Types of matrices

Every named matrix in the syllabus — row, column, null, diagonal, scalar, identity, triangular, symmetric, skew-symmetric, orthogonal, idempotent, involutory, nilpotent, periodic, Hermitian, unitary and more — each with its defining test and a solved check.

Module contents

Matrices get names from the shape they have or from an identity they satisfy. Both kinds appear in exams, and each name comes with a one-line test. Learn the test, not the picture — a question will ask you to prove a matrix is orthogonal, not to recognise it on sight.

The names below are grouped by what they depend on.

Group 1 — Names that depend only on shape

Rectangular matrix

Any matrix with $m \neq = n$ .

$[142536]_{2 \times 3}$

Square matrix

$m = n$ . Only square matrices have a determinant, an inverse, a trace, or eigenvalues, so most of this module lives here. A square matrix of order $n \times n$ is called a square matrix of order $n$ .

$[1324]_{2 \times 2}$

Row matrix (row vector)

Exactly one row: order $1 \times n$ .

$[5 - 2 07]_{1 \times 4}$

Column matrix (column vector)

Exactly one column: order $m \times 1$ .

$3 - 1 8_{3 \times 1}$

Null (zero) matrix — $O$

Every element is $0$ . It exists in every order, and $A + O = A$ , $A \cdot O = O$ .

$O_{2 \times 3} = [000000]$

A null matrix is the additive identity. It is not the multiplicative identity — that is $I$ . And unlike numbers, $A B = O$ does not force $A = O$ or $B = O$ .

Comparable matrices

Two matrices of the same order. Addition and subtraction need comparable matrices; multiplication does not.

Group 2 — Names for where the non-zero elements sit

For a square matrix $A = [a_{ij}]$ of order $n$ , the elements $a_{11}, a_{22}, \dots, a_{nn}$ form the principal (main) diagonal. Their sum is the trace:

$tr (A) = i = 1 \sum n a_{ii}$

Diagonal matrix

Square, and every element off the principal diagonal is zero: $a_{ij} = 0$ whenever $i \neq = j$ . Diagonal elements themselves may be anything, including zero.

$D = 200 0 - 5 0 007 = diag (2, - 5, 7)$

Scalar matrix

A diagonal matrix in which all the diagonal elements are equal: $a_{ij} = 0$ for $i \neq = j$ and $a_{ii} = k$ for all $i$ . Equivalently $A = k I$ .

$400040004 = 4 I_{3}$

Identity (unit) matrix — $I_{n}$

A scalar matrix with $k = 1$ : ones down the diagonal, zeros elsewhere. It is the multiplicative identity, $A I = I A = A$ .

$I_{3} = 100010001$

Formally $I = [δ_{ij}]$ where $δ_{ij}$ is the Kronecker delta: $δ_{ij} = 1$ if $i = j$ , else $0$ .

The nesting. Every identity matrix is scalar; every scalar matrix is diagonal; every diagonal matrix is both upper and lower triangular. The converses all fail.

Upper triangular matrix

Square with every element below the diagonal zero: $a_{ij} = 0$ for $i > j$ .

$100430 9 - 2 5$

Lower triangular matrix

Square with every element above the diagonal zero: $a_{ij} = 0$ for $i < j$ .

$162 03 - 7 005$

A matrix that is either is called triangular. For any triangular matrix the determinant is simply the product of the diagonal elements — a fact the Gauss elimination method exploits.

Sparse matrix

Most elements are zero. Not a formal exam definition, but a critical one in computing: a sparse matrix is stored by listing only its non-zero entries (as triples $(i, j, value)$ ), which turns an $n^{2}$ storage cost into something proportional to the number of non-zeros.

Sub-matrix

Any matrix obtained by deleting some rows and/or columns of $A$ . Deleting row $i$ and column $j$ of a square matrix leaves the sub-matrix whose determinant is the minor $M_{ij}$ .

Group 3 — Names defined by the transpose

The transpose $A^{T}$ (also written $A^{'}$ ) is obtained by turning rows into columns: if $A = [a_{ij}]_{m \times n}$ then $A^{T} = [a_{j i}]_{n \times m}$ .

$A = [142536] ⟹ A^{T} = 123456$

Symmetric matrix

$A^{T} = A i.e. a_{ij} = a_{j i}$

The matrix is a mirror image about the principal diagonal. Must be square.

$17 - 3 740 - 3 06$

Skew-symmetric matrix

$A^{T} = - A i.e. a_{ij} = - a_{j i}$

Setting $i = j$ gives $a_{ii} = - a_{ii}$ , so $2 a_{ii} = 0$ : every diagonal element of a skew-symmetric matrix is zero. That is the fastest test to apply.

$0 - 2 5 20 - 3 - 5 30$

Two standard results worth memorising:

Every square matrix splits uniquely as symmetric $+$ skew-symmetric:
$A = symmetric \frac{1}{2} (A + A^{T}) + skew-symmetric \frac{1}{2} (A - A^{T})$
The determinant of a skew-symmetric matrix of odd order is always $0$ .

Orthogonal matrix

$A A^{T} = A^{T} A = I equivalently A^{- 1} = A^{T}$

Its determinant must be $\pm 1$ . Rotation matrices are orthogonal, which is why graphics code can invert a rotation by transposing it — far cheaper than a general inverse.

$[cos θ sin θ - sin θ cos θ]$

Group 4 — Names defined by powers of the matrix

Idempotent matrix

$A^{2} = A$
Applying it twice does nothing more than applying it once — the algebra of a projection. $I$ and $O$ are both idempotent.

Involutory matrix

$A^{2} = I$
The matrix is its own inverse, $A^{- 1} = A$ . $I$ is involutory.

Nilpotent matrix

$A^{k} = O for some positive integer k$
The smallest such $k$ is the index of nilpotency. A nilpotent matrix always has $∣ A ∣ = 0$ .

Periodic matrix

$A^{k + 1} = A for some positive integer k$
The smallest such $k$ is the period. An idempotent matrix is exactly a periodic matrix of period $1$ .

Singular and non-singular

$∣ A ∣ = 0 \Rightarrow singular ∣ A ∣ \neq = 0 \Rightarrow non-singular$
Only non-singular matrices have an inverse. This is the single most consequential classification in the module.

Group 5 — Names for complex matrices

Let $A ˉ$ be the conjugate of $A$ (replace each entry by its complex conjugate), and let $A^{θ} = A ˉ^{T}$ be the transposed conjugate (also written $A^{*}$ or $A^{H}$ ).

Hermitian matrix

$A^{θ} = A i.e. a_{ij} = a_{j i}$
The diagonal elements must be real. The real-valued case of Hermitian is exactly symmetric.

$[3 2 + i 2 - i 5]$

Skew-Hermitian matrix

$A^{θ} = - A$
Diagonal elements must be zero or purely imaginary. The real case is exactly skew-symmetric.

$[2 i - 1 + i 1 + i 0]$

Unitary matrix

$A A^{θ} = A^{θ} A = I$
The complex counterpart of orthogonal.

Quick reference

Type	Test	Must be square?
Row	order $1 \times n$	No
Column	order $m \times 1$	No
Null $O$	all $a_{ij} = 0$	No
Diagonal	$a_{ij} = 0$ for $i \neq = j$	Yes
Scalar	diagonal with all $a_{ii} = k$	Yes
Identity $I$	diagonal with all $a_{ii} = 1$	Yes
Upper triangular	$a_{ij} = 0$ for $i > j$	Yes
Lower triangular	$a_{ij} = 0$ for $i < j$	Yes
Symmetric	$A^{T} = A$	Yes
Skew-symmetric	$A^{T} = - A$ , diagonal all $0$	Yes
Orthogonal	$A A^{T} = I$	Yes
Idempotent	$A^{2} = A$	Yes
Involutory	$A^{2} = I$	Yes
Nilpotent	$A^{k} = O$	Yes
Periodic	$A^{k + 1} = A$	Yes
Singular	$∣ A ∣ = 0$	Yes
Non-singular	$∣ A ∣ \neq = 0$	Yes
Hermitian	$A ˉ^{T} = A$	Yes
Skew-Hermitian	$A ˉ^{T} = - A$	Yes
Unitary	$A A ˉ^{T} = I$	Yes

Worked examples

5 solved

Every step is shown, in the order you would write it in an answer book.

Example 1

Classifying a matrix under every applicable name

Question

Name every type that $A = 500050005$ belongs to, and state its trace and determinant.

Shape
$3$ rows, $3$ columns $\Rightarrow$ square matrix of order 3.
Position of non-zeros
Every off-diagonal element is $0$ $\Rightarrow$ diagonal. All diagonal elements equal $5$ $\Rightarrow$ scalar, and $A = 5 I_{3}$ .

Since nothing lies below the diagonal and nothing lies above it, $A$ is also both upper triangular and lower triangular.
Transpose test
Transposing a diagonal matrix leaves it unchanged, so $A^{T} = A$ $\Rightarrow$ symmetric. It is not skew-symmetric, because a skew-symmetric matrix needs zero diagonal.
Determinant and singularity
For a triangular matrix, $∣ A ∣$ is the product of the diagonal:

$∣ A ∣ = 5 \times 5 \times 5 = 125 \neq = 0$

so $A$ is non-singular.
Trace
$tr (A) = 5 + 5 + 5 = 15$
Power tests
$A^{2} = 25 I \neq = A$ , so not idempotent. $A^{2} = 25 I \neq = I$ , so not involutory. No power is ever $O$ , so not nilpotent.

Answer

Square, diagonal, scalar, upper triangular, lower triangular, symmetric, non-singular. $tr (A) = 15$ , $∣ A ∣ = 125$ .

NoteOne matrix can carry many names at once. Exam questions asking to “identify the type” expect all applicable names.

Example 2

Splitting a matrix into symmetric and skew-symmetric parts

Question

Express $A = 271 43 - 2 - 6 50$ as the sum of a symmetric and a skew-symmetric matrix.

Write down the transpose
$A^{T} = 24 - 6 735 1 - 2 0$
Form $A + A^{T}$
$A + A^{T} = 411 - 5 1163 - 5 30$
Halve it — this is the symmetric part $P$
$P = \frac{1}{2} (A + A^{T}) = 2 \frac{11}{2} - \frac{5}{2} \frac{11}{2} 3 \frac{3}{2} - \frac{5}{2} \frac{3}{2} 0$

Check: $P^{T} = P$ ✓ (mirror about the diagonal).
Form $A - A^{T}$
$A - A^{T} = 037 - 3 0 - 7 - 7 70$
Halve it — this is the skew-symmetric part $Q$
$Q = \frac{1}{2} (A - A^{T}) = 0 \frac{3}{2} \frac{7}{2} - \frac{3}{2} 0 - \frac{7}{2} - \frac{7}{2} \frac{7}{2} 0$

Check: diagonal is all zero and $Q^{T} = - Q$ ✓
Verify the sum
$P + Q = 2 + 0 \frac{11}{2} + \frac{3}{2} - \frac{5}{2} + \frac{7}{2} \frac{11}{2} - \frac{3}{2} 3 + 0 \frac{3}{2} - \frac{7}{2} - \frac{5}{2} - \frac{7}{2} \frac{3}{2} + \frac{7}{2} 0 = 271 43 - 2 - 6 50 = A ✓$

Answer

$A = P + Q$ with $P = \frac{1}{2} (A + A^{T})$ symmetric and $Q = \frac{1}{2} (A - A^{T})$ skew-symmetric, as computed above.

NoteAlways finish with the P + Q = A check. It costs 30 seconds and catches every sign error.

Example 3

Proving a matrix is orthogonal

Question

Show that $A = \frac{1}{3} 12 - 2 212 2 - 2 - 1$ is orthogonal, and hence write down $A^{- 1}$ .

State what must be shown
$A$ is orthogonal $⟺ A A^{T} = I$ . Let $B = 3 A = 12 - 2 212 2 - 2 - 1$ , so $A = \frac{1}{3} B$ and

$A A^{T} = \frac{1}{3} B \cdot \frac{1}{3} B^{T} = \frac{1}{9} B B^{T}$

so it is enough to show $B B^{T} = 9 I$ .
Write $B^{T}$
$B^{T} = 122 21 - 2 - 2 2 - 1$
Compute $B B^{T}$ row by row
Row 1 of $B$ with the columns of $B^{T}$ (i.e. dotted against the rows of $B$ ):

$(1, 2, 2) \cdot (1, 2, 2) = 1 + 4 + 4 = 9$
$(1, 2, 2) \cdot (2, 1, - 2) = 2 + 2 - 4 = 0$
$(1, 2, 2) \cdot (- 2, 2, - 1) = - 2 + 4 - 2 = 0$

Row 2:
$(2, 1, - 2) \cdot (1, 2, 2) = 2 + 2 - 4 = 0 (2, 1, - 2) \cdot (2, 1, - 2) = 4 + 1 + 4 = 9$
$(2, 1, - 2) \cdot (- 2, 2, - 1) = - 4 + 2 + 2 = 0$

Row 3:
$(- 2, 2, - 1) \cdot (1, 2, 2) = - 2 + 4 - 2 = 0 (- 2, 2, - 1) \cdot (2, 1, - 2) = - 4 + 2 + 2 = 0$
$(- 2, 2, - 1) \cdot (- 2, 2, - 1) = 4 + 4 + 1 = 9$
Assemble
$B B^{T} = 900090009 = 9 I$

Therefore $A A^{T} = \frac{1}{9} \cdot 9 I = I$ , and $A$ is orthogonal.
Read off the inverse
For an orthogonal matrix $A^{- 1} = A^{T}$ , so

$A^{- 1} = \frac{1}{3} 122 21 - 2 - 2 2 - 1$

Answer

$A A^{T} = I$ , so $A$ is orthogonal and $A^{- 1} = A^{T} = \frac{1}{3} 122 21 - 2 - 2 2 - 1$

NoteNotice the shortcut: pulling the 1/3 out front turns fraction arithmetic into integer arithmetic. Do this whenever a common factor is visible.

Example 4

Idempotent, involutory or nilpotent?

Question

Classify each by computing the required power.
(a) $A = 2 - 1 1 - 2 3 - 2 - 4 4 - 3$
(b) $B = [412 - 1 - 3]$ — find $B^{2}$ and comment.
(c) $C = 15 - 2 12 - 1 36 - 3$

(a) Compute $A^{2}$
Row 1 of $A$ against each column:

$(2) (2) + (- 2) (- 1) + (- 4) (1) = 4 + 2 - 4 = 2$
$(2) (- 2) + (- 2) (3) + (- 4) (- 2) = - 4 - 6 + 8 = - 2$
$(2) (- 4) + (- 2) (4) + (- 4) (- 3) = - 8 - 8 + 12 = - 4$

Row 2:
$(- 1) (2) + (3) (- 1) + (4) (1) = - 2 - 3 + 4 = - 1$
$(- 1) (- 2) + (3) (3) + (4) (- 2) = 2 + 9 - 8 = 3$
$(- 1) (- 4) + (3) (4) + (4) (- 3) = 4 + 12 - 12 = 4$

Row 3:
$(1) (2) + (- 2) (- 1) + (- 3) (1) = 2 + 2 - 3 = 1$
$(1) (- 2) + (- 2) (3) + (- 3) (- 2) = - 2 - 6 + 6 = - 2$
$(1) (- 4) + (- 2) (4) + (- 3) (- 3) = - 4 - 8 + 9 = - 3$

$A^{2} = 2 - 1 1 - 2 3 - 2 - 4 4 - 3 = A$

$A$ is idempotent.
(b) Compute $B^{2}$
$B^{2} = [412 - 1 - 3] [412 - 1 - 3] = [16 - 12 48 - 36 - 4 + 3 - 12 + 9] = [412 - 1 - 3] = B$

$B$ is also idempotent (equivalently, periodic with period $1$ ).
(c) Compute $C^{2}$
Row 1: $(1) (1) + (1) (5) + (3) (- 2) = 1 + 5 - 6 = 0$ ; $(1) (1) + (1) (2) + (3) (- 1) = 1 + 2 - 3 = 0$ ; $(1) (3) + (1) (6) + (3) (- 3) = 3 + 6 - 9 = 0$

Row 2: $(5) (1) + (2) (5) + (6) (- 2) = 5 + 10 - 12 = 3$ ; $(5) (1) + (2) (2) + (6) (- 1) = 5 + 4 - 6 = 3$ ; $(5) (3) + (2) (6) + (6) (- 3) = 15 + 12 - 18 = 9$

Row 3: $(- 2) (1) + (- 1) (5) + (- 3) (- 2) = - 2 - 5 + 6 = - 1$ ; $(- 2) (1) + (- 1) (2) + (- 3) (- 1) = - 2 - 2 + 3 = - 1$ ; $(- 2) (3) + (- 1) (6) + (- 3) (- 3) = - 6 - 6 + 9 = - 3$

$C^{2} = 03 - 1 03 - 1 09 - 3$
(c) Go one power further
$C^{3} = C^{2} C = 03 - 1 03 - 1 09 - 3 15 - 2 12 - 1 36 - 3$

Row 2: $3 + 15 - 18 = 0$ ; $3 + 6 - 9 = 0$ ; $9 + 18 - 27 = 0$
Row 3: $- 1 - 5 + 6 = 0$ ; $- 1 - 2 + 3 = 0$ ; $- 3 - 6 + 9 = 0$

$C^{3} = O$

$C$ is nilpotent of index 3.

Answer

(a) idempotent; (b) idempotent; (c) nilpotent of index $3$ .

NoteIf A² is neither A nor I nor O, compute A³ before concluding anything — nilpotency often shows up only at the third power.

Example 5

Hermitian and skew-Hermitian

Question

Verify that $A = 3 2 + i - 1 - 3 i 2 - i 04 - 1 + 3 i 4 - 2$ is Hermitian, and that $i A$ would then be skew-Hermitian.

Take the conjugate $A ˉ$
Flip the sign of every imaginary part:

$A ˉ = 3 2 - i - 1 + 3 i 2 + i 04 - 1 - 3 i 4 - 2$
Transpose it to get $A^{θ} = A ˉ^{T}$
$A^{θ} = 3 2 + i - 1 - 3 i 2 - i 04 - 1 + 3 i 4 - 2$
Compare
$A^{θ} = A$ element for element, so $A$ is Hermitian.

Note the two structural signs of a Hermitian matrix, both present here: the diagonal $3, 0, - 2$ is entirely real, and each pair across the diagonal is a conjugate pair ( $2 - i$ against $2 + i$ ).
Now consider $i A$
Using $(k A)^{θ} = k ˉ A^{θ}$ with $k = i$ , so $k ˉ = - i$ :

$(i A)^{θ} = i ˉ A^{θ} = - i A = - (i A)$

which is exactly the definition of skew-Hermitian.

Answer

$A^{θ} = A$ , so $A$ is Hermitian; and $(i A)^{θ} = - (i A)$ , so $i A$ is skew-Hermitian.

NoteGeneral result: if A is Hermitian then iA is skew-Hermitian, and vice versa. Worth quoting directly in an exam.

Practice problems

8 with solutions

Work each one on paper first. The full solution — not just the answer — is one click away.

Problem 1Basic

State, with a one-line reason, whether each is true or false.
(a) Every scalar matrix is a diagonal matrix.
(b) Every diagonal matrix is a scalar matrix.
(c) Every identity matrix is a scalar matrix.
(d) A skew-symmetric matrix can have a non-zero diagonal element.
(e) A null matrix of order $2 \times 3$ is a square matrix.

Show solution

(a) True. A scalar matrix is defined as a diagonal matrix with all diagonal entries equal, so it satisfies the diagonal condition by definition.

(b) False. $diag (1, 2)$ is diagonal but its diagonal entries differ, so it is not scalar.

(c) True. $I = 1 \cdot I$ , a scalar matrix with $k = 1$ .

(d) False. $A^{T} = - A$ forces $a_{ii} = - a_{ii}$ , hence $2 a_{ii} = 0$ , hence $a_{ii} = 0$ for every $i$ .

(e) False. Square requires $m = n$ ; here $2 \neq = 3$ . A null matrix exists in any order and is square only when the order is square.

Answer(a) T (b) F (c) T (d) F (e) F

Problem 2Basic

Find $x$ and $y$ so that $A = 0 - 3 - 2 30 y x - 4 0$ is skew-symmetric.

Show solution

Skew-symmetric requires $a_{ij} = - a_{j i}$ .

The diagonal is already $0, 0, 0$ ✓

Position $(1, 3)$ and $(3, 1)$ : $a_{13} = - a_{31} \Rightarrow x = - (- 2) = 2$ .

Position $(2, 3)$ and $(3, 2)$ : $a_{23} = - a_{32} \Rightarrow - 4 = - y \Rightarrow y = 4$ .

Check $(1, 2)$ against $(2, 1)$ : $3 = - (- 3)$ ✓

$A = 0 - 3 - 2 304 2 - 4 0$

Answer

x = 2, y = 4

Problem 3Exam level

Express $A = [35 - 1 7]$ as the sum of a symmetric and a skew-symmetric matrix.

Show solution

$A^{T} = [3 - 1 57]$

Symmetric part:
$P = \frac{1}{2} (A + A^{T}) = \frac{1}{2} [64414] = [3227]$

Skew-symmetric part:
$Q = \frac{1}{2} (A - A^{T}) = \frac{1}{2} [06 - 6 0] = [03 - 3 0]$

Check: $P + Q = [35 - 1 7] = A$ ✓, $P^{T} = P$ ✓, $Q^{T} = - Q$ ✓

Answer

A = [3227] + [03 - 3 0]

Problem 4Exam level

Show that $A = [524 - 1 - 5]$ is involutory, and hence find $A^{- 1}$ and $A^{99}$ .

Show solution

$A^{2} = [524 - 1 - 5] [524 - 1 - 5] = [25 - 24 120 - 120 - 5 + 5 - 24 + 25] = [1001] = I$

So $A$ is involutory.

Inverse. $A^{2} = I$ means $A \cdot A = I$ , so by definition $A^{- 1} = A = [524 - 1 - 5]$ .

High power. Since $A^{2} = I$ , even powers give $I$ and odd powers give $A$ :

$A^{99} = A^{98} \cdot A = (A^{2})^{49} A = I^{49} A = A = [524 - 1 - 5]$

Answer

A^{- 1} = A

and

A^{99} = A = [524 - 1 - 5]

Problem 5Exam level

Show that $A = 15 - 2 12 - 1 36 - 3$ is nilpotent and state its index. Deduce $∣ A ∣$ without expanding a determinant.

Show solution

Compute $A^{2}$ .

Row 1: $1 + 5 - 6 = 0$ , $1 + 2 - 3 = 0$ , $3 + 6 - 9 = 0$
Row 2: $5 + 10 - 12 = 3$ , $5 + 4 - 6 = 3$ , $15 + 12 - 18 = 9$
Row 3: $- 2 - 5 + 6 = - 1$ , $- 2 - 2 + 3 = - 1$ , $- 6 - 6 + 9 = - 3$

$A^{2} = 03 - 1 03 - 1 09 - 3 \neq = O$

Compute $A^{3} = A^{2} A$ .

Row 1 is zero, so stays zero.
Row 2: $3 + 15 - 18 = 0$ , $3 + 6 - 9 = 0$ , $9 + 18 - 27 = 0$
Row 3: $- 1 - 5 + 6 = 0$ , $- 1 - 2 + 3 = 0$ , $- 3 - 6 + 9 = 0$

$A^{3} = O$

$A$ is nilpotent of index 3.

Determinant. Taking determinants of $A^{3} = O$ :

$∣ A ∣^{3} = ∣ A^{3} ∣ = ∣ O ∣ = 0 \Rightarrow ∣ A ∣ = 0$

So $A$ is singular — as every nilpotent matrix must be.

AnswerNilpotent of index

3

;

∣ A ∣ = 0

Problem 6Exam level

Prove that for any square matrix $A$ , both $A + A^{T}$ and $A A^{T}$ are symmetric, while $A - A^{T}$ is skew-symmetric.

Show solution

Use the two transpose rules $(X + Y)^{T} = X^{T} + Y^{T}$ , $(X Y)^{T} = Y^{T} X^{T}$ , and $(X^{T})^{T} = X$ .

$A + A^{T}$ is symmetric:
$(A + A^{T})^{T} = A^{T} + (A^{T})^{T} = A^{T} + A = A + A^{T}$
Equal to itself, so symmetric. ∎

$A - A^{T}$ is skew-symmetric:
$(A - A^{T})^{T} = A^{T} - (A^{T})^{T} = A^{T} - A = - (A - A^{T})$
Equal to its own negative, so skew-symmetric. ∎

$A A^{T}$ is symmetric:
$(A A^{T})^{T} = (A^{T})^{T} A^{T} = A A^{T}$
Equal to itself, so symmetric. ∎

(The same argument gives $A^{T} A$ symmetric, and it holds even for non-square $A$ , where $A A^{T}$ is $m \times m$ and $A^{T} A$ is $n \times n$ .)

AnswerShown by applying

(X Y)^{T} = Y^{T} X^{T}

and

(X^{T})^{T} = X

to each expression.

Problem 7Challenge

If $A$ is a skew-symmetric matrix of odd order $n$ , prove $∣ A ∣ = 0$ .

Show solution

Start from the definition and take determinants.

$A^{T} = - A$
$\Rightarrow ∣ A^{T} ∣ = ∣ - A ∣$

Now use two determinant properties:

$∣ A^{T} ∣ = ∣ A ∣$ — transposing does not change the determinant.
$∣ k A ∣ = k^{n} ∣ A ∣$ for a matrix of order $n$ — each of the $n$ rows contributes a factor $k$ .

With $k = - 1$ :

$∣ A ∣ = (- 1)^{n} ∣ A ∣$

Since $n$ is odd, $(- 1)^{n} = - 1$ :

$∣ A ∣ = - ∣ A ∣ \Rightarrow 2∣ A ∣ = 0 \Rightarrow ∣ A ∣ = 0 ■$

Every skew-symmetric matrix of odd order is therefore singular. (For even $n$ the argument gives $∣ A ∣ = ∣ A ∣$ , which says nothing — and indeed $[0 - 2 20]$ has determinant $4 \neq = 0$ .)

Answer

∣ A ∣ = (- 1)^{n} ∣ A ∣ = - ∣ A ∣ \Rightarrow ∣ A ∣ = 0

for odd

n

Problem 8Challenge

If $A$ is idempotent, prove that $B = I - A$ is also idempotent, and that $A B = B A = O$ .

Show solution

Given: $A^{2} = A$ .

$B$ is idempotent.
$B^{2} = (I - A) (I - A) = I \cdot I - I \cdot A - A \cdot I + A^{2}$
$= I - A - A + A^{2}$
Substituting $A^{2} = A$ :
$= I - A - A + A = I - A = B ■$

$A B = O$ .
$A B = A (I - A) = A I - A^{2} = A - A = O ■$

$B A = O$ .
$B A = (I - A) A = I A - A^{2} = A - A = O ■$

Interpretation: $A$ and $I - A$ are complementary projections. Whatever $A$ keeps, $I - A$ discards, so applying one after the other destroys everything.

Answer

B^{2} = I - A = B

, and

A B = B A = O

Group 1 — Names that depend only on shape

Rectangular matrix

Square matrix

Row matrix (row vector)

Column matrix (column vector)

Null (zero) matrix — O

Comparable matrices

Group 2 — Names for where the non-zero elements sit

Diagonal matrix

Scalar matrix

Identity (unit) matrix — In​

Upper triangular matrix

Lower triangular matrix

Sparse matrix

Sub-matrix

Group 3 — Names defined by the transpose

Symmetric matrix

Skew-symmetric matrix

Orthogonal matrix

Group 4 — Names defined by powers of the matrix

Idempotent matrix

Involutory matrix

Nilpotent matrix

Periodic matrix

Singular and non-singular

Group 5 — Names for complex matrices

Hermitian matrix

Skew-Hermitian matrix

Unitary matrix

Quick reference

Worked examples

Practice problems

Null (zero) matrix — $O$

Identity (unit) matrix — $I_{n}$