# Linear algebra : Doing a proof with a square matrix

1. Feb 15, 2016

### Physicaa

1. The problem statement, all variables and given/known data
Show that all square matrix (A whatever) can be written as the sum of a symmetric matrix and a anti symmetric matrix.

2. Relevant equations

I think this relation might be relevant : $$A=\frac{1}{2}*(A+A^{T})+\frac{1}{2}*(A-A^{T})$$
3. The attempt at a solution

I know that we have the following theorems : A square matrix "A" is symmetric if and only if $$A^T=A$$

and

A square matrix "A" is anti symmetric if and only if $$A^T=-A$$

But besides that I'm not sure where to go.... Any hints ?

2. Feb 15, 2016

### LCKurtz

Why don't you check to see if your proposed equation does it?

3. Feb 15, 2016

### Physicaa

Sorry I'm not following you. You mean I must reduce my equation ? We get for example :

$$2A=(A+A^T)+(A-A^T)$$

4. Feb 15, 2016

### Staff: Mentor

That's not what LC said. Read it again, carefully. It contains already all that can be said.

5. Feb 15, 2016

### Physicaa

To be honest I'm not sure of understanding the problem. I think that I must get this as a result : $$A = A + A^T$$

Am I right or not ?

6. Feb 15, 2016

### Staff: Mentor

Write it $A=B+C$ and see if you already have $B$ and $C$ with the desired properties.

7. Feb 15, 2016

### Physicaa

Ok, so if I use my equation I set the following equations which : $$B = \frac{1}{2}*(A+A^{T})$$

and $$C=\frac{1}{2}*(A-A^{T})$$

Is this where you go ?

8. Feb 15, 2016

### Staff: Mentor

Yep, just collect it.

9. Feb 15, 2016

### Physicaa

Ok, sorry but I'm not understanding the problem at all right now. I'm not even sure what I'm supposed to get as a final result... It says that I must have a sum of a symmetric and anti symmetric matrix. So... how does what I just did helpt exactly ?

10. Feb 15, 2016

### Staff: Mentor

Is $B$ symmetric and $C$ anti-symmetric as defined above?

11. Feb 15, 2016

### Physicaa

Ok I understand what you'Re saying. You'Re saying that B and C are respectively symmetric and anti symmetric. But how do I know that they are ? Must I prove that the sum in B is symmetric while the difference in C is anti symmetric ?

12. Feb 15, 2016

### Staff: Mentor

Yes. Put a big bracket around them, right a big T on the right one above and resolve the brackets.

13. Feb 15, 2016

### Physicaa

You want this, for example :

$$B=\frac{1}{2}*(A+A^{T})$$

then

$$B^T=(\frac{1}{2}*(A+A^{T}))^T$$

and I do the same for C.

14. Feb 15, 2016

### Staff: Mentor

Yes, but what is $B^T=(\frac{1}{2}*(A+A^{T}))^T$? It must equal $B$. And the version with $C^T$ has to be $-C$. Do you know why this is the case?

15. Feb 15, 2016

### Physicaa

I didn't get B though.

I got this : $$B^T=1/2*(A^T + A)$$

That was my end result.

16. Feb 15, 2016

### Staff: Mentor

And if you put all together you have for $A=B+C= 1/2(A+A^T)+1/2(A-A^T)$ (in very detailed form):
$B^T=(1/2*(A + A^T))^T= 1/2*(A^T + (A^T)^T)=1/2*(A^T+A)=1/2*(A+A^T) = B$
and similar with $C$.

17. Feb 15, 2016

### Physicaa

Ok yeah I ended up with $$C^T=-C$$

18. Feb 15, 2016

### Physicaa

Ok,

so i did the following $$A=1/2(A + A^T) + 1/2(A-A^T)$$

I applied the T on the whole of this...

I got $$A^T=1/2*(A+A^T) - 1/2 (A-A^T)$$

19. Feb 15, 2016

### Staff: Mentor

Ok then, you're probably blocked. That happens from time to time. (I have asked one of the silliest questions at all myself a couple of hours ago.)

You have $A= 1/2*(A+A^T) + 1/2*(A-A^T)$.
You need to show that $A$ is a sum of a symmetric and a anti-symmetric matrix.
All you need to show is that the first term is symmetric, i.e. $[1/2*(A+A^T)]^T=1/2*(A+A^T)$.
We did this for $B$ which was only another name for $1/2*(A+A^T)$.
And you need to show that the second term is anti-symmetric, i.e. $[1/2*(A-A^T)]^T=1/2*(A^T-A)= - 1/2*(A-A^T)$
We did this for $C$ which was only another name for $1/2*(A-A^T)$.

It's nothing left to do.

And if you still don't see what it is about, then let it rest for a day and have another look at it tomorrow.

20. Feb 15, 2016

### Physicaa

I think I understood. Just another question for C. We know that C is anti symmetric because when we tranpose it we obtain $$C^T=-C$$ and we know there's a theorem which says if we obtian this then our square matrix is anti symmetric, right ?

After Ive tranposed the B and C I know that my sum is a sum of symmetric and anti symmetric.