Transpose — definition, properties: Exercises

This page is a self-test ladder. Each problem builds on the parent note and climbs from "can you spot it?" to "can you invent with it?". Every symbol used here was defined there — but we re-anchor each idea so you can follow from line one.

Reminder of our one core rule, in plain words: the transpose $A^{T}$ is the matrix you get by turning every row of $A$ into a column. Formally, the entry in row $i$ , column $j$ of $A^{T}$ is the entry in row $j$ , column $i$ of $A$ : $[A^{T}]_{ij} = [A]_{j i}$

The figure below shows this rule in action on the very first exercise — study it before you start.

Figure — Transpose — definition, properties

L1 — Recognition

Problem 1.1

Write the transpose of $A = [24 - 1 7 05] .$

Recall Solution 1.1

$A$ has $2$ rows and $3$ columns, so $A^{T}$ has $3$ rows and $2$ columns.

Row 1 of $A$ is $[2, - 1, 0]$ → this becomes column 1 of $A^{T}$ (the blue block in the figure).
Row 2 of $A$ is $[4, 7, 5]$ → this becomes column 2 of $A^{T}$ (the green block). $A^{T} = 2 - 1 0 475 .$ Check the rule: $[A^{T}]_{31} = 0$ should equal $[A]_{13} = 0$ . ✓

Problem 1.2

Is this matrix symmetric, skew-symmetric, or neither? $S = 0 - 3 2 30 - 4 - 2 40 .$

Recall Solution 1.2

A matrix is symmetric if flipping it gives back the same matrix ( $S^{T} = S$ ), and skew-symmetric if flipping it gives the negative ( $S^{T} = - S$ ).

First, the diagonal is all zeros — that's the tell-tale sign of skew-symmetric (a symmetric matrix can have any diagonal, but a skew one must have zeros because $a_{ii} = - a_{ii}$ forces $a_{ii} = 0$ ).

Flip it: $S^{T} = 03 - 2 - 3 04 2 - 4 0 = - S .$ So $S$ is skew-symmetric. See Symmetric Matrices.

L2 — Application

Problem 2.1

For $A = [1023]$ and $B = [42 - 1 5]$ , verify $(A B)^{T} = B^{T} A^{T}$ by computing both sides.

Recall Solution 2.1

Left side — multiply then flip. Recall matrix product: entry $(i, j)$ = (row $i$ of first) · (column $j$ of second). $A B = [1 (4) + 2 (2) 0 (4) + 3 (2) 1 (- 1) + 2 (5) 0 (- 1) + 3 (5)] = [86915] .$ Flip it: $(A B)^{T} = [89615] .$ Right side — flip each, then multiply in reversed order. $B^{T} = [4 - 1 25], A^{T} = [1203] .$ $B^{T} A^{T} = [4 (1) + 2 (2) - 1 (1) + 5 (2) 4 (0) + 2 (3) - 1 (0) + 5 (3)] = [89615] .$ They match. ✓ See Matrix Operations.

Problem 2.2

Given $A = [3 - 2 14]$ , compute $(3 A)^{T}$ and confirm it equals $3 A^{T}$ .

Recall Solution 2.2

$3 A = [9 - 6 312] ⟹ (3 A)^{T} = [93 - 6 12] .$ Separately: $A^{T} = [31 - 2 4] ⟹ 3 A^{T} = [93 - 6 12] .$ Equal. ✓ Scalars pass straight through the flip — the mirror doesn't care about a stretch.

L3 — Analysis

Problem 3.1

A square matrix $M$ satisfies $M^{T} = M$ (symmetric). If $M = [15 x 3]$ , find $x$ .

Recall Solution 3.1

Symmetric means $[M]_{ij} = [M]_{j i}$ . Compare position $(1, 2)$ with position $(2, 1)$ : $[M]_{12} = x, [M]_{21} = 5.$ For symmetry they must be equal, so $x = 5$ . $M = [1553] .$ Picture: the entries $x$ and $5$ sit as mirror images across the diagonal. Symmetry forces the mirror image to be identical.

Problem 3.2

Every square matrix $A$ can be split into a symmetric part $S$ and a skew-symmetric part $K$ via $S = \frac{1}{2} (A + A^{T}), K = \frac{1}{2} (A - A^{T}), A = S + K .$ Verify this decomposition for $A = [2064]$ , and check $S$ is symmetric and $K$ is skew.

Recall Solution 3.2

$A^{T} = [2604] .$ Symmetric part: $S = \frac{1}{2} ([2064] + [2604]) = \frac{1}{2} [4668] = [2334] .$ Skew part: $K = \frac{1}{2} ([2064] - [2604]) = \frac{1}{2} [0 - 6 60] = [0 - 3 30] .$ Check symmetry: $S^{T} = S$ ✓ (mirror image identical). Check skew: $K^{T} = [03 - 3 0] = - K$ ✓ (zero diagonal, off-diagonals negate). Sum: $S + K = [2064] = A$ ✓

Why this works: $S^{T} = \frac{1}{2} (A^{T} + (A^{T})^{T}) = \frac{1}{2} (A^{T} + A) = S$ using the double-transpose rule $(A^{T})^{T} = A$ . The $\frac{1}{2}$ shares $A$ equally between the "mirror-friendly" and "mirror-flipping" halves.

L4 — Synthesis

Problem 4.1

Prove that for any matrix $A$ , the product $A^{T} A$ is always symmetric.

Recall Solution 4.1

Goal: show $(A^{T} A)^{T} = A^{T} A$ . Apply the product-reversal rule $(X Y)^{T} = Y^{T} X^{T}$ with $X = A^{T}$ and $Y = A$ : $(A^{T} A)^{T} = A^{T} (A^{T})^{T} .$ Now use double-transpose $(A^{T})^{T} = A$ : $(A^{T} A)^{T} = A^{T} A .$ Since the flip returns the same matrix, $A^{T} A$ is symmetric. $■$

Why this matters: $A^{T} A$ appears everywhere — least squares, inner products, and the definition of orthogonal matrices where $A^{T} A = I$ . Its guaranteed symmetry is the reason its eigenvalues are always real.

Problem 4.2

$A$ is invertible. Prove $(A^{- 1})^{T} = (A^{T})^{- 1}$ , and use it to compute $(A^{T})^{- 1}$ for $A = [1324] .$

Recall Solution 4.2

Proof. Start from the defining identity $A A^{- 1} = I$ (the inverse is what multiplies $A$ to give the identity $I$ ). Flip both sides and use the reversal rule: $(A A^{- 1})^{T} = I^{T} ⟹ (A^{- 1})^{T} A^{T} = I .$ (We used $I^{T} = I$ : the identity is symmetric.) This says $(A^{- 1})^{T}$ is the matrix that multiplies $A^{T}$ to give $I$ — i.e. it is $(A^{T})^{- 1}$ . So $(A^{- 1})^{T} = (A^{T})^{- 1}$ . $■$

Computation. First $A^{- 1}$ . For a $2 \times 2$ matrix $[a c b d]$ the inverse is $\frac{1}{a d - b c} [d - c - b a]$ . Here $a d - b c = 1 (4) - 2 (3) = - 2$ : $A^{- 1} = \frac{1}{- 2} [4 - 3 - 2 1] = [- 2 1.5 1 - 0.5] .$ By the theorem, $(A^{T})^{- 1} = (A^{- 1})^{T}$ — just flip: $(A^{T})^{- 1} = [- 2 1 1.5 - 0.5] .$ See Matrix Inverse and Determinants (the $- 2$ is $det A$ ).

L5 — Mastery

Problem 5.1

Show that if $A$ is skew-symmetric ( $A^{T} = - A$ ) and $n \times n$ , then the diagonal is all zeros, and check the degenerate $1 \times 1$ case.

Recall Solution 5.1

Skew means $[A]_{ij} = - [A]_{j i}$ for all $i, j$ . Set $i = j$ (a diagonal entry): $[A]_{ii} = - [A]_{ii} ⟹ 2 [A]_{ii} = 0 ⟹ [A]_{ii} = 0.$ So every diagonal entry is $0$ . ✓

Degenerate $1 \times 1$ case: $A = [a]$ . Its transpose is itself, $A^{T} = [a]$ . Skew requires $[a] = - [a]$ , forcing $a = 0$ . So the only $1 \times 1$ skew-symmetric matrix is $[0]$ — consistent with "diagonal must be zero", since a $1 \times 1$ matrix is all diagonal.

Problem 5.2

Let $A$ be $2 \times 3$ . Determine the sizes of $A^{T} A$ and $A A^{T}$ , decide which (if either) can be invertible, and compute both for $A = [100121] .$

Recall Solution 5.2

Sizes. $A$ is $2 \times 3$ , so $A^{T}$ is $3 \times 2$ .

$A^{T} A$ : $(3 \times 2) (2 \times 3) = 3 \times 3$ .
$A A^{T}$ : $(2 \times 3) (3 \times 2) = 2 \times 2$ .

Both are square and (by Problem 4.1's argument) symmetric.

Compute $A A^{T}$ (the smaller one):

A A^T = \begin{bmatrix} 1{\cdot}1+0{\cdot}0+2{\cdot}2 & 1{\cdot}0+0{\cdot}1+2{\cdot}1 \\ 0{\cdot}1+1{\cdot}0+1{\cdot}2 & 0{\cdot}0+1{\cdot}1+1{\cdot}1 \end{bmatrix} = \begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}.$$ Its determinant is $5(2)-2(2) = 6 \ne 0$, so $A A^T$ **is invertible**. **Compute $A^T A$:** $$A^T A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 2 & 1 & 5 \end{bmatrix}.$$ **Can it be invertible?** No. $A$ has only $2$ rows, so its rows can span at most a $2$-dimensional space — the $3\times 3$ result cannot have full rank $3$. Indeed row 3 $=2\cdot(\text{row 1}) + 1\cdot(\text{row 2})$, so $\det(A^T A) = 0$: **not invertible**. **The lesson:** for a "wide" matrix ($m<n$), only $A A^T$ (the small square) can be invertible; the big square $A^T A$ is always singular. This governs which normal-equation form you can solve. See [[Determinants]].

Recall Self-audit checklist

Can you now, without notes... Flip a non-square matrix and state its new size? ::: Yes — rows become columns, $m \times n \to n \times m$ . Recall why $(A B)^{T} = B^{T} A^{T}$ reverses order? ::: The summed inner index swaps sides, forcing the factors to swap. Prove $A^{T} A$ is always symmetric? ::: $(A^{T} A)^{T} = A^{T} (A^{T})^{T} = A^{T} A$ . Say when $A^{T} = A^{- 1}$ ? ::: Only for orthogonal matrices, where $A^{T} A = I$ .