For AI agents: use /llms.txt for the Nakafa content index.
A matrix transpose is a new matrix obtained by interchanging the rows and columns of the original matrix. The elements of the rows become the elements of the columns, and conversely, the elements of the columns become the elements of the rows.
If we have a matrix A A A A A A A T A^T A T
Formally, if matrix A A A m × n m \times n m × n a i j a_{ij} a ij i i i j j j A T A^T A T n × m n \times m n × m a j i T = a i j a_{ji}^T = a_{ij} a j i T = a ij
This means that the element in the j j j i i i A T A^T A T i i i j j j A A A
To obtain the matrix transpose, follow these steps:
Write the first row of the original matrix as the first column of the transpose matrix.
Write the second row of the original matrix as the second column of the transpose matrix.
Continue this process for all rows in the original matrix.
Suppose we have matrix A A A
Then, the transpose of matrix A A A
Notice how the first row [ a b ] \begin{bmatrix} a & b \end{bmatrix} [ a b ] [ a b ] \begin{bmatrix} a \\ b \end{bmatrix} [ a b ] [ c d ] \begin{bmatrix} c & d \end{bmatrix} [ c d ] [ c d ] \begin{bmatrix} c \\ d \end{bmatrix} [ c d ]
Given matrix B B B 2 × 3 2 \times 3 2 × 3
The transpose of matrix B B B B T B^T B T 3 × 2 3 \times 2 3 × 2
The first row of B B B [ 1 2 3 ] \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} [ 1 2 3 ] B T B^T B T
The second row of B B B [ 4 5 6 ] \begin{bmatrix} 4 & 5 & 6 \end{bmatrix} [ 4 5 6 ]
If C C C
Then its transpose, C T C^T C T
Given a square matrix D D D
Then its transpose, D T D^T D T
Some important properties of matrix transpose are:
( A T ) T = A (A^T)^T = A ( A T ) T = A ( A + B ) T = A T + B T (A + B)^T = A^T + B^T ( A + B ) T = A T + B T ( A − B ) T = A T − B T (A - B)^T = A^T - B^T ( A − B ) T = A T − B T ( k A ) T = k A T (kA)^T = kA^T ( k A ) T = k A T k k k ( A B ) T = B T A T (AB)^T = B^T A^T ( A B ) T = B T A T Determine the transpose of the following matrices and state the type of the resulting matrix (e.g., row matrix, column matrix, square matrix).
A = [ 1 3 − 5 ] A = \begin{bmatrix} 1 & 3 & -5 \end{bmatrix} A = [ 1 3 − 5 ]
B = [ 9 − 1 3 0 1 5 ] B = \begin{bmatrix} 9 & -1 \\ 3 & 0 \\ 1 & 5 \end{bmatrix} B = 9 3 1 − 1 0 5
C = [ 2 3 1 1 6 − 2 4 5 − 7 ] C = \begin{bmatrix} 2 & 3 & 1 \\ 1 & 6 & -2 \\ 4 & 5 & -7 \end{bmatrix} C = 2 1 4 3 6 5 1 − 2 − 7
A T = [ 1 3 − 5 ] A^T = \begin{bmatrix} 1 \\ 3 \\ -5 \end{bmatrix} A T = 1 3 − 5 A T A^T A T
B T = [ 9 3 1 − 1 0 5 ] B^T = \begin{bmatrix} 9 & 3 & 1 \\ -1 & 0 & 5 \end{bmatrix} B T = [ 9 − 1 3 0 1 5 ]
C T = [ 2 1 4 3 6 5 1 − 2 − 7 ] C^T = \begin{bmatrix} 2 & 1 & 4 \\ 3 & 6 & 5 \\ 1 & -2 & -7 \end{bmatrix} C T = 2 3 1 1 6 − 2 4 5 − 7
B T B^T B T
C T C^T C T
