Determine whether the given set S is a subspace of the vector space V. A. V=Mn(R

Question

Determine whether the given set S is a subspace of the vector space V. A. V=Mn(R), and S is the subset of all upper triangular matrices. B. V=C2(I), and S is the subset of V consisting of those functions satisfying the differential equation y4y+3y=0. C. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=f(b). D. V=C3(I), and S is the subset of V consisting of those functions satisfying the differential equation y+4y=x2. E. V=R4, and S is the set of vectors of the form (0,x2,5,x4). F. V=P3, and S is the subset of P3 consisting of all polynomials of the form p(x)=ax3+bx. G. V=P5, and S is the subset of P5 consisting of those polynomials satisfying p(1)>p(0). Preview My AnswersSubmit Answers

Explanation / Answer

Solution:

"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.

a subspace inherits the algebraic structure of + and the field ,

The 0 element exists in the subspace.

Let's (b). 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, if 'v' is in the space then '-v' is too, and we can see that y = 0 is within the space.

comparing (d), which is a firm *no*, because that equation is *not* homogeneous.

So, if

y''' - y' = x2

k (y''' - y') = k

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

And this is 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 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.

(f),If we the exact complement of this problem in (g): Instead of we merely have =. You can 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.

(e), we again a situation where vectors in R4 are our concern. Again, if u and v are members of the subspace

ax3+bx = 0, then A (u + v) = A u + A v = 0. Thus u + v and k v both satisfy the closure.

Now 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?

Note: "matrices" are also "vectors" on the above definition of a vector and a vector space.)

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