Linear algebra ( symmetric matrix)

[Vijay Raghavan]

I am currently brushing on my linear algebra skills when i read this
For any Matrix A
1)A*At is symmetric , where At is A transpose ( sorry I tried using the super script option given in the editor and i couldn't figure it out )
2)(A + At)/2 is symmetric
Now my question is , why should it be divided by 2? doesn't just A + At alone give a symmetric matrix

 

[fresh_42]

I am currently brushing on my linear algebra skills when i read this
For any Matrix A
1)A*At is symmetric , where At is A transpose ( sorry I tried using the super script option given in the editor and i couldn't figure it out )
2)(A + At)/2 is symmetric
Now my question is , why should it be divided by 2? doesn't just A + At alone give a symmetric matrix

It goes like this $\text{ A^t }$ or $\text{$ A^t $}$.

The relevant formulas are $(A \cdot B)^t = B^t \cdot A^t \, , \, (A+B)^t = A^t + B^t$ and $(A^t)^t=A$.
You are correct, the factor $\frac{1}{2}$ isn't necessary here. It usually is taken when $A$ is written as $A = \frac{1}{2}(A+A^t) + \frac{1}{2}(A-A^t)$, i.e. as a sum of a symmetric matrix $B=B^t=\frac{1}{2}(A+A^t)$ and a skew-symmetric matrix $C=-C^t=\frac{1}{2}(A-A^t)$. Here it is needed to get back $A$, instead of $2A$.

 

[Vijay Raghavan]

It goes like this $\text{ A^t }$ or $\text{$ A^t $}$.

The relevant formulas are $(A \cdot B)^t = B^t \cdot A^t \, , \, (A+B)^t = A^t + B^t$ and $(A^t)^t=A$.
You are correct, the factor $\frac{1}{2}$ isn't necessary here. It usually is taken when $A$ is written as $A = \frac{1}{2}(A+A^t) + \frac{1}{2}(A-A^t)$, i.e. as a sum of a symmetric matrix $B=B^t=\frac{1}{2}(A+A^t)$ and a skew-symmetric matrix $C=-C^t=\frac{1}{2}(A-A^t)$. Here it is needed to get back $A$, instead of $2A$.

Thank you for the clarification.