2.5: The Transpose
Another important operation on matrices is that of taking the transpose . For a matrix \(A\), we denote the transpose of \(A\) by \(A^T\). Before formally defining the transpose, we explore this operation on the following matrix.
\[\left[ \begin{array}{cc} 1 & 4 \\ 3 & 1 \\ 2 & 6 \end{array} \right] ^{T}= \ \ \left[ \begin{array}{ccc} 1 & 3 & 2 \\ 4 & 1 & 6 \end{array} \right] \nonumber\]
What happened? The first column became the first row and the second column became the second row. Thus the \(3\times 2\) matrix became a \(2\times 3\) matrix. The number \(4\) was in the first row and the second column and it ended up in the second row and first column.
The definition of the transpose is as follows.
Let \(A\) be an \(m\times n\) matrix. Then \(A^{T}\), the transpose of \(A\), denotes the \(n\times m\) matrix given by
\[A^{T} = \left[ a _{ij}\right] ^{T}= \left[ a_{ji} \right]\nonumber \]
The \(\left( i, j \right)\)-entry of \(A\) becomes the \(\left( j,i \right)\)-entry of \(A^T\).
Consider the following example.
Calculate \(A^T\) for the following matrix
\[A = \left[ \begin{array}{rrr} 1 & 2 & -6 \\ 3 & 5 & 4 \end{array} \right] \nonumber\]
Solution
By Definition \(\PageIndex{1}\) , we know that for \(A = \left[ a_{ij} \right]\), \(A^T = \left[ a_{ji} \right]\). In other words, we switch the row and column location of each entry. The \(\left( 1, 2 \right)\)-entry becomes the \(\left( 2,1 \right)\)-entry.
Thus, \[A^T = \left[ \begin{array}{rr} 1 & 3 \\ 2 & 5 \\ -6 & 4 \end{array} \right] \nonumber\]
Notice that \(A\) is a \(2 \times 3\) matrix, while \(A^T\) is a \(3 \times 2\) matrix.
The transpose of a matrix has the following important properties.
Let \(A\) be an \(m\times n\) matrix, \(B\) an \(n\times p\) matrix, and \(r\) and \(s\) scalars. Then
- \[\left(A^{T}\right)^{T} = A\nonumber \]
- \[\left( AB\right) ^{T}=B^{T}A^{T} \nonumber\]
- \[\left( rA+ sB\right) ^{T}=rA^{T}+ sB^{T} \nonumber\]
- Proof
-
First we prove 2. From Definition \(\PageIndex{1}\) ,
\[ \begin{aligned} \left(AB\right)^{T} &= \left[ (AB) _{ij} \right] ^{T}=\left[ (AB)_{ji} \right]=\sum_{k}a_{jk}b_{ki}= \sum_{k}b_{ki}a_{jk} \\[4pt] &= \sum_{k}\left[ b_{ik}\right]^{T}\left[ a_{kj}\right]^{T}=\left[ b_{ij}\right] ^{T} \left[ a_{ij}\right]^{T} = B^{T}A^{T} \end{aligned}\]
The proof of Formula 3 is left as an exercise.
Below is a video on the transpose of a matrix.
The transpose of a matrix is related to other important topics. Consider the following definition.
An \(n\times n\) matrix \(A\) is said to be symmetric if \(A=A^{T}.\) It is said to be skew symmetric if \(A=-A^{T}.\)
We will explore these definitions in the following examples.
Let
\[A=\left[ \begin{array}{rrr} 2 & 1 & 3 \\ 1 & 5 & -3 \\ 3 & -3 & 7 \end{array} \right] \nonumber\]
Use Definition \(\PageIndex{2}\) to show that \(A\) is symmetric.
Solution
By Definition \(\PageIndex{2}\) , we need to show that \(A = A^T\). Now, using Definition \(\PageIndex{1}\) ,
\[A^{T} = \left[ \begin{array}{rrr} 2 & 1 & 3 \\ 1 & 5 & -3 \\ 3 & -3 & 7 \end{array} \right]\nonumber\]
Hence, \(A = A^{T}\), so \(A\) is symmetric.
Let
\[A=\left[ \begin{array}{rrr} 0 & 1 & 3 \\ -1 & 0 & 2 \\ -3 & -2 & 0 \end{array} \right] \nonumber \]
Show that \(A\) is skew symmetric.
Solution
By Definition \(\PageIndex{2}\) ,
\[A^{T} = \left[ \begin{array}{rrr} 0 & -1 & -3\\ 1 & 0 & -2\\ 3 & 2 & 0 \end{array} \right] \nonumber\]
You can see that each entry of \(A^T\) is equal to \(-1\) times the same entry of \(A\). Hence, \(A^{T} = - A\) and so by Definition \(\PageIndex{2}\) , \(A\) is skew symmetric.