Transpose — definition, properties: Worked examples

This page is the "hands dirty" companion to Transpose — definition, properties. We take the rules from the parent note and throw every kind of matrix at them — rectangular, square, zero, degenerate, symmetric, and an exam-style trap — so that no shape ever surprises you.

Before we compute anything, one reminder of the single rule everything rests on:

The scenario matrix

Here is the full landscape of cases a transpose problem can hand you. Every cell is covered by at least one worked example below.

Cell	Case class	What makes it tricky	Example
A	Rectangular $m \times n$ , $m \neq = n$	shape changes $m \times n \to n \times m$	Ex 1
B	Square, general	shape same, entries reflect over diagonal	Ex 2
C	Row/column vector (degenerate: $1 \times n$ )	a "flat" matrix becomes a "tall" one	Ex 3
D	Product with reversal $(A B)^{T}$	order flips, non-commutative	Ex 4
E	Chained product $(A B C)^{T}$	reversal cascades	Ex 5
F	Symmetric / skew split (sign cases)	$A^{T} = A$ vs $A^{T} = - A$ , zero diagonal	Ex 6
G	Zero matrix & identity (degenerate/limit)	fixed points of transpose	Ex 7
H	Inverse + transpose together	$(A^{- 1})^{T} = (A^{T})^{- 1}$ , real numbers	Ex 8
I	Word problem (real-world table flip)	reading rows-vs-columns correctly	Ex 9
J	Exam twist ( $x^{T} A x$ / trace / trap)	scalar output, dimension trap	Ex 10

Prerequisites we lean on: Matrix Operations, Symmetric Matrices, Orthogonal Matrices, Matrix Inverse, Inner Product Spaces.

Cell A — Rectangular, shape changes

Worked example Ex 1 — a

$2 \times 3$ becomes a $3 \times 2$ $A = [2503 - 1 4]$ Find $A^{T}$ and its size.

Forecast: Before reading on — will $A^{T}$ be wider or taller than $A$ ? Write down your guess.

Step 1. $A$ has $2$ rows and $3$ columns, so it is $2 \times 3$ . Why this step? The rule $[A^{T}]_{ij} = [A]_{j i}$ swaps the two size numbers, so $A^{T}$ must be $3 \times 2$ — taller than wide. (Guess check: taller.)

Step 2. Take row $1$ of $A$ , namely $[2, 0, - 1]$ , and stand it up as column 1 of $A^{T}$ . Why this step? "Row $k$ of $A$ becomes column $k$ of $A^{T}$ " is just the label swap read across a whole line.

Step 3. Take row $2$ of $A$ , namely $[5, 3, 4]$ , and stand it up as column 2 of $A^{T}$ . Why this step? Same rule for the second row. $A^{T} = 20 - 1 534_{3 \times 2}$

Verify: Pick one entry blind. $[A^{T}]_{31}$ should equal $[A]_{13}$ . We read $[A^{T}]_{31} = - 1$ and $[A]_{13} = - 1$ . ✓ Size is $3 \times 2$ . ✓

The negative sign in $- 1$ rode along untouched — transpose never changes a value, only its address.

Cell B — Square, general

Cell C — Degenerate: a vector

Cell D — Product with reversal

Worked example Ex 4 —

$(A B)^{T} = B^{T} A^{T}$ $A = [10 2 - 1], B = [3214]$ Compute $(A B)^{T}$ both ways and confirm the order really must reverse.

Forecast: Guess whether $A^{T} B^{T}$ (wrong order) would give the same answer here.

Step 1. Multiply $A B$ (row-times-column, from Matrix Operations): $A B = [1 \cdot 3 + 2 \cdot 2 0 \cdot 3 + (- 1) \cdot 2 1 \cdot 1 + 2 \cdot 4 0 \cdot 1 + (- 1) \cdot 4] = [7 - 2 9 - 4]$ Why this step? We need the product first so we can transpose it as one block.

Step 2. Transpose that block (swap off-diagonals): $(A B)^{T} = [79 - 2 - 4]$ Why this step? Direct application of the rule to $A B$ .

Step 3. Now the reversed product $B^{T} A^{T}$ : $B^{T} = [3124], A^{T} = [12 0 - 1]$ $B^{T} A^{T} = [3 \cdot 1 + 2 \cdot 2 1 \cdot 1 + 4 \cdot 2 3 \cdot 0 + 2 \cdot (- 1) 1 \cdot 0 + 4 \cdot (- 1)] = [79 - 2 - 4]$ Why this step? The reversal rule predicts this equals $(A B)^{T}$ .

Step 4. For contrast, the wrong order $A^{T} B^{T}$ : $A^{T} B^{T} = [12 0 - 1] [3124] = [3520]$ Why this step? To see that keeping the order gives a genuinely different matrix.

Verify: $(A B)^{T} = B^{T} A^{T} = [79 - 2 - 4]$ ✓, but $A^{T} B^{T} = [3520] \neq =$ that. The reversal is not optional. ✓

Cell E — Chained reversal

Worked example Ex 5 —

$(A B C)^{T} = C^{T} B^{T} A^{T}$ With diagonal matrices (so arithmetic stays clean): $A = [2001], B = [1011], C = [1301]$ Show $(A B C)^{T} = C^{T} B^{T} A^{T}$ .

Forecast: In what order will $A, B, C$ appear on the right? Write it before computing.

Step 1. $A B = [2001] [1011] = [2021]$ . Why this step? Build the product left-to-right.

Step 2. $A B C = [2021] [1301] = [2 + 6 0 + 3 0 + 2 0 + 1] = [8321]$ . Why this step? Finish the triple product.

Step 3. Transpose it: $(A B C)^{T} = [8231]$ . Why this step? Apply the rule to the whole block.

Step 4. Apply reversal twice: $(A B C)^{T} = (B C)^{T} A^{T} = C^{T} B^{T} A^{T}$ . Compute $C^{T} = [1031], B^{T} = [1101], A^{T} = [2001]$ . $C^{T} B^{T} = [1031] [1101] = [4131]$ , then $C^{T} B^{T} A^{T} = [4131] [2001] = [8231]$ . Why this step? Each transpose peels the outermost factor off and lands it in front — the order fully reverses.

Verify: $[8231] = [8231]$ ✓

Cell F — Sign cases: symmetric vs skew

Worked example Ex 6 — split a matrix into symmetric + skew parts

$M = [4172]$ Write $M = S + K$ where $S^{T} = S$ (symmetric) and $K^{T} = - K$ (skew-symmetric).

Forecast: What must the diagonal of the skew part $K$ be? (Reread the parent's skew definition.)

Step 1. Use the standard split $S = \frac{1}{2} (M + M^{T})$ , $K = \frac{1}{2} (M - M^{T})$ . Why this step? $(M + M^{T})^{T} = M^{T} + M = M + M^{T}$ , so $S$ is symmetric; $(M - M^{T})^{T} = M^{T} - M = - (M - M^{T})$ , so $K$ is skew. This uses the sum rule and $(A^{T})^{T} = A$ .

Step 2. $M^{T} = [4712]$ , so $S = \frac{1}{2} [8884] = [4442], K = \frac{1}{2} [0 - 6 60] = [0 - 3 30]$ Why this step? Plug $M$ and $M^{T}$ into the formulas.

Step 3. Notice the zero diagonal of $K$ : since $k_{ii} = - k_{ii} \Rightarrow k_{ii} = 0$ . This is the degenerate sign case where a number must equal its own negative. Why this step? It confirms $K$ is genuinely skew and not just anti-off-diagonal.

Verify: $S + K = [4172] = M$ ✓. $S^{T} = S$ ✓ (off-diagonals both $4$ ). $K^{T} = [03 - 3 0] = - K$ ✓. Every square matrix splits this way — see Symmetric Matrices.

Cell G — Degenerate fixed points: zero and identity

Worked example Ex 7 — matrices that are their own transpose

Find $O^{T}$ for the $2 \times 3$ zero matrix and $I^{T}$ for the $3 \times 3$ identity.

Forecast: One of these keeps its shape, one changes it. Which is which?

Step 1. Zero matrix $O = [000000]$ is $2 \times 3$ , so $O^{T}$ is $3 \times 3$ … no — it is $3 \times 2$ , and every entry is still $0$ . Why this step? The values are all $0$ so nothing changes value, but the shape flips $2 \times 3 \to 3 \times 2$ . A rectangular zero matrix is not equal to its transpose (different shapes).

Step 2. Identity $I = 100010001$ . Every $1$ is on the diagonal ( $i = j$ ), and diagonal entries are fixed by transpose. Off-diagonals are all $0$ , which stay $0$ . Why this step? $I$ is symmetric: $I^{T} = I$ . This is the fact that makes Property 5 (transpose of inverse) work, since $I^{T} = I$ .

Verify: $O^{T}$ is the $3 \times 2$ zero matrix (shape changed, still all zeros) ✓. $I^{T} = I$ exactly ✓. Limit intuition: a square zero matrix is its own transpose; a rectangular one is not — shape is what decides equality.

Cell H — Inverse and transpose together (real numbers)

Worked example Ex 8 — verify

$(A^{- 1})^{T} = (A^{T})^{- 1}$ $A = [2111]$ Compute both sides numerically.

Forecast: Do you expect messy fractions or clean integers here? ( $det A$ is the clue.)

Step 1. $det A = 2 \cdot 1 - 1 \cdot 1 = 1$ (see Determinants). Since it is $1$ , the inverse is clean: $A^{- 1} = \frac{1}{1} [1 - 1 - 1 2] = [1 - 1 - 1 2]$ Why this step? Matrix Inverse for $2 \times 2$ : swap the diagonal, negate the off-diagonal, divide by $det$ .

Step 2. Transpose it: $(A^{- 1})^{T} = [1 - 1 - 1 2]$ (it happens to be symmetric). Why this step? Left-hand side of the identity.

Step 3. Now the other side. $A^{T} = [2111] = A$ (symmetric!), so $(A^{T})^{- 1} = A^{- 1} = [1 - 1 - 1 2]$ . Why this step? Right-hand side of the identity.

Verify: Both equal $[1 - 1 - 1 2]$ ✓. Extra check: $A^{T} (A^{- 1})^{T} = [2111] [1 - 1 - 1 2] = [1001] = I$ ✓ — confirming $(A^{- 1})^{T}$ really inverts $A^{T}$ .

Cell I — Real-world table flip

Worked example Ex 9 — the sales spreadsheet

A shop records units sold. Rows are stores (North, South), columns are months (Jan, Feb, Mar): $S = [1281511914] \leftarrow North \leftarrow South$ The head office wants a table where rows are months and columns are stores. Produce it and read off "February, South".

Forecast: Which single operation reorganises the whole table? (You already know it.)

Step 1. "Rows↔columns" is precisely the transpose. Compute $S^{T}$ : $S^{T} = 1215981114 \leftarrow Jan \leftarrow Feb \leftarrow Mar$ Why this step? Each store-row becomes a store-column; each month-column becomes a month-row.

Step 2. "February, South" in the new table is row Feb, column South $= [S^{T}]_{22} = 11$ . Why this step? In the original that same number lived at row South ( $= 2$ ), column Feb ( $= 2$ ): $[S]_{22} = 11$ . The label swap $[S^{T}]_{22} = [S]_{22}$ — identical because both indices are $2$ .

Verify: North-March in original is $[S]_{13} = 9$ ; in the flipped table it is $[S^{T}]_{31} = 9$ ✓. The data never changed, only the way we address it — which is the whole point of transpose in data work.

Cell J — Exam twist: scalar output & a dimension trap

The classic exam object is the quadratic form $x^{T} A x$ , which turns a vector and a matrix into a single number. It shows up in Eigenvalues and Eigenvectors and geometry.

Figure — Transpose — definition, properties

Worked example Ex 10 — evaluate

$x^{T} A x$ and dodge the shape trap $x = [12], A = [3012]$ Compute the scalar $x^{T} A x$ . Then explain why $A x^{T}$ is illegal.

Forecast: What size is the final answer — a matrix or a single number? Guess before Step 1.

Step 1. Sizes: $x^{T}$ is $1 \times 2$ , $A$ is $2 \times 2$ , $x$ is $2 \times 1$ . Chain: $(1 \times 2) (2 \times 2) (2 \times 1) \to 1 \times 1$ . A single number. Why this step? Multiplication needs inner sizes to match; here $2$ – $2$ and $2$ – $2$ match, and the surviving outer sizes are $1$ and $1$ . Look at the figure: the row on the left and column on the right pinch down to one cell.

Step 2. Compute inner product first: $A x = [3012] [12] = [3 + 2 0 + 4] = [54]$ . Why this step? Grouping $A x$ first keeps arithmetic simple.

Step 3. Then $x^{T} (A x) = [12] [54] = 1 \cdot 5 + 2 \cdot 4 = 13$ . Why this step? A row times a column is exactly the dot product — it collapses to one number, $13$ .

Step 4. The trap: $A x^{T}$ tries $(2 \times 2) (1 \times 2)$ . Inner sizes $2$ and $1$ do not match, so the product does not exist. Why this step? Transposing $x$ turned a legal column into an illegal-position row. Order and orientation both matter.

Verify: $x^{T} A x = 13$ (a scalar). Cross-check via the entry formula $\sum_{i, j} x_{i} A_{ij} x_{j} = 1 \cdot 3 \cdot 1 + 1 \cdot 1 \cdot 2 + 2 \cdot 0 \cdot 1 + 2 \cdot 2 \cdot 2 = 3 + 2 + 0 + 8 = 13$ ✓.

Recall check

Recall Every cell in one breath

Rectangular flips shape ::: $m \times n \to n \times m$ (Ex 1, 3) Diagonal entries under transpose ::: stay fixed, since $i = j$ (Ex 2, 7) $(A B)^{T}$ equals ::: $B^{T} A^{T}$ — order reverses (Ex 4) $(A B C)^{T}$ equals ::: $C^{T} B^{T} A^{T}$ (Ex 5) Skew-symmetric diagonal must be ::: all zeros (Ex 6) $(A^{- 1})^{T}$ equals ::: $(A^{T})^{- 1}$ (Ex 8) Shape of $x^{T} A x$ ::: a single number $1 \times 1$ (Ex 10)