Homework Help: 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.

21. Feb 15, 2016

Staff: Mentor

It's the definition of anti-symmetric matrices and not a theorem but yes, that's right.

22. Feb 15, 2016

Physicaa

In my book it's written as a theorem. Anyway, thank you again. The reason why I blocked is because we never saw this,really. All we ever did was computational stuff with matrix and we suddenly got these to do. And I'm not in a math program or univeristy. What was this problem that you had btw ?

23. Feb 15, 2016

Staff: Mentor

I asked why we can't use the sun's gravity to travel to mercury.
The answer is: we are already in the system of the sun and all her gravity already affects us.
Falling into the sun could only happen if we were not moving with the earth. Jumping into an earth orbit doesn't take away this motion.