# Thread: Skew Symmetric Matrix

1. ## Skew Symmetric Matrix

Hi,

I would like some guidance on how to approach this question please:

Question: Let B be an n*n skew symmetric matrix and let A = I + B. Show that $\displaystyle AA^T = (I+B)(I-B)= A^TA$

2. If $\displaystyle B$ is a skew symmetric matrix, then $\displaystyle B^T=-B$. And $\displaystyle AA^T=(I+B)(I+B)^T=(I+B)(I^T+B^T)$. Using the property of skew symmetric matrix as stated, you should be able to work it out now.

I am still trying to figure out how you got $\displaystyle AA^T = (I + B)(I + B)^T$

Here is what I understand so far:

$\displaystyle B^T= -B$ (this is the definition of a skew symmetric matrix)

$\displaystyle A = I + B$ (this is given)

If $\displaystyle A = I + B$, then $\displaystyle A^T = -A = I - B$ (does this make any sense?)

Therefore, $\displaystyle AA^T = (I+B)(I-B)$ (is this correct?)

4. Originally Posted by sparky

If $\displaystyle A = I + B$, then $\displaystyle A^T = -A = I - B$
This final result is correct, but not the intermediate step. It comes as follows: $\displaystyle A^T=(I+B)^T=(I^T+B^T)=(I+B^T)=(I-B)$ But note that $\displaystyle A^T \neq -A$ since A is not mentioned as a skew-symmetric matrix.

5. Thanks Sambit