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Transposition of matrices

Definition 4.89 (Related exercises: Exercise 4.23, Exercise 4.24)

If \(A\) is an \(m \times n\)-matrix, then the transpose (denoted \(A^T\)) is the \(n \times m\)-matrix obtained by \(A\) by reflecting the entries along the main diagonal. More formally, if \(A = (a_{ij})\), then

\[ A^T := (a_{ji}). \]

Example 4.90

For \(A = \left ( \begin{array}{cc} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{array} \right )\),

\[ A^T = \left ( \begin{array}{ccc} 1 & 3 & 5 \\ 2 & 4 & 6 \end{array} \right ). \]

The transpose of a column vector is a row vector and vice versa. For example, for \(v = \left ( \begin{array}{c} x \\ y \end{array} \right )\),

\[ v^T = \left ( \begin{array}{cc} x & y \end{array} \right ). \]

We have the following basic computation rules involving the transpose.

Lemma 4.91 (Related exercises: Exercise 4.23)

Let \(A\) be an \(m \times n\)-matrix and \(r \in {\bf R}\) a real number.

  1. \((A^T)^T = A\), i.e., the transpose of the transpose equals the original matrix,

  2. For another matrix \(B\) of the same size as \(A\), \((A + B)^T = A^T + B^T\).

  3. For an \(n \times k\)-matrix \(B\), the transpose of the matrix product \(AB\) is the products of the transposes in the opposite order:

\[ (AB)^T = B^T A^T. \]

(4.92)

  1. If a square matrix \(A\) is invertible, then \(A^T\) is also invertible with inverse
\[ (A^T)^{-1} = (A^{-1})^T. \]

(4.93)

Proof. The first two rules are quite immediate to check (and hardly surprising). The first one can also be seen by noting that doing twice the reflection of the entries along the main diagonal gives back the original matrix.

The equation is also directly following from the definitions: let \(A = (a_{ij})\), \(B = (b_{ij})\). Let us write \(C = AB = (c_{ij})\). Then \(c_{ij} = \sum_{e=1}^n a_{ie} b_{ej}\). Thus \(C^T = (c_{ji}) = \sum_{e=1}^n a_{je} b_{ei}\). This equals the \((i,j)\)-entry of \(B^T A^T\).

For , we compute using :

\[ \begin{align*} A^T (A^{-1})^T & = (A^{-1} A)^T \\ & = {\mathrm {id}}^T \\ & = {\mathrm {id}}. \end{align*} \]

Similarly, again using by :

\[ \begin{align*} (A^{-1})^T A^T & = (A A^{-1})^T \\ & = {\mathrm {id}}^T \\ & = {\mathrm {id}}. \end{align*} \]

Thus the product of \(A^T\) and \((A^{-1})^T\) (in the two possible orders) equals \({\mathrm {id}}\), so they are inverse to each other. ◻

The usage of transposes helps us prove another set of equivalent characterizations:

Corollary 4.94 (Related exercises: Exercise 4.14)

Let \(A \in {\mathrm {Mat}}_{n \times n}\) be a square matrix. Then the following are equivalent:

  1. \(A\) is invertible.

  2. The \(n\) columns of \(A\) are linearly independent.

  3. The rank of \(A\) is \(n\).

  4. The \(n\) rows of \(A\) are linearly independent.

Proof. 1. \(\Leftrightarrow\) 2.: According to Theorem 4.81, \(A\) is invertible precisely if the only solution to the system \(Ax = 0\) is the zero vector \(x = 0\). Recalling that for \(x = \left ( \begin{array}{c} x_1 \\ \vdots \\ x_n \end{array} \right )\) we have

\[ Ax = x_1 c_1 + \dots + x_n c_n, \]

where \(A = (c_1 \ \dots \ c_n)\) are the columns of \(A\), we see that the above condition is equivalent to the columns being linearly independent.

2. \(\Leftrightarrow\) 3.: The rank is, by definition, the dimension of the column space, i.e., the subspace of \({\bf R}^n\) generated by the columns \(c_1, \dots, c_n\). In order to show that these vectors span \({\bf R}^n\), let \(b \in {\bf R}^n\). By the invertibility of \(A\), we know that the system \(Ax = b\) has a (unique) solution \(x\). Therefore \(Ax = \sum_{k=1}^n x_i c_i = b\).

1. \(\Leftrightarrow\) 4.: \(A\) is invertible if and only if the transpose \(A^T\) is invertible. Now use that the rows of \(A^T\) are the columns of \(A\), and apply the (already proved) equivalence 1. \(\Leftrightarrow\) 2.. ◻