Is the set of all continuous functions on the interval [0,1] a vector space?

[Defennder]

Ok, I can't tell if you can see why the sum of two symmetic matrices is itself symmetric, or if you can see that it is so but can't think of a formal or acceptable way to prove it. Consider this then:

A matrix A is symmetric if for all its entries a_{ij}=a_{ji} Suppose there's another symmetric matrix B with the same property.

The sum of the 2 matrices is C and a typical matrix entry of C is c_{ij} = a_{ij} + b_{ij}. Now can you show if c_{ij} = c_{ji}?

 

[Saladsamurai]

Ok, I can't tell if you can see why the sum of two symmetic matrices is itself symmetric, or if you can see that it is so but can't think of a formal or acceptable way to prove it. Consider this then:

A matrix A is symmetric if for all its entries a_{ij}=a_{ji} Suppose there's another symmetric matrix B with the same property.

The sum of the 2 matrices is C and a typical matrix entry of C is c_{ij} = a_{ij} + b_{ij}. Now can you show if c_{ij} = c_{ji}?

How about this:
c_{ij} = a_{ij} + b_{ij}



c_{ji} = a_{ji} + b_{ji}

But since a_{ij}=a_{ji} and b_{ij}=b_{ji}

then c_{ij} = a_{ij} + b_{ij} =a_{ji} + b_{ji}=c_{ji}


Does that work? I think it does if I got my indexes right

 

[Defennder]

Yep that should do it. I don't know how formal you need that to be, though. I've never been a fan of mathematical formalism.

 

[Saladsamurai]

Thanks!

 

[nicksauce]

Hold on a second guys. I don't like to hijack what looks completed, but I am a little befuddled for part c. The polynomial 0 is not a polynomial of degree n, so how can we say a zero element exists? Furthermore it is not closed under addition, for example a=(x^2 + 1), b = -x^2... a+b is not a polynomial of degree 2. According to wikipedia, some sources choose to include the axioms of closure as additional vector space axioms.

 

[Saladsamurai]

Anyone else want to chime in? This is all new to me

You can't just say that a+b is polynomial of degree n with zero coefficients?

But like I said, this is new to me.

 

[Dick]

Anyone else want to chime in? This is all new to me

You can't just say that a+b is polynomial of degree n with zero coefficients?

But like I said, this is new to me.

nicksauce makes a good point. The question does say "degree EXACTLY n". The 'exactly' is likely there for a reason.

 

[Saladsamurai]

Okee-dokee. So since there is some polynomial of degree exactly n that when added to some other polynomial of exactly degree n does not YIELD a polynomial of exactly degree n, then the set of all polynomials of exactly degree n IS NOT a vector space.

So my coefficients of zero thing in post #36 is valid.

 

[Dick]

Right. You could also think of cases like p=x^2 and q=x^2-x, so p-q=x. Not a polynomial of degree EXACTLY two. If they had said polynomial of degree two or less, then it would be a vector space.