Properties of transpose

Flip a matrix across its main diagonal and you get its transpose — but the syllabus rarely stops at computing a single transpose. It asks how transpose behaves when matrices are added, scaled, or multiplied together, because those four rules turn transpose from a bookkeeping trick into a genuine proof tool. The star rule, and the one everybody forgets first, is that transposing a product reverses the order of multiplication: $(AB)^T = B^TA^T$ , not $A^TB^T$ . That single reversal is what lets you prove, in a couple of lines and without touching a single number, that $AA^T$ and $A^TA$ are symmetric for any matrix $A$ — a fact this chapter leans on constantly when it later defines symmetric and skew-symmetric matrices. This section builds the four properties, proves the reversal rule is forced by matrix shapes, and applies them.

The core insight

Transpose distributes over sums and scalars normally but reverses order over products, $(AB)^T = B^TA^T$ , and this reversal is exactly what proves $AA^T$ is always symmetric.

The core idea

Let $A$ and $B$ be matrices for which the stated operation is defined, and let $k$ be a scalar. Transpose obeys four standard properties.

(A^T)^T = A

where: flipping across the diagonal twice returns the original matrix — the double transpose undoes itself.

(A+B)^T = A^T + B^T

where: $A$ , $B$ must have the same order for $A+B$ to exist; transposing a sum equals the sum of the transposes.

(kA)^T = k\,A^T

where: $k$ is any scalar; it stays outside and simply multiplies the transposed matrix.

(AB)^T = B^T A^T

where: $A$ is $m\times n$ , $B$ is $n\times p$ , so $AB$ is $m\times p$ and $(AB)^T$ is $p\times m$ ; the order of the two factors flips. The fourth rule is the one that matters most in practice. It extends to a chain of any length, so for three matrices $(ABC)^T = C^TB^TA^T$ — the whole order reverses, not just the last pair. Why must it reverse? Track the shapes. If $A$ is $m\times n$ and $B$ is $n\times p$ , then $AB$ is $m\times p$ , so $(AB)^T$ is $p\times m$ . Now $B^T$ is $p\times n$ and $A^T$ is $n\times m$ , so the product $B^TA^T$ is $p\times m$ — matching. The product $A^TB^T$ , by contrast, is $n\times m$ times $p\times n$ , which is not even defined unless $m=p$ . So the reversed order isn't a memory trick to be forgotten under pressure; it's the only order that keeps the matrices multiplication-compatible in general. This topic sits directly after "Transpose of a matrix" in the CBSE Class 12 Matrices chapter and supplies the algebraic toolkit the chapter needs next, when it defines symmetric matrices ( $A^T = A$ ) and skew-symmetric matrices ( $A^T = -A$ ). Almost every proof about symmetric or skew-symmetric matrices — including the standard result that any square matrix can be written as the sum of a symmetric part and a skew-symmetric part — leans on exactly these four properties, particularly the reversal rule, rather than on entry-by-entry computation. Getting comfortable manipulating transposes abstractly, without plugging in numbers, is therefore the real skill this topic is building toward.

Figure

Shape check showing that reversed order B transpose A transpose matches the shape of AB transpose while the unreversed order fails to be defined

Derivation AProving AA transpose is always symmetric

Using only the four transpose properties above — with no reference to the specific numerical entries of $A$ — show that for any matrix $A$ (square or rectangular), the product $AA^T$ satisfies $(AA^T)^T = AA^T$ , which is precisely the definition of a symmetric matrix, so $AA^T$ is symmetric for every choice of $A$ .

Step 1. Treat $AA^T$ as the product of two matrices, $A$ and $A^T$ , and apply the reversal rule $(XY)^T = Y^TX^T$ with $X=A$ , $Y=A^T$ :

(AA^T)^T = (A^T)^T\,A^T

Step 2. Apply the double-transpose rule $(A^T)^T = A$ to simplify the first factor:

(AA^T)^T = A\,A^T

Step 3. Compare this result to the matrix we started with. The right-hand side, $AA^T$ , is exactly the original matrix — so $(AA^T)^T = AA^T$ .

Step 4. By definition, a matrix that equals its own transpose is symmetric. Since Steps 1–3 used only the general transpose properties (never any specific entries of $A$ ), this holds for every matrix $A$ , whatever its order — even if $A$ itself is not square, $AA^T$ is always a square, symmetric matrix. The identical argument, starting from $A^TA$ instead, proves $A^TA$ is symmetric too.

Worked example

Verify the reversal rule $(AB)^T = B^TA^T$ numerically for $A = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}$ , then use the same $A$ to confirm $AA^T$ is symmetric.

Step 1. Compute $AB$ entry by entry: $(1,1) = (1)(2)+(2)(1) = 4$ ; $(1,2) = (1)(1)+(2)(0) = 1$ ; $(2,1) = (0)(2)+(3)(1) = 3$ ; $(2,2) = (0)(1)+(3)(0) = 0$ . So $AB = \begin{bmatrix} 4 & 1 \\ 3 & 0 \end{bmatrix}$ , and transposing gives $(AB)^T = \begin{bmatrix} 4 & 3 \\ 1 & 0 \end{bmatrix}$ .

Step 2. Write $B^T = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}$ and $A^T = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}$ , then multiply $B^TA^T$ : $(1,1) = (2)(1)+(1)(2) = 4$ ; $(1,2) = (2)(0)+(1)(3) = 3$ ; $(2,1) = (1)(1)+(0)(2) = 1$ ; $(2,2) = (1)(0)+(0)(3) = 0$ . So $B^TA^T = \begin{bmatrix} 4 & 3 \\ 1 & 0 \end{bmatrix}$ , matching Step 1 exactly and confirming $(AB)^T = B^TA^T$ .

Step 3. Now compute $AA^T$ using the same $A$ : $A A^T = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix}$ , giving $(1,1)=1+4=5$ , $(1,2)=0+6=6$ , $(2,1)=0+6=6$ , $(2,2)=0+9=9$ . So $AA^T = \begin{bmatrix} 5 & 6 \\ 6 & 9 \end{bmatrix}$ .

Step 4. Check symmetry directly: swapping rows and columns of $AA^T$ leaves the off-diagonal entries both equal to $6$ , so $(AA^T)^T = AA^T$ . This matches the general proof above — the reversal rule verifies numerically, and $AA^T = \begin{bmatrix} 5 & 6 \\ 6 & 9 \end{bmatrix}$ is confirmed symmetric.

Competitive stretch

For a $3\times2$ (non-square) matrix $A$ , $AA^T$ is a $3\times3$ symmetric matrix while $A^TA$ is a different $2\times2$ symmetric matrix — both symmetric by the same proof, even though $A$ itself has no transpose-equals-itself property to speak of, since it isn't even square. The same reversal trick handles a related favourite: showing $B = \frac{1}{2}(A+A^T)$ is symmetric and $C = \frac{1}{2}(A-A^T)$ is skew-symmetric for any square $A$ , so that $A = B + C$ splits any square matrix into a symmetric part plus a skew-symmetric part — apply $(A+B)^T=A^T+B^T$ , $(kA)^T=kA^T$ and $(A^T)^T=A$ to $B^T$ and $C^T$ directly, and both facts fall out in two lines, exactly as in the derivation above.

Common traps

Writing $(AB)^T = A^TB^T$ without reversing the order — the single most common transpose slip, and it silently breaks any proof that relies on it.
Applying $(A+B)^T = A^T+B^T$ when $A$ and $B$ don't share the same order, so $A+B$ was never defined to begin with.
Dropping the scalar in $(kA)^T = kA^T$ , or writing it as just $A^T$ .
Trying to prove $AA^T$ is symmetric by checking one numerical example — that shows it's true for that matrix, not for every matrix; a real proof must run abstractly through the properties, as in the derivation above.

Ask the tutor: Want to walk through proving that $A - A^T$ is always skew-symmetric using these same properties, or verify another product transpose with your own numbers?

Why does the transpose of a product reverse order instead of staying in the same order

Because the shapes force it. If $A$ is $m\times n$ and $B$ is $n\times p$ , then $(AB)^T$ must be $p\times m$ . Only $B^TA^T$ (shapes $p\times n$ times $n\times m$ ) produces a $p\times m$ result in general; $A^TB^T$ usually isn't even a valid multiplication. So the reversal isn't arbitrary — it's the only order that keeps the matrices compatible.

Do I need to memorise all four transpose properties separately, or can I derive them

Memorise them — they're used constantly and re-deriving from scratch under exam time pressure wastes minutes. But understanding why the product rule reverses (via the shape argument) makes it far less likely you'll forget it or misremember the order, unlike the other three which are simpler and rarely get flipped.

Is AA transpose always the same matrix as A transpose A

Not generally — for a non-square $A$ , they even have different sizes ( $AA^T$ is $m\times m$ , $A^TA$ is $n\times n$ ). Both are symmetric by the transpose properties, but they are typically different matrices with different entries; only in special cases, like when $A$ itself is symmetric or orthogonal, do extra relationships appear.

Properties of transpose

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads