Chapter 3 · Matrices

3.3 Transpose, Symmetric and Skew-Symmetric Matrices

The transpose $A'$ turns rows into columns. Matrices equal to their transpose are symmetric, those equal to minus their transpose are skew-symmetric — and every square matrix is a sum of one of each.

In Section 3.1 we wrote a class's marks with one row per student. A subject teacher would rather have one row per subject. Swapping the roles of rows and columns like this is called taking the transpose, and it is far more than book-keeping: it interacts with multiplication in a surprising “reversed” way, $(AB)'=B'A'$, which you will use again with inverses in Section 3.4.

Some tables do not change at all when you transpose them. A table of road distances between cities is one: the distance from Delhi to Agra is the distance from Agra to Delhi. Such matrices are symmetric. Others change only by a sign, and are called skew-symmetric. This section shows that these two kinds are the building blocks of all square matrices: every square matrix splits, in exactly one way, into a symmetric part and a skew-symmetric part.

The Transpose of a Matrix

📘 Definition 3.9 — Transpose
If $A=[a_{ij}]$ is an $m\times n$ matrix, its transpose is the $n\times m$ matrix $A'=[a_{ji}]$ obtained by writing the rows of $A$ as columns (equivalently, the columns as rows). So the $(i,j)$ entry of $A'$ is the $(j,i)$ entry of $A$. The transpose is also written $A^T$; in this course we use $A'$.

For a square matrix the transpose is a reflection in the main diagonal: the diagonal entries stay where they are, and each entry above the diagonal swaps places with its mirror image below it.

✏️ Example 1 — Writing transposes
Find the transpose of $A=\begin{bmatrix}2&-1&5\\0&3&4\end{bmatrix}$ and of the row matrix $R=\begin{bmatrix}1&-2&7\end{bmatrix}$, and check that $(A')'=A$.
A′ Row 1 of $A$, $(2,-1,5)$, becomes column 1; row 2, $(0,3,4)$, becomes column 2. $A$ is $2\times3$, so $A'$ is $3\times2$: $$A'=\begin{bmatrix}2&0\\-1&3\\5&4\end{bmatrix}$$
R′ The transpose of a row matrix is a column matrix: $R'=\begin{bmatrix}1\\-2\\7\end{bmatrix}$.
(A′)′ Writing the rows of $A'$ — $(2,0)$, $(-1,3)$, $(5,4)$ — as columns gives back $\begin{bmatrix}2&-1&5\\0&3&4\end{bmatrix}=A$ ✓. Flipping twice undoes the flip.
Interactive 1 — The Transpose Flipper

Play flips the matrix over its main diagonal: every entry $a_{ij}$ glides to position $(j,i)$, so row $i$ becomes column $i$. The green diagonal entries do not move. Choose a matrix and watch what the flip does: a $2\times3$ matrix turns into a $3\times2$ one, a symmetric matrix looks exactly the same afterwards, and a skew-symmetric one comes back with every sign changed.

order of A → A′—
tracked entry—
A′ compared with A—

How does the transpose interact with the operations of Section 3.2? For sums and scalar multiples it behaves as simply as possible. For products it reverses the order of the factors.

📐 Theorem 3.5 — Properties of the transpose
For matrices $A$, $B$ and a real number $k$, whenever the operations are defined:
  • $(A')'=A$;
  • $(kA)'=kA'$;
  • $(A+B)'=A'+B'$ (for $A$, $B$ of the same order);
  • (reversal law) $(AB)'=B'A'$ (for $A$ of order $m\times n$ and $B$ of order $n\times p$).

Why. The first three hold entry by entry: for instance the $(i,j)$ entry of $(A+B)'$ is the $(j,i)$ entry of $A+B$, namely $a_{ji}+b_{ji}$, which is the $(i,j)$ entry of $A'+B'$. For the reversal law, first check orders: $AB$ is $m\times p$, so $(AB)'$ is $p\times m$; and $B'A'$ is $(p\times n)(n\times m)=p\times m$. Then compare the $(k,i)$ entries: $$\big((AB)'\big)_{ki}=(AB)_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}=\sum_{j=1}^{n}(B')_{kj}(A')_{ji}=(B'A')_{ki}$$ The same orders and the same entries give $(AB)'=B'A'$. Note that $A'B'$ is usually not even defined, as the next example shows.

✏️ Example 2 — Checking the reversal law
For $$A=\begin{bmatrix}1&0&2\\-1&3&1\end{bmatrix}$$ $$B=\begin{bmatrix}2&1\\0&-1\\4&3\end{bmatrix}$$ verify that $(AB)'=B'A'$, and find the order of $A'B'$.
AB $(2\times3)(3\times2)$: $AB=\begin{bmatrix}2+0+8&1+0+6\\-2+0+4&-1-3+3\end{bmatrix}=\begin{bmatrix}10&7\\2&-1\end{bmatrix}$, so $(AB)'=\begin{bmatrix}10&2\\7&-1\end{bmatrix}$.
B′A′ $B'=\begin{bmatrix}2&0&4\\1&-1&3\end{bmatrix}$ and $A'=\begin{bmatrix}1&-1\\0&3\\2&1\end{bmatrix}$. Row 1 of $B'$ times the columns of $A'$: $2+0+8=10$, $-2+0+4=2$. Row 2: $1+0+6=7$, $-1-3+3=-1$. So $B'A'=\begin{bmatrix}10&2\\7&-1\end{bmatrix}=(AB)'$ ✓.
A′B′ $(3\times2)(2\times3)$ is defined but is $3\times3$ — it cannot equal the $2\times2$ matrix $(AB)'$. Reversing the order is essential.
✏️ Example 3 — Using the rules to save work
Let $A=\begin{bmatrix}2&-1\\0&3\\1&4\end{bmatrix}$ and $B=\begin{bmatrix}1&0&-2\\3&1&5\end{bmatrix}$. Find $(A'+2B)'$ in two ways.
Directly $A'=\begin{bmatrix}2&0&1\\-1&3&4\end{bmatrix}$, so $A'+2B=\begin{bmatrix}4&0&-3\\5&5&14\end{bmatrix}$ and its transpose is $\begin{bmatrix}4&5\\0&5\\-3&14\end{bmatrix}$.
By Theorem 3.5 $(A'+2B)'=(A')'+(2B)'=A+2B'$. With $2B'=\begin{bmatrix}2&6\\0&2\\-4&10\end{bmatrix}$, $A+2B'=\begin{bmatrix}4&5\\0&5\\-3&14\end{bmatrix}$ — the same, with one transpose fewer.

Symmetric and Skew-Symmetric Matrices

📘 Definition 3.10 — Symmetric and skew-symmetric matrices
A square matrix $A=[a_{ij}]$ is
  • symmetric if $A'=A$, that is, $a_{ji}=a_{ij}$ for all $i,j$ — each entry equals its mirror image in the main diagonal;
  • skew-symmetric if $A'=-A$, that is, $a_{ji}=-a_{ij}$ for all $i,j$ — each entry is the negative of its mirror image.
Only square matrices can be either: if $A$ is $m\times n$ with $m\ne n$, then $A'$ has a different order from $A$.
📐 Theorem 3.6 — Building symmetric and skew-symmetric matrices
(a) Every diagonal entry of a skew-symmetric matrix is $0$.
(b) For every square matrix $A$, the matrix $A+A'$ is symmetric and the matrix $A-A'$ is skew-symmetric.

Why. (a) Put $j=i$ in $a_{ji}=-a_{ij}$: $a_{ii}=-a_{ii}$, so $2a_{ii}=0$ and $a_{ii}=0$. (b) By Theorem 3.5, $(A+A')'=A'+(A')'=A'+A=A+A'$, so $A+A'$ is symmetric; and $(A-A')'=A'-(A')'=A'-A=-(A-A')$, so $A-A'$ is skew-symmetric.

✏️ Example 4 — Recognising the type, and finding unknowns
(a) Classify $P=\begin{bmatrix}1&4&-2\\4&0&5\\-2&5&3\end{bmatrix}$, $Q=\begin{bmatrix}0&3&-1\\-3&0&2\\1&-2&0\end{bmatrix}$ and $R=\begin{bmatrix}1&2\\-2&0\end{bmatrix}$ as symmetric, skew-symmetric or neither.
(b) Find $x$, $y$, $z$ if $\begin{bmatrix}0&x&3\\2&0&y\\z&-1&0\end{bmatrix}$ is skew-symmetric.
P Mirror pairs: $p_{12}=p_{21}=4$, $p_{13}=p_{31}=-2$, $p_{23}=p_{32}=5$. Symmetric. (Its diagonal $1,0,3$ can be anything.)
Q Diagonal all $0$, and $q_{21}=-3=-q_{12}$, $q_{31}=1=-q_{13}$, $q_{32}=-2=-q_{23}$. Skew-symmetric.
R $r_{21}=-2=-r_{12}$ looks skew, but $r_{11}=1\ne0$, so by Theorem 3.6(a) it is not skew-symmetric; and $r_{21}\ne r_{12}$, so it is not symmetric. Neither.
(b) The diagonal is already $0$. Mirror pairs: $x=-a_{21}=-2$, $y=-a_{32}=-(-1)=1$, $z=-a_{13}=-3$. Check: the matrix $\begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix}$ equals minus its transpose ✓.
✏️ Example 5 — A proof with the reversal law
Let $A$ be any $m\times n$ matrix. Show that $A'A$ and $AA'$ are both symmetric. Show also that if $B$ is a symmetric $n\times n$ matrix, then $A B A'$ is symmetric.
Orders $A'A$ is $(n\times m)(m\times n)=n\times n$ and $AA'$ is $m\times m$ — both square, so the question makes sense.
A′A $(A'A)'=A'(A')'=A'A$ by the reversal law and $(A')'=A$. So $A'A$ is symmetric. The same two steps give $(AA')'=(A')'A'=AA'$.
ABA′ For three factors the reversal law reverses all three: $(ABA')'=\big((AB)A'\big)'=(A')'(AB)'=AB'A'$. Since $B'=B$, this is $ABA'$. Symmetric. ∎
Check $A=\begin{bmatrix}1&2\end{bmatrix}$: $A'A=\begin{bmatrix}1&2\\2&4\end{bmatrix}$ — symmetric ✓ — and $AA'=\begin{bmatrix}5\end{bmatrix}$.

Every Square Matrix Is Symmetric + Skew-Symmetric

Theorem 3.6(b) hands us a symmetric matrix $A+A'$ and a skew-symmetric matrix $A-A'$, and their sum is $2A$. Halving both gives the decomposition.

📐 Theorem 3.7 — Symmetric plus skew-symmetric decomposition
Every square matrix $A$ can be written as $A=P+Q$ with $P$ symmetric and $Q$ skew-symmetric, namely $$P=\tfrac12\left(A+A'\right)$$ $$Q=\tfrac12\left(A-A'\right)$$ and this is the only way to do it.

Why. Existence: $P+Q=\tfrac12A+\tfrac12A'+\tfrac12A-\tfrac12A'=A$; $P$ is symmetric and $Q$ skew-symmetric by Theorem 3.6(b) and $(kA)'=kA'$. Uniqueness: suppose also $A=R+S$ with $R$ symmetric and $S$ skew-symmetric. Transposing, $A'=R'+S'=R-S$. Adding and subtracting the two equations gives $R=\tfrac12(A+A')=P$ and $S=\tfrac12(A-A')=Q$. So there is no other split.

🧭 Problem-solving strategy — the decomposition
  1. Write $A'$ carefully (rows of $A$ become columns).
  2. Compute $A+A'$ and halve it: $P$. Its diagonal is the diagonal of $A$; each off-diagonal pair is the average of $a_{ij}$ and $a_{ji}$.
  3. Compute $A-A'$ and halve it: $Q$. Its diagonal is $0$; $q_{ij}=\tfrac12(a_{ij}-a_{ji})$.
  4. Check: $P'=P$, $Q'=-Q$ and $P+Q=A$. Board examiners give marks for writing these checks.
✏️ Example 6 — Splitting a 3 × 3 matrix
Express $A=\begin{bmatrix}3&-2&4\\6&1&-1\\0&5&2\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
A′ $A'=\begin{bmatrix}3&6&0\\-2&1&5\\4&-1&2\end{bmatrix}$.
P $A+A'=\begin{bmatrix}6&4&4\\4&2&4\\4&4&4\end{bmatrix}$, so $$P=\begin{bmatrix}3&2&2\\2&1&2\\2&2&2\end{bmatrix}$$
Q $A-A'=\begin{bmatrix}0&-8&4\\8&0&-6\\-4&6&0\end{bmatrix}$, so $$Q=\begin{bmatrix}0&-4&2\\4&0&-3\\-2&3&0\end{bmatrix}$$
Check $P'=P$ (mirror pairs $2,2,2$) ✓; $Q'=-Q$ with zero diagonal ✓; $P+Q=\begin{bmatrix}3&-2&4\\6&1&-1\\0&5&2\end{bmatrix}=A$ ✓. Enter $A$ in Interactive 2 to see the split.
Interactive 2 — Splitting A into Symmetric + Skew-Symmetric

Type any $3\times3$ matrix (or let Play try one random integer matrix after another). The widget computes $P=\tfrac12(A+A')$ in blue and $Q=\tfrac12(A-A')$ in red. Watch a mirror pair: $p_{ij}$ is the average of $a_{ij}$ and $a_{ji}$, and $q_{ij}$ is half their difference, so $p_{ij}+q_{ij}=a_{ij}$.

P′ = P ?—
Q′ = −Q ?—
P + Q = A ?—
tracked pair—
⚠️ Common mistakes when studying alone

1. $(AB)'\ne A'B'$. The order reverses: $(AB)'=B'A'$, and for three factors $(ABC)'=C'B'A'$. Check orders — $A'B'$ is often not even defined.

2. Transposing the wrong way. Rows of $A$ become columns of $A'$. Read off row 1 of $A$ and write it down the first column — do not reverse the order of the entries.

3. Forgetting the diagonal. A skew-symmetric matrix must have zeros on the diagonal; $\begin{bmatrix}1&2\\-2&1\end{bmatrix}$ is not skew-symmetric.

4. Forgetting the ½. $A+A'$ is symmetric, but $A=(A+A')+(A-A')$ is false — the right side is $2A$.

5. Only square matrices. “Symmetric” and “skew-symmetric” make sense only for square matrices, and the decomposition theorem is about square matrices only.

Exercises

I
Transposes and their properties.
  • 1 Find the transpose of $\begin{bmatrix}1&-2\\3&0\\5&4\end{bmatrix}$.
  • 2 Find the transpose of $\begin{bmatrix}7&-3&0\end{bmatrix}$.
  • 3 $A$ has order $4\times2$. What is the order of $A'$, of $A'A$ and of $AA'$?
  • 4 Verify $(A')'=A$ for the matrix of part 1.
  • 5 Verify $(A+B)'=A'+B'$ for $A=\begin{bmatrix}1&2\\3&-1\end{bmatrix}$, $B=\begin{bmatrix}0&-4\\2&5\end{bmatrix}$.
  • 6 For $A=\begin{bmatrix}2&1&0\\-1&4&3\end{bmatrix}$, find $(-3A)'$ and check that it equals $-3A'$.
  • 7 Verify $(AB)'=B'A'$ for $A=\begin{bmatrix}1\\-2\\3\end{bmatrix}$ and $B=\begin{bmatrix}2&-1&4\end{bmatrix}$.
  • 8 $A$ is $2\times3$ and $B$ is $3\times4$. What is the order of $(AB)'$? Is $A'B'$ defined?
1 $\begin{bmatrix}1&3&5\\-2&0&4\end{bmatrix}$ (order $2\times3$).
2 $\begin{bmatrix}7\\-3\\0\end{bmatrix}$
3 $A'$ is $2\times4$; $A'A$ is $(2\times4)(4\times2)=2\times2$; $AA'$ is $4\times4$.
4 The rows of $\begin{bmatrix}1&3&5\\-2&0&4\end{bmatrix}$ written as columns give $\begin{bmatrix}1&-2\\3&0\\5&4\end{bmatrix}$ ✓.
5 $A+B=\begin{bmatrix}1&-2\\5&4\end{bmatrix}$, so $(A+B)'=\begin{bmatrix}1&5\\-2&4\end{bmatrix}$. $A'+B'=\begin{bmatrix}1&3\\2&-1\end{bmatrix}+\begin{bmatrix}0&2\\-4&5\end{bmatrix}=\begin{bmatrix}1&5\\-2&4\end{bmatrix}$ ✓.
6 $-3A=\begin{bmatrix}-6&-3&0\\3&-12&-9\end{bmatrix}$, so $(-3A)'=\begin{bmatrix}-6&3\\-3&-12\\0&-9\end{bmatrix}$; and $-3A'=-3\begin{bmatrix}2&-1\\1&4\\0&3\end{bmatrix}$ is the same ✓.
7 $AB$ is $3\times3$: $\begin{bmatrix}2&-1&4\\-4&2&-8\\6&-3&12\end{bmatrix}$, so $(AB)'=\begin{bmatrix}2&-4&6\\-1&2&-3\\4&-8&12\end{bmatrix}$. $B'A'=\begin{bmatrix}2\\-1\\4\end{bmatrix}\begin{bmatrix}1&-2&3\end{bmatrix}$ gives the same matrix ✓.
8 $AB$ is $2\times4$, so $(AB)'$ is $4\times2$ (and so is $B'A'$). $A'B'$ is $(3\times2)(4\times3)$: $2\ne4$, not defined.
II
Symmetric or skew-symmetric? (Parts 1–5: classify.)
  • 1 $\begin{bmatrix}2&-3\\-3&5\end{bmatrix}$
  • 2 $\begin{bmatrix}0&7\\-7&0\end{bmatrix}$
  • 3 $\begin{bmatrix}0&1&-2\\-1&0&4\\2&-4&0\end{bmatrix}$
  • 4 $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$
  • 5 $\begin{bmatrix}1&2\\-2&1\end{bmatrix}$
  • 6 Find $a,b,c$ so that $\begin{bmatrix}2&a&3\\-1&4&b\\c&5&6\end{bmatrix}$ is symmetric.
  • 7 Find $x,y,z$ so that $\begin{bmatrix}0&x&-4\\3&y&2\\z&-2&0\end{bmatrix}$ is skew-symmetric.
  • 8 Which square matrices are both symmetric and skew-symmetric?
  • 9 For $A=\begin{bmatrix}1&5\\6&7\end{bmatrix}$, verify that $A+A'$ is symmetric and $A-A'$ is skew-symmetric.
1 Symmetric ($a_{12}=a_{21}=-3$).
2 Skew-symmetric (zero diagonal, $a_{21}=-a_{12}$).
3 Skew-symmetric: zero diagonal and $-1=-1$, $2=-(-2)$, $-4=-4$ for the three mirror pairs.
4 Symmetric — every diagonal matrix is symmetric (its off-diagonal entries are all $0$). Not skew, since the diagonal is non-zero.
5 Neither: $a_{21}\ne a_{12}$, and the diagonal is not zero.
6 Mirror pairs must agree: $a=-1$, $b=5$, $c=3$.
7 Diagonal: $y=0$. Pairs: $x=-3$, $z=-(-4)=4$; and $a_{23}=2=-a_{32}$ ✓.
8 If $A'=A$ and $A'=-A$ then $A=-A$, so $2A=O$ and $A=O$: only the zero matrices.
9 $A'=\begin{bmatrix}1&6\\5&7\end{bmatrix}$; $A+A'=\begin{bmatrix}2&11\\11&14\end{bmatrix}$, symmetric ✓; $A-A'=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, skew-symmetric ✓.
III
Proofs. All matrices are square of the same order.
  • 1 If $A$ and $B$ are symmetric, prove that $A+B$ is symmetric.
  • 2 If $A$ is skew-symmetric and $k$ is real, prove that $kA$ is skew-symmetric.
  • 3 If $A$ and $B$ are symmetric, prove that $AB$ is symmetric if and only if $AB=BA$.
  • 4 If $A$ is skew-symmetric, prove that $A^2$ is symmetric.
  • 5 If $A$ is symmetric, prove that $B'AB$ is symmetric for every $B$.
  • 6 If $A$ is skew-symmetric and $B$ is symmetric, prove that $AB+BA$ is skew-symmetric.
  • 7 Prove that $(ABC)'=C'B'A'$.
1 $(A+B)'=A'+B'=A+B$.
2 $(kA)'=kA'=k(-A)=-(kA)$.
3 $(AB)'=B'A'=BA$. So $(AB)'=AB$ exactly when $BA=AB$.
4 $(A^2)'=(AA)'=A'A'=(-A)(-A)=A^2$.
5 $(B'AB)'=B'A'(B')'=B'AB$, using part 7 and $A'=A$.
6 $(AB+BA)'=B'A'+A'B'=B(-A)+(-A)B=-(BA+AB)=-(AB+BA)$.
7 $(ABC)'=\big((AB)C\big)'=C'(AB)'=C'B'A'$, applying the reversal law twice.
IV
Decomposition and modelling. (Check parts 1–3 with Interactive 2.)
  • 1 Express $\begin{bmatrix}4&1\\-3&2\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
  • 2 Do the same for $\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$.
  • 3 Do the same for $\begin{bmatrix}0&5&-3\\1&1&1\\4&0&2\end{bmatrix}$.
  • 4 Hill roads. Bus times in minutes between three hill towns are $T=\begin{bmatrix}0&50&70\\40&0&30\\60&36&0\end{bmatrix}$ ($t_{ij}$ = time from town $i$ to town $j$). Split $T$ into its symmetric and skew-symmetric parts and say what each part means.
  • 5 How many entries can be chosen freely in a $3\times3$ symmetric matrix? In a $3\times3$ skew-symmetric matrix?
  • 6 True or false: (a) the transpose of a symmetric matrix is symmetric; (b) a zero matrix is skew-symmetric; (c) the identity matrix is skew-symmetric; (d) every diagonal matrix is symmetric.
1 $P=\begin{bmatrix}4&-1\\-1&2\end{bmatrix}$, $Q=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$; $P+Q=\begin{bmatrix}4&1\\-3&2\end{bmatrix}$ ✓.
2 $$P=\begin{bmatrix}1&3&5\\3&5&7\\5&7&9\end{bmatrix}$$ $$Q=\begin{bmatrix}0&-1&-2\\1&0&-1\\2&1&0\end{bmatrix}$$
3 $$P=\begin{bmatrix}0&3&\tfrac12\\3&1&\tfrac12\\\tfrac12&\tfrac12&2\end{bmatrix}$$ $$Q=\begin{bmatrix}0&2&-\tfrac72\\-2&0&\tfrac12\\\tfrac72&-\tfrac12&0\end{bmatrix}$$
4 $$P=\begin{bmatrix}0&45&65\\45&0&33\\65&33&0\end{bmatrix}$$ $$Q=\begin{bmatrix}0&5&5\\-5&0&-3\\-5&3&0\end{bmatrix}$$ $P$ gives the average time of the two directions between each pair of towns; $Q$ gives how far each direction is above or below that average. For example $q_{12}=5$: going from town 1 to town 2 takes $5$ minutes more than the average $45$ — perhaps the road climbs.
5 Symmetric: the $3$ diagonal entries and the $3$ entries above the diagonal — $6$ free entries. Skew-symmetric: the diagonal is $0$, so only the $3$ entries above it — $3$ free entries.
6 (a) True: $(A')'=A=A'$. (b) True: $O'=O=-O$. (c) False: its diagonal is not zero. (d) True.
🎯 Exam corner Board · 1 markBoard · 3 marksBoard · 5 marksJEE MainJEE Advanced

How this is examined: Board papers almost always contain a 1-mark question on symmetric or skew-symmetric matrices and a longer question asking you to express a matrix as the sum of the two, with the checks written out. JEE tests the reversal law and the type of products such as $AB-BA$.

E1. Board · 1 mark If $\begin{bmatrix}0&2&b\\a&0&-4\\5&c&0\end{bmatrix}$ is skew-symmetric, then $a+b+c$ equals  (A) $-3$  (B) $3$  (C) $-11$  (D) $0$

Idea $a_{ji}=-a_{ij}$: $a=-2$, $b=-5$, $c=-(-4)=4$.
Answer (A) $-3$. Trap (B) uses symmetric instead ($a=2$, $b=5$, $c=-4$); (C) forgets the sign of $c$.

E2. Board · 1 mark Assertion–Reason. Assertion (A): every diagonal entry of a skew-symmetric matrix is zero. Reason (R): if $A$ is skew-symmetric then $a_{ii}=-a_{ii}$ for each $i$.  (A) Both A and R are true and R is the correct explanation of A  (B) Both A and R are true but R is not the correct explanation of A  (C) A is true but R is false  (D) A is false but R is true

Idea R is the case $j=i$ of $a_{ji}=-a_{ij}$, and $a_{ii}=-a_{ii}$ gives $a_{ii}=0$, which is A (Theorem 3.6(a)).
Answer (A).

E3. Board · 3 marks Express $A=\begin{bmatrix}3&5\\1&-1\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix, and verify your answer.

P $A'=\begin{bmatrix}3&1\\5&-1\end{bmatrix}$; $P=\tfrac12(A+A')=\tfrac12\begin{bmatrix}6&6\\6&-2\end{bmatrix}=\begin{bmatrix}3&3\\3&-1\end{bmatrix}$, and $P'=P$. (1 mark)
Q $Q=\tfrac12(A-A')=\tfrac12\begin{bmatrix}0&4\\-4&0\end{bmatrix}=\begin{bmatrix}0&2\\-2&0\end{bmatrix}$, and $Q'=-Q$. (1 mark)
Verify $P+Q=\begin{bmatrix}3&5\\1&-1\end{bmatrix}=A$. (1 mark)

E4. Board · 5 marks Express $A=\begin{bmatrix}2&4&-6\\7&3&5\\1&-2&4\end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

A′ $A'=\begin{bmatrix}2&7&1\\4&3&-2\\-6&5&4\end{bmatrix}$. (1 mark)
P $A+A'=\begin{bmatrix}4&11&-5\\11&6&3\\-5&3&8\end{bmatrix}$, so $$P=\begin{bmatrix}2&\tfrac{11}{2}&-\tfrac52\\\tfrac{11}{2}&3&\tfrac32\\-\tfrac52&\tfrac32&4\end{bmatrix}$$ which satisfies $P'=P$. (1½ marks)
Q $A-A'=\begin{bmatrix}0&-3&-7\\3&0&7\\7&-7&0\end{bmatrix}$, so $$Q=\begin{bmatrix}0&-\tfrac32&-\tfrac72\\\tfrac32&0&\tfrac72\\\tfrac72&-\tfrac72&0\end{bmatrix}$$ which satisfies $Q'=-Q$. (1½ marks)
Verify Adding entry by entry, e.g. $p_{12}+q_{12}=\tfrac{11}{2}-\tfrac32=4=a_{12}$ and $p_{31}+q_{31}=-\tfrac52+\tfrac72=1=a_{31}$; all nine positions give $P+Q=A$. (1 mark)

E5. JEE Main If $A$ and $B$ are symmetric matrices of the same order, then $AB-BA$ is  (A) symmetric  (B) skew-symmetric  (C) always the zero matrix  (D) the identity matrix

Method $(AB-BA)'=B'A'-A'B'=BA-AB=-(AB-BA)$.
Answer (B). Trap (C): $AB-BA=O$ only if $A$ and $B$ commute, which symmetric matrices need not do (try $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&1\\1&0\end{bmatrix}$).

E6. JEE Advanced (One or more options correct.) Let $A$ be a skew-symmetric and $B$ a symmetric matrix, both $3\times3$. Which of the following are always symmetric?  (A) $A^2$  (B) $AB-BA$  (C) $ABA$  (D) $BAB$

(A) Yes $(A^2)'=A'A'=(-A)(-A)=A^2$.
(B) Yes $(AB-BA)'=B'A'-A'B'=-BA+AB=AB-BA$.
(C) Yes $(ABA)'=A'B'A'=(-A)B(-A)=ABA$.
(D) No $(BAB)'=B'A'B'=-BAB$: it is skew-symmetric, and symmetric only when it is $O$. (For $A=\begin{bmatrix}0&2&-1\\-2&0&3\\1&-3&0\end{bmatrix}$ and $B=\begin{bmatrix}1&4&2\\4&-1&0\\2&0&5\end{bmatrix}$ it is not $O$.)
Answer (A), (B), (C). ⏱ Fast route: each of these products reads the same backwards, so transposing just replaces every $A$ by $-A$. An even number of $A$'s gives a symmetric matrix, an odd number a skew-symmetric one.
🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. In your town, is travelling from $X$ to $Y$ as quick as travelling from $Y$ to $X$? Use the symmetric/skew-symmetric split to measure the “one-way-ness” of local journeys.

Why it matters. City planners study origin–destination tables to place bus lanes and one-way streets. The symmetric part of a travel-time matrix shows how far apart places are on average; the skew-symmetric part isolates direction effects — one-way systems, hills, the morning rush into the centre.

What to do.

  1. Choose four places (home, school, market, station). With a map app, record the driving or bus time in minutes for all twelve ordered pairs, at the same time of day. Enter them as a $4\times4$ matrix $T$ with zero diagonal.
  2. In a spreadsheet compute $T'$ (TRANSPOSE), then $P=\tfrac12(T+T')$ and $Q=\tfrac12(T-T')$. Check $P+Q=T$.
  3. Find the largest entry of $Q$ in absolute value. Which journey is most one-sided, and why might that be?
  4. Repeat at a different time of day. Does $P$ change more, or $Q$?

Tools: Spreadsheet (TRANSPOSE, array arithmetic); a map app.

Hand in: $T$, $P$ and $Q$ with units, a sentence explaining the meaning of one entry of $P$ and one of $Q$, and your explanation for the most one-sided journey.

Going further. A matrix of road distances is usually symmetric. Find a pair of places where it is not, and explain why.

Instructor note: the hill-road matrix of Exercise IV.4 is a worked model. Expect $Q$ to be small compared with $P$ off-peak and larger at rush hour; one-way streets give asymmetric distances.

🧪 Self-check quiz

0/5 correct

Q1. If $A$ is a $3\times2$ matrix, the order of $A'$ is

Q2. For matrices whose product $AB$ is defined, $(AB)'$ equals

Q3. Which matrix is skew-symmetric?

Q4. The diagonal entries of a skew-symmetric matrix are

Q5. The symmetric part $\tfrac12(A+A')$ of $A=\begin{bmatrix}1&4\\2&3\end{bmatrix}$ is

← 3.2 Operations on Matrices3.4 Invertible Matrices →