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

- 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:
\[
  {\left( {A + B} \right)^T} = {B^T} + {A^T} \\
  {\left( {{A^T}} \right)^T} = A \\
  {\left( {AB} \right)^T} = {B^T}{A^T} \\
\]