Determine whether the given set S is a subspace of the vector space V Determine

Question

Determine whether the given set S is a subspace of the vector space V

Determine whether the given set S is a subspace of the vector space V. V = Rn, and S is the set of solutions to the homogeneous linear system Ax = 0 where A is a fixed m X n matrix. V = R4, and S is the set of vectors of the form (0, x2, 7, x4). V = C1 (R), and S is the subset of V consisting of those functions satisfying f'(0) ge 0. V = C2 (I), and S is the subset of V consisting of those functions satisfying the differential equation y" - 4y' + 3y = 0. V = P5, and S is the subset of P5 consisting of those polynomials satisfying p(1) > p(0). V is the vector space of all real-valued functions defined on the interval [a, 6], and S is the subset of V consisting of those functions satisfying f (a) = f(b). V = P4, and S is the subset of P4 consisting of all polynomials of the form p(x) = ax3 + bx.

Explanation / Answer

First, before anything else, a discourse on what "vector space" means.

A vector space is a set of objects, called "vectors," which is closed under a + operation which works like addition does, and is closed under multiplication by an associated scalar field which distributes over addition, and has an additive identity 0, and has additive inverses. Anything which fulfils these criteria is a vector space.

A vector subspace is a vector space embedded in another with the same + and scalar field operations. For example, we have the vector space R², which has a plus operation:

[a, b] + [c, d] = [a + c, b + d]

And has a scalar field R associated with it. Given this, here is a vector subspace: the vectors [x, y] for which x = y. Here is something which is *not* a subspace: The vectors [x, y] where x * y 0 (aka the first and third quadrants). It is not a subspace because [-1, -1] is an element and [2, 0] is an element, but their sum, [1, -1] is not.

In other words, a subspace inherits the algebraic structure of + and the field, so that you don't have to check "do they commute, do they distribute" et cetera. But you *do* have to check closure, and you might also need to check that the 0 element exists in the subspace, and that additive inverses exist in the subspace. (Especially if the field which backs the vector space is weird.)

Still with me?

Let's first tackle (a). You have two functions f(x) and g(x) which reside within your subspace S, and both satisfy:

y'' 4 y' + 3 y = 0.

Since this is a linear homogeneous equation, if both f and g solve it, then the linear combination also solves it:

(a f + b g)'' 4 (a f + b g)' + 3 (a f + b g) = a (f'' 4 f' + 3 f) + b (g'' 4 g' + g) = 0.

So that any linear combination (a f + b g) is within the space if f and g are in it. This proves closure of the subspace under addition and scalar multiplication. And, we can see that if v is in the space then -v is too, and we can see that y = 0 is within the space.

So, yes it's a subspace. Done.

Contrast this with (c), which is a firm *no*, because that equation is *not* homogeneous. So, if:

y''' - y' = 1
k (y''' - y') = k
(ky)''' - (k y)' = k

And thus the system *cannot* be closed under scalar multiplication or addition in general.

We skip to (g). This is a similar problem, but now the condition is the value of f'(0). We immediately see that 0 is allowed, but additive inverses are not: -v is not in the subspace even though v is, as long as v'(0) is positive.

For (f), we have the exact complement of this problem in (g): Instead of we merely have =. You can immediately see that this polynomial subspace is closed under addition and scalar multiplication, because if f and g are polynomials in the subspace then [f + g](0) = f(0) + g(0) = 0, and k f(0) = 0. If f is in the subspace then so is -f, and 0 is in the subspace.

For (e), we again have a situation where vectors in R are our concern. Again, if u and v are members of the subspace A x = 0, then A (u + v) = A u + A v = 0. Thus u + v and k v both satisfy the closure.

Now I leave you to think about nonsingular and symmetric matrices among the n × n real matrices. When you add symmetric matrices, is the result symmetric? When you add nonsingular matrices, is the result nonsingular? (Notice that these "matrices" are also "vectors" on the above definition of a vector and a vector space.)

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.