Properties of Matrices Transpose

A collection of numbers arranged in the fixed number of rows and columns is called a matrix. It is a rectangular array of rows and columns. When we swap the rows into columns and columns into rows of the matrix, the resultant matrix is called the Transpose of the given matrix.

This interchanging of rows and columns of the actual matrix is Matrices Transposing.

If M[ ij ] is a m x n matrix, and we want to find the transpose of this matrix, we need to interchange the rows to columns and columns to rows. It would be denoted by MT or M’. So if M = [M[ ij ] ]m x n is the original matrix, then M’ = [M[ ji ] ]n x m is the transpose of it.

For example: M = \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\)

the M’ = \(\begin{bmatrix} 2 & 5\\ 3 & 6\\ 4& 7 \end{bmatrix}\)

Properties of matrices transpose with examples

  1. Transpose of transpose of a matrix is the matrix itself. [MT]T = M
  2. For example: M = \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\)

    the M’ = \(\begin{bmatrix} 2 & 5\\ 3 & 6\\ 4& 7 \end{bmatrix}\)

    and [M’]’ = \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\)

  1. If there’s a scalar a, then the transpose of the matrix M times the scalar (a) is equal to the constant times the transpose of the matrix M’
  2. I.e (aM)T = aMT.

    For example:

    if M = \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\) and constant a = 2 ,then

    LHS : [aM]T = (2 \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\))T

    I.e \(\begin{bmatrix} 4 & 6 & 8\\ 10 & 12 & 14 \end{bmatrix}\)T

    \(\begin{bmatrix} 4 & 10\\ 6 & 12\\ 8 & 14 \end{bmatrix}\)

    RHS: a[M]T = 2 (\(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\))T

    = 2 (\(\begin{bmatrix} 2 & 5\\ 3 & 6\\ 4& 7 \end{bmatrix}\))

    = \(\begin{bmatrix} 4 & 10\\ 6 & 12\\ 8 & 14 \end{bmatrix}\)

    So, LHS = RHS

  1. The sum of transposes of matrices is equal to the transpose of the sum of two
  2. matrices.

    (M + N )T = MT + NT

    M = \(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}\)

    N = \(\begin{bmatrix} 8 & 9 & 10\\ 11 & 12 & 13 \end{bmatrix}\)

    Proof :

    (M + N )T = MT + NT

    LHS = (\(\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix}+\begin{bmatrix} 8 & 9 & 10\\ 11 & 12 & 13 \end{bmatrix}\))T

    \((\begin{bmatrix}2 + 8 & 3 + 9 & 4 + 10\\ 5 + 11 & 6 + 12 & 7 + 13\end{bmatrix})\)T

    ( \(\begin{bmatrix} 10 & 12 & 14\\ 16 & 18 & 20 \end{bmatrix}\))T

    \(\begin{bmatrix} 10 & 16\\ 12 & 18\\ 14 & 20 \end{bmatrix}\)

    RHS = \((\begin{bmatrix} 2 & 3 & 4\\ 5 & 6 & 7 \end{bmatrix})^{T} + (\begin{bmatrix} 8 & 9 & 10\\ 11 & 12 & 13 \end{bmatrix})^{T}\)

    = (\(\begin{bmatrix} 2 & 5\\ 3 & 6\\ 4& 7 \end{bmatrix}\)) +

    (\(\begin{bmatrix} 8 & 11\\ 9 & 12\\ 10 & 13 \end{bmatrix}\))

    = (\(\begin{bmatrix} 2 + 8 & 5 + 11\\ 3 + 9& 6 + 12\\ 4 + 10& 7 + 13\end{bmatrix}\))

    =\(\begin{bmatrix} 10 & 16\\ 12 & 18\\ 14 & 20 \end{bmatrix}\)

    LHS = RHS

  1. the product of the transposes of two matrices in reverse order is equal to the
  2. transpose of the product of them. (MN)T = NT MT

    The above property is true for any product of any number of matrices.

    LHS = (MN)T = \((\begin{bmatrix} 1 & 2\\ 3 & 4\\ 5 & 6 \end{bmatrix} X \begin{bmatrix} 7 & 8\\ 9 & 10\\ 11 & 12 \end{bmatrix}) ^{T}\)

    = (\(\begin{bmatrix} 1 X 7 & 2 X 8\\ 3 X 9 & 4 X 10\\ 5 X 11 & 6 X 12\end{bmatrix}\))T

    =(\(\begin{bmatrix} 7 & 16\\ 27 & 40\\ 55 & 72 \end{bmatrix}\))T

    = \(\begin{bmatrix} 7 & 27 & 55\\ 16 & 40 & 72 \end{bmatrix}\)

    RHS = \((\begin{bmatrix} 7 & 8\\ 9 & 10\\ 11 & 12 \end{bmatrix})^{T} X (\begin{bmatrix} 1 & 2\\ 3 & 4\\ 5 & 6 \end{bmatrix})^{T}\)

    = \((\begin{bmatrix} 7 & 9 & 11\\ 8 & 10 & 12 \end{bmatrix}) \, X (\begin{bmatrix} 1 & 3 & 5\\ 2 & 4 & 6 \end{bmatrix})\)

    = (\(\begin{bmatrix} 7 X 1 & 9 X 3& 11 X 5\\ 8 X 2 & 10 X 4 & 12 X 6\end{bmatrix}\))

    = (\(\begin{bmatrix} 7 & 27 & 55\\ 16 & 40 & 72 \end{bmatrix}\))

    LHS = RHS

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