Linear algebra : Doing a proof with a square matrix

[Physicaa]

Homework Statement

Show that all square matrix (A whatever) can be written as the sum of a symmetric matrix and a anti symmetric matrix.

Homework Equations

I think this relation might be relevant : $$
A=\frac{1}{2}*(A+A^{T})+\frac{1}{2}*(A-A^{T})
$$

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 ?

 

[LCKurtz]

Homework Statement

Show that all square matrix (A whatever) can be written as the sum of a symmetric matrix and a anti symmetric matrix.

Homework Equations

I think this relation might be relevant : $$
A=\frac{1}{2}*(A+A^{T})+\frac{1}{2}*(A-A^{T})
$$

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 ?

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

 

[Physicaa]

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

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

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

 

[fresh_42]

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

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

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

 

[Physicaa]

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

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 ?

 

[fresh_42]

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

 

[Physicaa]

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

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 ?

 

[fresh_42]

Yep, just collect it.

 

[Physicaa]

Yep, just collect it.

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 ?

 

[fresh_42]

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 ?

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

 

[Physicaa]

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

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 ?

 

[fresh_42]

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

 

[Physicaa]

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

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.

 

[fresh_42]

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.

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?

 

[Physicaa]

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?

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?

I didn't get B though.

I got this : $$B^T=1/2*(A^T + A)$$That was my end result.

 

[fresh_42]

I didn't get B though.

I got this : $$B^T=1/2*(A^T + A)$$That was my end result.

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 $C$.

 

[Physicaa]

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 $C$.

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

 

[Physicaa]

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 $C$.

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

 

[fresh_42]

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.
You already have a sum (the plus sign in the middle).
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.

 

[Physicaa]

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.
You already have a sum (the plus sign in the middle).
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.

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

 

[fresh_42]

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

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

 

[Physicaa]

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

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 ?

 

[fresh_42]

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.