BA + B also has order m×nm \times n. Matrix addition is commutative and associative, has identity OO (the zero matrix), and every AA has inverse A-A.

If AA and BB have different orders, A+BA + B is undefined.

Scalar multiplication

For a scalar λR\lambda \in \mathbb{R} and matrix A=(aij)A = (a_{ij}), the product λA\lambda A is defined by multiplying every entry by λ\lambda:

(λA)ij=λaij.(\lambda A)_{ij} = \lambda \cdot a_{ij}.

So λA\lambda A has the same order as AA.

Properties: λ(A+B)=λA+λB\lambda(A + B) = \lambda A + \lambda B, (λ+μ)A=λA+μA(\lambda + \mu) A = \lambda A + \mu A, λ(μA)=(λμ)A\lambda(\mu A) = (\lambda \mu) A, 1A=A1 \cdot A = A.

Matrix multiplication

For AA of order m×pm \times p and BB of order p×np \times n, the product ABAB is the m×nm \times n matrix with

(AB)ij=k=1paikbkj.(AB)_{ij} = \sum_{k = 1}^{p} a_{ik} b_{kj}.

The number of columns of AA must equal the number of rows of BB. Otherwise ABAB is undefined.

In words: to compute the (i,j)(i, j) entry of ABAB, take the ii-th row of AA, the jj-th column of BB, multiply corresponding entries and add.

Worked computation

Let A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} and B=(5678)B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}.

(AB)11=15+27=19(AB)_{11} = 1 \cdot 5 + 2 \cdot 7 = 19. (AB)12=16+28=22(AB)_{12} = 1 \cdot 6 + 2 \cdot 8 = 22. (AB)21=35+47=43(AB)_{21} = 3 \cdot 5 + 4 \cdot 7 = 43. (AB)22=36+48=50(AB)_{22} = 3 \cdot 6 + 4 \cdot 8 = 50.

So AB=(19224350)AB = \begin{pmatrix} 19 & 22 \\ 43 & 50 \end{pmatrix}. Compute BABA the same way: BA=(23343146)BA = \begin{pmatrix} 23 & 34 \\ 31 & 46 \end{pmatrix}. Note ABBAAB \neq BA.

Algebraic laws

Matrix addition and multiplication satisfy:

  • Associativity of addition: (A+B)+C=A+(B+C)(A + B) + C = A + (B + C).
  • Commutativity of addition: A+B=B+AA + B = B + A.
  • Zero matrix: A+O=AA + O = A.
  • Associativity of multiplication: A(BC)=(AB)CA(BC) = (AB)C whenever all products are defined.
  • Distributivity: A(B+C)=AB+ACA(B + C) = AB + AC, (A+B)C=AC+BC(A + B)C = AC + BC.
  • Identity matrix: AI=IA=AAI = IA = A for square AA of the right order.

Non-commutativity: in general ABBAAB \neq BA.

Zero divisors exist: ABAB can be OO even if A,BOA, B \neq O. Example: A=(1000)A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, B=(0001)B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, then AB=OAB = O. This makes cancellation laws fail.

Powers of a square matrix

For a square AA, A2=AAA^2 = AA, A3=AA2A^3 = A \cdot A^2, and so on. A0=IA^0 = I by convention.

A matrix AA is called

  • idempotent if A2=AA^2 = A,
  • involutory if A2=IA^2 = I,
  • nilpotent if Ak=OA^k = O for some positive integer kk.

Worked examples

Example 1. Compute A+BA + B for A=(1234),B=(1102)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}, B = \begin{pmatrix} -1 & 1 \\ 0 & 2 \end{pmatrix}.

(0336)\begin{pmatrix} 0 & 3 \\ 3 & 6 \end{pmatrix}.

Example 2. Compute 3A2B3A - 2B for the same A,BA, B.

3A=(36912),2B=(2204)3A = \begin{pmatrix} 3 & 6 \\ 9 & 12 \end{pmatrix}, 2B = \begin{pmatrix} -2 & 2 \\ 0 & 4 \end{pmatrix}. Difference: (5498)\begin{pmatrix} 5 & 4 \\ 9 & 8 \end{pmatrix}.

Example 3. Multiply A=(123)A = \begin{pmatrix} 1 & 2 & 3 \end{pmatrix} and B=(456)B = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}.

ABAB is 1×11 \times 1: 14+25+36=321 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 = 32. So AB=(32)AB = (32). And BABA is 3×33 \times 3: BA=(48125101561218)BA = \begin{pmatrix} 4 & 8 & 12 \\ 5 & 10 & 15 \\ 6 & 12 & 18 \end{pmatrix}.

Example 4. Show A=(2312)A = \begin{pmatrix} 2 & -3 \\ 1 & -2 \end{pmatrix} is involutory.

A2=(2(2)+(3)(1)2(3)+(3)(2)1(2)+(2)(1)1(3)+(2)(2))=(1001)=IA^2 = \begin{pmatrix} 2(2) + (-3)(1) & 2(-3) + (-3)(-2) \\ 1(2) + (-2)(1) & 1(-3) + (-2)(-2) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I.

Example 5. Show A=(0100)A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} is nilpotent.

A2=(0000)=OA^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O. So AA is nilpotent of index 22.

Example 6. Let A=(1101)A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}. Show An=(1n01)A^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} by induction.

Base: A1=AA^1 = A has (1,2)(1, 2)-entry equal to 11. Step: An+1=AAn=(1101)(1n01)=(1n+101)A^{n+1} = A \cdot A^n = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & n + 1 \\ 0 & 1 \end{pmatrix}. \blacksquare

Try it yourself

  1. Compute A+BA + B for A=(12),B=(34)A = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, B = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.
  2. Compute 2A2A for A=(0110)A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
  3. Multiply (1234)(5006)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ 0 & 6 \end{pmatrix}.
  4. Multiply (1234)(11)\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 \\ -1 \end{pmatrix}.
  5. Find A2A^2 for A=(2003)A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.
  6. Find A2A^2 for A=(0110)A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
  7. Verify: A(B+C)=AB+ACA(B + C) = AB + AC for matrices A=(1,1),B=(1),C=(2)A = (1, 1), B = (1), C = (2). (Choose orders.)
  8. Find a 2×22 \times 2 matrix AOA \ne O with A2=OA^2 = O.
  9. Find a 2×22 \times 2 matrix AIA \ne I with A2=IA^2 = I.
  10. Find A10A^{10} for A=(1101)A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.
  11. Show that the sum of two symmetric matrices is symmetric.
  12. Find ABAB and BABA for A=(1000)A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} and B=(0100)B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}.
  13. Show A2=3AA^2 = 3A for A=(3000)A = \begin{pmatrix} 3 & 0 \\ 0 & 0 \end{pmatrix}.
  14. Multiply (cosθsinθsinθcosθ)(cosϕsinϕsinϕcosϕ)\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} \cos\phi & -\sin\phi \\ \sin\phi & \cos\phi \end{pmatrix} and identify the result.

Pitfalls / Tricks

  • Matrix multiplication is not commutative: always check the order before declaring AB=BAAB = BA.
  • Always confirm dimensions before multiplying. A2×3A_{2 \times 3} times B2×3B_{2 \times 3} is undefined.
  • AB=OAB = O does not imply A=OA = O or B=OB = O. There exist zero divisors.
  • Powers of a matrix can be computed efficiently using induction (as in Example 6) or eigenvalues.
  • The rotation matrix in problem 14 satisfies R(θ)R(ϕ)=R(θ+ϕ)R(\theta) R(\phi) = R(\theta + \phi) , a useful structural identity.

Next we look at the transpose and the symmetric/skew-symmetric decomposition.

Practice quiz

Quick check on this topic.

Quiz
Quick check : Operations on matrices
6 questions · pick the best answer
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