Question

For any $n \times n$ matrix, prove that A can be uniquely expressed as a sum of symmetric and skew symmetric matrices.

Answer 1

- Hint: In this problem, we need to use the property of the transpose of a matrix to prove that any square matrix can be expressed as the sum of symmetric and skew symmetric matrices. The transpose of a matrix is the original matrix.

Complete step-by-step solution -
Consider, A be a $n \times n$ square matrix.
Now, multiply and divide by 2 in matrix A as shown below.

$$\,\,\,\,\,\,\,{A_{n \times n}} = \dfrac{2}{2}A \\\ \Rightarrow {A_{n \times n}} = \dfrac{1}{2}\left[ {2A} \right] \\\ \Rightarrow {A_{n \times n}} = \dfrac{1}{2}\left[ {A + A} \right] \\\$$

Now, add and subtract the transpose of matrix A as shown below.

$$\,\,\,\,\,\,\,{A_{n \times n}} = \dfrac{1}{2}\left[ {A + A + {A^T} - {A^T}} \right] \\\ \Rightarrow {A_{n \times n}} = \dfrac{1}{2}\left[ {A + {A^T} + A - {A^T}} \right] \\\ \Rightarrow {A_{n \times n}} = \dfrac{1}{2}\left[ {A + {A^T}} \right] + \dfrac{1}{2}\left[ {A - {A^T}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( 1 \right) \\\$$

Now, from equation (1), take the transpose of the matrix $\left[ {A + {A^T}} \right]$ as shown below.

$$\,\,\,\,\,\,\,{\left[ {A + {A^T}} \right]^T} \\\ \Rightarrow {A^T} + {\left( {{A^T}} \right)^T} \\\ \Rightarrow {A^T} + A \\\$$

Thus, the matrix $\left[ {A + {A^T}} \right]$ is a symmetric matrix.
Similarly, from equation (1), take the transpose of the matrix $\left[ {A - {A^T}} \right]$ as shown below.

$$\,\,\,\,\,\,\,{\left[ {A - {A^T}} \right]^T} \\\ \Rightarrow {A^T} - {\left( {{A^T}} \right)^T} \\\ \Rightarrow {A^T} - A \\\ \Rightarrow - \left( {A - {A^T}} \right) \\\$$

Thus, the matrix $\left[ {A - {A^T}} \right]$ is a skew-symmetric matrix.

Hence, a square matrix of any order can be expressed as a sum of symmetric and skew symmetric matrices.

Note: The properties of the transpose of a matrix are shown below:

$${\left( {A + B} \right)^T} = {B^T} + {A^T} \\\ {\left( {{A^T}} \right)^T} = A \\\ {\left( {AB} \right)^T} = {B^T}{A^T} \\\$$