Prove any square matrix is sum of symmetric and skew-symmetric

[thomas49th]

Homework Statement

Prove that any square matrix can be written as the sum of a symmetric and a skew-symmetric matrix

Homework Equations

For symmetric $$A=A^{T}$$
For scew-symmetric $$A=-A^{T}$$

The Attempt at a Solution

Not sure where to begin. Using algebra didn't work. Got powers and nothing cancelled.

p.s is there an inbuilt template for matrix in latex?

Thanks
Tom

 

[Hurkyl]

Not sure where to begin.

Have you tried some explicit examples?

Using algebra didn't work. Got powers and nothing cancelled.

Could you explain what you wanted to try with algebra? And demonstrate what you actually tried?

 

[thomas49th]

well i set up the matrix
A =
a b
c d

and
$$A^{T} = \frac{a c}{b d}$$


and i multiplied the two together, but that doesn't get you anywhere?

I've been reading through a book, and the last question presented in that topic. I have the answer on the following page

A= 0.5(A+A^T) + 0.5(A-A^T)

but I'm not sure how they come to that answer. I mean I can see that the transforms cancel each other out, but how did they get there?

Thanks :)

 

[HallsofIvy]

If f(x) is any function at all, then g(x)= f(x)+ f(-x) is an "even" function, because g(-x)= (-x)= f(-x)= f(-(-x))= f(x)+ f(-x)= f_e(x), and h(x)= f(x)- f(-x) is an odd function, because h(-x)= f(-x)-f(-(-x))= -(f(x)- f(-x)). Adding those two functions, g(x)+ h(x)= 2f(x) so g(x)/2= (f(x)+ f(-x))/2 and h(x)/2= (f(x)-f(-x))/2 are even and odd functions that sum to f- they are the "even" and "odd" parts of f. Do you see the similarity with your problem?

 

[clamtrox]

How many free parameters are there in a M times M matrix? How many are there if the matrix is symmetric or antisymmetric?

 

[D H]

I've been reading through a book, and the last question presented in that topic. I have the answer on the following page

A= 0.5(A+A^T) + 0.5(A-A^T)

but I'm not sure how they come to that answer. I mean I can see that the transforms cancel each other out, but how did they get there?

Denote B and C to represent the symmetric and skew-symmetric parts of the given matrix A: A=B+C

Taking the transpose of A, A^T=B^T+C^T. Now using the fact that B is symmetric and C is skew-symmetric, A^T=B-C. The problem here is to solve for B and C. Writing the equations for A and A^T as

$$\aligned<br /> B+C&=A \\<br /> B-C&=A^T<br /> \endaligned$$

So, two linear equations in two unknowns, and a particularly easy one to solve at that.

$$\aligned<br /> B&=(A+A^T)/2 \\<br /> C&=(A-A^T)/2<br /> \endaligned$$

 

[thomas49th]

Ahh I think I see DH!
If we add B and C, we get A!
$$A^{T} = B^{T}+C^{T}$$

As we have square matrices we can say $$B^{T} = B$$ and $$C^{T} = -C$$
this means
$$A^{T} = B-C$$
solving that with
A = B + C

gives us your bottom 2 equations (the symmetric and skew-symmetric matrices), which we add up to give a matrice.


Hallofivy, I can see what your kinda getting at!

clamtrox I don't see what you mean by parameters. A matrix holds elements, are these the parameters? Can you expand?

Thanks
Tom

 

[clamtrox]

Yes yes, I was just thinking about a more high-flying explanation. It obviously works like DH explained.

But yeah, a physicist looks immediately how many degrees of freedom there are. Since a M times M matrix contains M^2 independent functions (independent elements, parameters, degrees of freedom, DOF's), then one would expect that a symmetric M times M and an antisymmetric M times M matrix would have the same number of DOF's combined.

 

[thomas49th]

If f(x) is any function at all, then g(x)= f(x)+ f(-x) is an "even" function, because g(-x)= (-x)= f(-x)= f(-(-x))= f(x)+ f(-x)= f_e(x), and h(x)= f(x)- f(-x) is an odd function, because h(-x)= f(-x)-f(-(-x))= -(f(x)- f(-x)). Adding those two functions, g(x)+ h(x)= 2f(x) so g(x)/2= (f(x)+ f(-x))/2 and h(x)/2= (f(x)-f(-x))/2 are even and odd functions that sum to f- they are the "even" and "odd" parts of f. Do you see the similarity with your problem?

I was just looking back over this and I was wondering if you could expand on it

I know that an even function is that which f(x) = f(-x) - like a quadratic
An odd function is f(-x) = -f(x) - like x³

When hitting this line, my brain turns to mush:
g(-x)= (-x)= f(-x)= f(-(-x))= f(x)+ f(-x)= f_e(x)

so a negative value x is inputted into function g. G is defined as f(x) + f(-x). If f(x) is odd then f(x) + f(-x) = 0, but I'm not sure what
g(-x)= (-x)= f(-x)= f(-(-x))= f(x)+ f(-x)= f_e(x)
is telling me. I don't even know what the e does. Can you say a function in number e.

Never really taught such maths in that kind of "comprehensive" manner. I'm a sucker for pretty pictures ;)

Thanks
Tom