Properties of Transpose of Matrix

Last Updated : 23 Jul, 2025

The transpose of a matrix is a new matrix that is formed by flipping the original matrix over its diagonal. This operation essentially switches the rows and columns of the matrix.
If you have a matrix A, the transpose of matrix A is denoted as AT.

Example: For a matrix A its transpose AT is given as :

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ \end{pmatrix} A^T = \begin{pmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \\ a_{13} & a_{23} \\ \end{pmatrix}

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Matrix Transpose Properties

Let's learn about the important properties of the transpose of a matrix:

  • A square matrix "A" of order "n × n" is said to be an orthogonal matrix, if AAT = ATA = I, where "I" is an identity matrix of order "n × n."
  • A square matrix "A" of order "n × n" is said to be a symmetric matrix if its transpose is the same as the original matrix, i.e., AT = A.
  • A square matrix "A" of order "n × n" is said to be a skew-symmetric matrix if its transpose is equal to the negative of the original matrix, i.e., AT = –A.
  • Double Transpose of a Matrix: Transpose of the transpose matrix is the original matrix itself. (At)t = A
  • Transpose of Product of Matrices: This property says that  (AB)t = BtAt

Proof:

If matrices A and B are of orders m × n and n × p, respectively.

At and Bt are the transpose of matrices A and B of orders n × m and p × n respectively (from the product rule of matrices).
It implies, if A = [a(ij)], and At = [c(ji)]

Then, [c(ji)] = [a(ij)]

and, If B = [b(jk)], and Bt = [d(kj)]
Then, [d(kj)] = [b(jk)]

Now, from the product rule of matrices, we can write,

AB is m × p matrix and (AB)t is p × m matrix.
Also, Bt is a p × n matrix, and At is an n × m matrix.

This implies that, (Bt)(At) is a p × m matrix.

Therefore, (AB)t and (Bt)(At) are both p × m matrices.

Now we can write,
((k, i)th element of (AB)t = (i, k)th element of AB 
\sum_{j=1}^{n}  a_{ij}  b_{jk}     \sum_{j=1}^{n}  c_{ji}  d_{kj}

\sum_{j=1}^{n}  d_{kj}  c_{ji}

(k, i)th element of (Bt)(At)
Therefore, the elements of (AB)t and (Bt)(At) are equal.

Therefore, (AB)t = (Bt)(At)

  • Multiplication by Constant: If a matrix is multiplied by a scalar value and its transpose is taken, then the resultant matrix will be equal to the transpose of the original matrix multiplied by the scalar value, i.e., (kA)t = kAt, where k is a scalar value.

Proof:

Let us consider a matrix A = [aij]m × n and a scalar k.

The order of the given matrix A is m × n.

If matrix A is multiplied by the scalar value k, then all the elements of the matrix are multiplied with this scalar constant k, however, the order of matrix kA remain same, i.e., m × n.

Now, the order of the transpose of the matrix kA, i.e., (kA)t will be n × m.
As the order of the matrix A is m × n, the order of its transpose matrix, i.e., At will be n × m.

If matrix At is multiplied by the scalar value k, then the order of the matrix kAt will also be n × m.
So, the order of the matrices (kA)t and kAt is the same, i.e., n × m.

Now, let us prove that the corresponding elements of (kA)t and kAt are equal.
The (i, j)th element of (kA)t will be equal to the (j, i)th element of kA.

(i, j)th element of (kA)t = (j, i)th element of kA ⇒ (i, j)th element of (kA)t = (i, j)th element of kAt

So, we say that the corresponding elements of (kA)t and kAt are equal.

As the order and corresponding elements of (kA)t and kAt are equal, 

Therefore, we can conclude that (kA)t = kAt.

  • Transpose of Addition of Matrices: This property says that, (A + B)t = At + Bt

Proof:

Here A and B are two matrices of order m × n
Let, A = [a(ij)] and B = [b(ij)] of order m × n.

So, (A + B) is also of order m × n matrix
Also, At and Bt are of order n × m matrices.

So, the Transpose of matrix (A + B) or (A + B)t is an n × m matrix.
Now we can say, At + Bt is also an n × m matrix.

Now, from the transpose rule,
(j, i)th element of (A + B)t = (i, j)th element of (A + B)

= (i, j)th element of A + (i, j)th element of B 
= (j, i)th element of At + (j, i)th element of Bt
= (j, i)th element of (At + Bt)

Therefore,
(A + B)t = At + Bt

  • If "A" is a square matrix of any order and is invertible, then the inverse of its transpose is equal to the transpose of the inverse of the original matrix, i.e.,

(At)-1 = (A-1)t

To prove that (At)-1 = (A-1)t

Proof:
Let us consider a non-singular square matrix A.
RHS = (A-1)t

Now, multiply (A-1)t by At
= (A-1)t × At

We know that (AB)t = BtAt
So, (A-1)tAt = (AA-1)t

We know that the AA-1 = I, where "I" is an identity matrix.
So, (A-1)tAt = It

⇒ (A-1)tAt = I (Since, It = I)
⇒ (A-1)t = (At)-1 = LHS

Hence proved.
Therefore, (At)-1 = (A-1)t

Solved Examples of Properties of Transpose of Matrices

Example 1: For matrices, A =  \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix}      and  B = \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}
Prove that for these matrices hold the property, (AB)t = (Bt)(At)
Solution:

Here A and B are 2 × 3 and 3 × 2 matrices respectively. So, by the product rule of a matrix, we can find their product and the final matrices would be of 2 × 2 matrix.

L.H.S

Now,
AB=  \begin{bmatrix} -2 & 1 & 3\\ 0 & 4 & -1 \end{bmatrix} \times  \begin{bmatrix} 2 & 1 \\-3 & 0 \\ 4 & -5 \end{bmatrix}   

AB =\begin{bmatrix} (-2)×2+1×(-3)+3×4 & (-2)×1+1×0+3×(-5) \\ 0×2+4×(-3)+(-1)×4 & 0×1+4×0+(-1)×(-5) \end{bmatrix}

AB= \begin{bmatrix} 5 & -17 \\ -16 & 5 \end{bmatrix}

So, Transpose of matrix AB is,

(AB)^{t} =  \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}     \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}

R.H.S
A^{t} =  \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}
and
B^{t} = \begin{bmatrix} 2 & -3 & 4 \\ 1 & 0 & -5 \end{bmatrix}

So,
B^{t}A^{t} =  \begin{bmatrix} 2 & -3 & 4\\ 1 & 0 & -5 \end{bmatrix} \times  \begin{bmatrix} -2 & 0 \\ 1 & 4 \\ 3 & -1 \end{bmatrix}

B^{t}A^{t} =  \begin{bmatrix} 2×(-2)+(-3)×1+4×3 & 2×0+(-3)×4+4×(-1) \\ 1×(-2)+0×1+(-5)×3 & 1×0+0×4+(-5)×(-1) \end{bmatrix}

B^{t}A^{t} =  \begin{bmatrix} 5 & -16 \\ -17 & 5 \end{bmatrix}

Therefore,
(AB)t = BtAt

Example 2: For matrices  A = \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix}      and  B = \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}
Prove that these matrices hold this property, (A + B)t = At + Bt
Solution:

L.H.S
(A+B)= \begin{bmatrix} -1 & 5 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\5 & 4 \end{bmatrix}  

= \begin{bmatrix} (-1)+3 & 5+(-2) \\ 3+5 & 2+4 \end{bmatrix}     = \begin{bmatrix} 2 & 3 \\ 8 & 6 \end{bmatrix}
So, 
(A+B)^{t} =  \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}

R.H.S
A^{t} =  \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix} 
and, 
B^{t} =  \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}
Now, 

A^{t} + B^{t} =  \begin{bmatrix} -1 & 3 \\ 5 & 2 \end{bmatrix} +  \begin{bmatrix} 3 & 5 \\ -2 & 4 \end{bmatrix}  

= \begin{bmatrix} (-1)+3 & 3+5 \\ 5+(-2) & 2+4 \end{bmatrix}

 A^{t} + B^{t} =  \begin{bmatrix} 2 & 8 \\ 3 & 6 \end{bmatrix}

Therefore, 
(A + B)t = At + Bt

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