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# If $A$ and $B$ are symmetric matrices, prove that $AB - BA$ is a skew symmetric matrix.

## 1 Answer

Toolbox:
• A square matrix A=[a$_{ij}$] is said to be skew symmetric if A'=-A that is $[a_{ij}]= -[a_{ji}]$ for all possible value of i and j.
• A square matrix A=[a$_{ij}$] is said to be symmetric if A'=A that is $[a_{ij}]=[a_{ji}]$ for all possible value of i and j.
Step 1: Given
A$\rightarrow$ symmetric matrix
B$\rightarrow$ symmetric matrix
$\Rightarrow$ A=A'
$\Rightarrow$ B=B'
From the property of transpose of matrices,we have
(AB)'=B'A'
Step :2 Now consider AB-BA and by taking transpose of it. we get
(AB-BA)'=(AB)'-(BA)'
$\;\;\;\qquad=B'A'-A'B'$
Replace A'=A And B'=B
$\;\;\;\qquad=BA-AB$
$\;\;\;\qquad=-(AB-BA)$ (By taking negative common)
We know that a matrix is said to be skew symmetric matrix if A = -A
Hence AB-BA is a skew symmetric matrices.
answered Mar 16, 2013
edited Mar 19, 2013

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