Visual walkthrough — Transpose — definition, properties

The parent note Transpose — definition, properties lists five properties. Four of them are gentle. But one is the deep one, the one everyone gets wrong: the reversal rule for a product. Here we build it from nothing but pictures — grids, arrows, and one careful sum. By the end you will see why the order has to flip.

We only need one earlier idea: a matrix is a grid of numbers, and a wikilink to Matrix Operations for how we multiply two grids. Everything else we build here.

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Step 1 — What a grid is, and what "row $i$, column $j$" means

The symbol $[A{]}_{ij}$ is read "the number living at row $i$, column $j$ of the grid $A$."

• $A$ ::: the whole grid of numbers
• $i$ ::: how many boxes down (the row)
• $j$ ::: how many boxes across (the column)
• $[A{]}_{ij}$ ::: the single number at that address

Term by term: on the left $i$ is a row and $j$ is a column; on the right those same two labels are handed to $A$ in the opposite slots. That single swap is the entire idea of transpose.

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Step 2 — What it looks like to flip one grid

$A=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\end{array}\right]\stackrel{\text{flip}}{\to }{A}^T=\left[\begin{array}{cc}{a}_{11}& a_{21}\\ a_{12}& a_{22}\\ a_{13}& a_{23}\end{array}\right]$

Notice the shape changed: a $2\times 3$ grid became a $3\times 2$ grid. Why does the shape flip too? Because rows and columns literally traded jobs — if $A$ had $m$ rows and $n$ columns, then $A^T$ has $n$ rows and $m$ columns. Hold this fact; it is the first hint that the product rule must reverse order.

• $a_{11}$ stays put ::: it is on the diagonal, its address $1,1$ is unchanged by swapping
• $a_{12}$ and $a_{21}$ ::: they trade places across the diagonal

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Step 3 — What multiplication actually does (the tool we need)

Before we can transpose a product, we must be crisp about what a product is. Here is the one tool, and why we use this one: multiplication is the only operation that pairs a row of the left grid with a column of the right grid, and that pairing is exactly what the transpose will scramble.

Read every symbol:

• $[AB{]}_{ij}$ ::: the number we are building, at row $i$, column $j$ of the answer
• $i$ ::: fixes which row of $A$ we slide across (highlighted mint in the figure)
• $j$ ::: fixes which column of $B$ we slide down (highlighted coral)
• $k$ ::: the running partner index — it walks $1,2,\dots ,p$ pairing the $k$-th entry of that row with the $k$-th entry of that column
• ${\sum }_{k=1}^p$ ::: "add up all $p$ of those little products"

The crucial observation for what follows: in $a_{ik}{b}_{kj}$, the letter $i$ sits on the left factor $a$, and the letter $j$ sits on the right factor $b$. Watch those two letters.

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Step 4 — WHAT: write down the transpose of the product

We want $(AB{)}^T$. Apply the transpose definition from Step 1 to the single grid $AB$: swap its two address labels.

WHY this step: the transpose rule says $[C^T{]}_{ij}=[C{]}_{ji}$ for any grid $C$; we just chose $C=AB$.

WHAT IT LOOKS LIKE: in the figure, asking for entry $(i,j)$ of the flipped grid means reaching into the unflipped grid at $(j,i)$ — so we pick row $j$ of $A$ and column $i$ of $B$.

Look hard at the sum on the right. Compared with Step 3, the labels moved:

• $a_{jk}$ ::: now $j$ (the column label of the answer) sits on $A$ — because we asked for row $j$ of $A$
• $b_{ki}$ ::: now $i$ (the row label of the answer) sits on $B$
• $k$ ::: still the running partner, still summed from $1$ to $p$

So after the flip, $i$ ended up glued to $B$ and $j$ ended up glued to $A$. This is the whole mystery in one line. Now we ask: what product naturally produces exactly this pattern?

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Step 5 — WHY: build $B^T{A}^T$ and watch the same sum appear

We guess the answer is $B^T{A}^T$ (right times left, reversed) and check by computing its $(i,j)$ entry with the Step 3 rule.

WHY multiply $B^T$ by $A^T$ in that order? Because for the shapes to even fit: $A$ was $m\times p$ and $B$ was $p\times n$, so $A^T$ is $p\times m$ and $B^T$ is $n\times p$. The only way to multiply them is $B^T(n\times p)$ times $A^T(p\times m)$ — the reversed order is the only legal order. The picture below shows the mismatched vs matched dimensions.

Now unflip the two transposes inside the sum using Step 1:

• $[B^T{]}_{ik}=b_{ki}$ ::: swap the labels of $B^T$'s address
• $[A^T{]}_{kj}=a_{jk}$ ::: swap the labels of $A^T$'s address

Substitute: $[B^T{A}^T{]}_{ij}={\sum }_{k=1}^p{b}_{ki}{a}_{jk}={\sum }_{k=1}^p{a}_{jk}{b}_{ki}$

The last step just reorders two ordinary numbers being multiplied ($b_{ki}{a}_{jk}=a_{jk}{b}_{ki}$) — plain number multiplication doesn't care about order.

WHAT IT LOOKS LIKE: this final sum is letter-for-letter identical to the one at the end of Step 4.

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Step 6 — The two sums are the same, so the rule is proved

Set the Step 4 result equal to the Step 5 result:

$[(AB{)}^T{]}_{ij}={\sum }_{k=1}^p{a}_{jk}{b}_{ki}=[B^T{A}^T{]}_{ij}$

Since this holds at every address $(i,j)$, the two whole grids are equal:

The order reverses because transposing sent the label $i$ from the left factor over to the right factor, and sent $j$ from the right factor over to the left. To keep every letter attached to the correct grid, the grids themselves had to swap sides. See Symmetric Matrices for the beautiful special case where $A^T=A$ and nothing needs flipping.

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Step 7 — Edge cases: don't let a corner surprise you

A derivation you can't stress-test is a derivation you don't trust. Four cases:

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The one-picture summary

The whole proof is one migration of two little letters: $i$ starts on $A$, $j$ starts on $B$; the flip trades them; the only grids that catch them correctly are $B^T$ then $A^T$.

Recall Feynman retelling — say it like you'd tell a friend

A matrix is a grid where every number has an address: row $i$, column $j$. Transposing means "swap the two parts of every address" — rows tip over into columns. Multiplying two grids means: take a row from the left grid, a column from the right grid, multiply matching entries, add them up. Now, the entry at address $(i,j)$ of a product secretly carries $i$ on the left grid and $j$ on the right grid. When you transpose the product, you swap $i$ and $j$ — so now $i$ wants to be on the right grid and $j$ on the left. To honor that, you flip both grids and swap their order: right-one-transposed first, left-one-transposed second. That's why $(AB{)}^T=B^T{A}^T$ reverses the order. Addition never does this, because addition never mixes rows with columns — so $(A+B{)}^T=A^T+B^T$ keeps its order. Simple as a pair of dance partners switching sides.

Recall Quick self-test

State the reversal rule and its three-factor version. ::: $(AB{)}^T=B^T{A}^T$ and $(ABC{)}^T=C^T{B}^T{A}^T$. Why can't the order stay the same? ::: because the sum ${\sum }_k{a}_{jk}{b}_{ki}$ from the transpose only matches $B^T{A}^T$, not $A^T{B}^T$.

Related paths from here: Determinants ($\det (A^T)=\det A$), Orthogonal Matrices ($A^{T}A=I$), and Eigenvalues and Eigenvectors where $A$ and $A^T$ share the same eigenvalues.