Thread: Question on subspace

I assume the question deals with vector spaces over real numbers. Is W closed under multiplication by a real number?

Yes, it is deal with vector spaces over real number. For a vector to be subspaces, it must be closed under addition and scalar multiplication.

(a) (a₁, b₁) + (a₂, b₂) = (a₁ + a₂, b₁ + b₂) => closed under addition
k(a, b) = (ka, kb) => closed under scalar multiplication

But i do think that if a>= 0, then k(a, b) = (ka, kb) is not closed under scalar multiplication when k is negative value so it is not subspace of V.


(b) (a₁, c1, c1) + (a₂, c₂, c₂) = (a1 + a2, c1 + c2, c1 + c2) => closed under addition
k(a, c, c) = (ka, kc, kc) => closed under scalar multiplication

Similarly, a^2 + b^2 + c^2 <= 1 could mean a, b, c might be 0, 0<= a <= 1 or -1 <= a <= 0. But i do not know why b is there since this question does not have value b.
and if a is 1, then a₁ + a₂ = 2 which is not closed under addition so it is not subspace too.

I hope that my working is correct. But i doubt that i explain it in correct way. Or maybe i shall prove that with contradiction.
(1, 0, 0) + (1, 0, 0) = (2, 0, 0) which is not closed under addition.

Could you further elaborate on that?

Originally Posted by whyyie

But i do think that if a>= 0, then k(a, b) = (ka, kb) is not closed under scalar multiplication when k is negative value so it is not subspace of V.

Yes. I would note that only a set can be closed under an operation. It is not correct to say that k(a, b) = (ka, kb), which is an equality and not a set, is closed under scalar multiplication.

Originally Posted by whyyie

Similarly, a^2 + b^2 + c^2 <= 1 could mean a, b, c might be 0, 1 or -1. But i do not know why b is there since this question does not have value b.

I am not sure what you mean. Why do you set apart b? Does the question have the value of a or c?

Originally Posted by whyyie

and if a is 1, then a₁ + a₂ = 2 which is not closed under addition

What are a₁ and a₂? Why is a₁ + a₂ = 2? Again, "... = ..., which is not closed" is incorrect.

The claim "W is closed under scalar multiplication" means "For all real numbers k and all vectors (a, b, c) ∈ W, (ka, kb, kc) ∈ W." To form its negation, switch the quantifiers and add the negation to the final part: "There exists a real number k and a vector (a, b, c) ∈ W such that (ka, kb, kc) ∉ W." Can you find specific k, a, b, c for which this negation is true?

it appears there is a typo in the second part of the problem, and i believe it should be triples of the form (a,b,c) instead of (a,c,c).

i also suggest using a k much larger than 1 as a scalar for a counter-example for the second part.

Originally Posted by Deveno

it appears there is a typo in the second part of the problem, and i believe it should be triples of the form (a,b,c) instead of (a,c,c).

Thanks for noticing.

Originally Posted by Deveno

i also suggest using a k much larger than 1 as a scalar for a counter-example for the second part.

This reminds me what my calculus teacher used to say: "Let ε be arbitrarily small: for example, 100."

one of my favorite jokes is:

1+1 = 3, for large values of 1.