Difference with transposition
Task number: 4532
Consider the linear mapping \(\mathbb R^{2\times 2} \to \mathbb R^{2\times 2} \) given by \(\boldsymbol A \to \boldsymbol A - \boldsymbol A^\mathsf T\).
First decide which of the following matrices belong to the kernel and which belong to the image:
\( \mathbf I_2, \boldsymbol J_2=\left(\begin{smallmatrix}1 & 1 \end{smallmatrix}\right), \mathbf 0_{2{,}2}=\left(\begin{smallmatrix}0 & 0\\ 0 & 0 \end{smallmatrix}\right), \begin{pmatrix}0 & 1 \\-1 & 0 \end{pmatrix}, \begin{pmatrix}0 & 1 \\1 & 0 \end{pmatrix} \)
Then, extend the characterization to all square matrices of order \(n\) and determine the dimensions of the kernel and the image of that mapping.
Solution
Lets denote the mapping \(f(\boldsymbol A)= \boldsymbol A - \boldsymbol A^\mathsf T\)We find that \(f(\mathbf I_2)=\mathbf I_2-\mathbf I_2^\mathsf T=\mathbf 0\), so \(\mathbf I_2\in \ker f\).
Similarly \(\boldsymbol J_2, \mathbf 0_{2{,}2}, \begin{pmatrix}0 & 1 \\1 & 0 \end{pmatrix} \in \ker f \) and \( \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} \notin \ker f \).
To determine the image of \(f\), we substitute the general matrix into \(f\) \(\left(\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}\right)\): \(f\left(\left(\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}\right)\right) = \left(\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}\right) - \left(\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}\right)^\mathsf T = \left(\begin{smallmatrix}a & b \\ c & d \end{smallmatrix}\right) - \left(\begin{smallmatrix}a & c \\ b & d \end{smallmatrix}\right) = \left(\begin{smallmatrix}a-a & b-c \\ c-b & d-d \end{smallmatrix}\right) = \left(\begin{smallmatrix}0 & b-c \\ c-b & 0 \end{smallmatrix}\right) \)In the image of \(f\) there can be matrices with zeros on the main diagonal and opposite elements on the antidiagonal.
From here we get:
\(\mathbf 0_{2{,}2}, \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix} \in \operatorname{Im} f \),
\(\mathbf I_2, \boldsymbol J_2, \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \notin \operatorname{Im} f \).
In general, all symmetric matrices form the kernel, because \(f(\boldsymbol A)= \boldsymbol A - \boldsymbol A^\mathsf T=\boldsymbol A - \boldsymbol A =\boldsymbol 0\).
An image is made up of all antisymmetric matrices, because every image is an antisymmetric matrix \(f(\boldsymbol A)_{i,j}=a_{i,j}-a_{j,i}=-(a_{j,i}-a_{i,j})=-f(\boldsymbol A)_{j,i}\), and also each antisymmetric matrix \(\boldsymbol A\) is the image of the upper triangular matrix \(\boldsymbol B\) with zeros on the diagonal, where the elements above the diagonal (i.e. for \(i \lt j\)) are given by \(b_{i,j}=a_{i,j}.\)
Symmetric matrices are uniquely determined by the elements on and above the diagonal, so the kernel has dimension \(\tfrac{n(n+1)}2\).
Similarly, antisymmetric matrices are uniquely determined by the elements above the diagonal, so the image has dimension \(\tfrac{n(n-1)}2\).
Their bases together generate the entire space of square matrices of order \(n\), since this space has dimension \(n^2\), and each matrix can be written as the sum of a symmetric and an antisymmetric one, see problem Partition of a matrix.
