Thread: need pointers for my HW. Linear subspaces

Ok here's the deal. I'm returning to Math after a break and I need all the help I can get. I'm not looking for answers just hints/pointers to help me address my HW and advance my understanding of Linear Algebra. About me...former B.S. Biochem major making the transition to ChemE. The book is the 10th Edition Elementary Linear Algebra by Anton.
I will be answering the prob and stating why I think it so or why I have no freaking idea

4.2 probs 4, 8 ac, 10 ac, 14, and T/F(k)

4) Which of the following are subspaces of F(-infinity, infinity)
a) all fxns f in F(-infin, infin) for which f(0)= 0
yes because that satisfies the zero vector
b) all fxns f in F (-infin, infin) for which f(0) = 1
no b/c doesn't satisfy zero vector
c) all fxns f in F(-infin, infin) for which f(-x) = f(x)
yes b/c needs to satisfy for negative scalar
d) all polynomials of degree 2
???????

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8) express the following as linear combos of u = (2,1,4), v = (1,-1,3) and w = (3,2,5)
a) (-9, -7, -15)
9 <2,1,4> + -7 <1,-1,3> + -15 <3,2,5> = <-34, -14, -92>

c) (0,0,0)
0 < > + 0 < > + 0 < > = 0

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10) express the vector as a linear combo of p₁= 2 + x + 4x² , p₂= 1 - x + 3x² , p₃= 3 + 2x + 5x²

a) -9 - 7x -15x²
-9 - 7x -15x² = (2a1 + a2 + 3a3) + x(a1 - a2 + 2a3) + x² (4a1 + 3a2 + 5a3)

2 1 3 -9
1 -1 2 -7
4 3 5 -15

c) 0
0 = (2a1 + a2 + 3a3) + x(a1 - a2 + 2a3) + x² (4a1 + 3a2 + 5a3)

2 1 3 0
1 -1 2 0
4 3 5 0

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14) let f = cos²x and g = sin²x. Which of the following lie in the space spanned by f and g ?
a) cos2x b) 3 + x² c) 1 d) sin x e) 0

I tried (f, g) = k₁f + k₂g = cos²x, sin²x = k1(cos²x) + k2(sin²x) .............. : (

I don't have the slightest. I know that I'm trying to prove that a-e are subspaces enclosed by f and g and must have zero vector but I'm at a loss. I thought maybe cos2x + sin2x = 1 might help but I got nothing

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True or False (K) and explain why

The polynomials (x-1), (x-1)² and (x-1)³ span P₃?
I started out p₁ = x -1, p₂ = (x - 1)² and p₃ = (x - 1)³

Please Help

a) unless I am mistaken, since the space does not have a unity it cannot be a subspace. edit: nm I think that a submodule need not have a unity so I think it is a subspace now

b) not closed under pointwise addition at x = 0 so not a vector space

c) Let V = {even functions}. ill get back to you on this one. edit: the guy below me gave a good answer for this one

d) as a subset of polynomial space, the set of V ={polynomials of degree 2} does NOT determine a vector space because there is no 0 vector in this space. the 0 vector ( f(x) = 0) is a constant function which has degree 0 and hence is not a member of the set of polynomials of deg 2. Also a subspace is a vector space in its own right, so it has to have a unity, i.e. an element 1 s.t. 1f = f for all f in the subspace. this would have to be f(x) = 1, which doesn't exist in V either.

try to post fewer questions at a time. i don't think i can adequately address ALL of these in a single post. it's OK to post different questions in separate threads.

it's not clear what F(-∞,∞) is supposed to be, but my guess is: all real-valued functions defined on all of R.

for (a): this is not sufficient (but it IS necessary!). you have to also prove closure under vector addition and scalar multiplication. although you haven't told us what these operations are defined to be, my guess is your text says they are:

(f+g)(x) = f(x) + g(x)
(cf)(x) = c(f(x))

now if f(0) = 0, and g(0) = 0, we must verify that (f+g)(0) = 0 and (cf)(0) = 0, as well. this isn't hard to do:

(f+g)(0) = f(0) + g(0) = 0 + 0 = 0
(cf)(0) = c(f(0)) = c(0) = 0.

(b) is correct. you need only falsify one of:

1) closure of addition
2) closure of scalar multiplication
3) the 0-vector is in the set

(c) what you wrote here doesn't make a lot of sense to me. let's go through our run-down:

is this set closed under vector addition?

suppose f(-x) = f(x), and g(-x) = g(x).

then (f+g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f+g)(x), so f+g is in this set whenever f and g are (closure under addition).

is this set closed under scalar multiplication?

suppose f(-x) = f(x).

then (cf)(-x) = c(f(-x)) = c(f(x)) = (cf)(x), so yes.

finally: is the 0-function in this set?

yes....0(-x) = 0 = 0(x), regardless of what x is.

(d) no. f(x) = x² + x is a polynomial of degree 2, and g(x) = -x² is a polynomial of degree 2, but (f+g)(x) = x² + x - x² = x, is only of degree 1, we do not have closure.

8a) no, not even close.

what we need is to find a,b,c in R (or whatever our underlying field F is) such that:

a(2,1,4) + b(1,-1,3) + c(3,2,5) = (-9,-7,-15)

we can re-write this as:

(2a,a,4a) + (b,-b,3b) + (3c,2c,5c) = (-9,-7,-15), or even better, as:

(2a+b+3c,a-b+2c,4a+3b+5c) = (-9,-7,-15)

this gives us a system of 3 equations, in 3 unknowns:

2a+b+3c = -9
a-b+2c = -7
4a+3b+5c = -15

linear algebra is all about solving such systems in an orderly fashion.

here is the "high-school method" (to which i often resort to out of laziness. my bad.).

if we add eqs. 1 & 2, we get:

3a+5c = -16

thus a = (1/3)(-16-5c).

if we add 3 times eq. 2 to eq. 3, we get:

7a+11c = -36. substituting the value we got for a in this:

(7/3)(-16-5c) + 11c = -36

-112/3 - 35c/c + 33c/3 = -36

-2c/3 = -36 + 112/3

c = (-3/2)(-36) + (-3/2)(112/3) = 3*18 - 56 = 54 - 56 = -2

thus a = (1/3)(-16 - 5(-2)) = (1/3)(-16 + 10) = -2

and finally:

2(-2) + b + 3(-2) = -9
-4 + b - 6 = -9
b = 1

always check your work!

(-2)(2,1,4) + (1,-1,3) + (-2)(3,2,5) = (-4,-2,-8) + (1,-1,3) + (-6,-4,-10) = (-9,-7,-15), as required.

8c) is trivial: (0,0,0) = 0(2,1,4) + 0(1,-1,3) + 0(3,2,5) = (0,0,0) + (0,0,0) + (0,0,0)

10a) this is the same problem as 8a! think about that for a bit, and if the light bulb doesn't go on in your head, we'll try again.

for 14):

a) cos(2x) = cos²(x) - sin²(x) = (1)f + (-1)g. looks like a linear combination to me.

b) this function is not periodic. how can we sum periodic functions and get a non-periodic function?

c) 1 = f + g, right?

d) this one is tricky. it IS periodic, but it's the square root of sin²(x), and square roots aren't linear.

e) 0 is always in ANY span (look at problem 8a again).

the last problem has an easy answer: P₃ has dimension 4, it cannot be spanned by merely 3 vectors.

thanks. For 4 how did you choose g(x)?

14) a) I don't get how we go from cos(2x) to the rest
b) what's periodic?
d) same

and T/F

how did you figure out it is in the 4th dimension?

for 4, "g" is just a symbol, meaning some OTHER function that f (or maybe the same one, but maybe not). you can use any two symbols you like, i could have used p(x) and q(x), as long as you understand i mean two elements of F(-∞,∞).

for 14 a)

a standard trigonometric identity is the "angle-sum identity for cosine":

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

using a = b = x gives the result.

a periodic function is a function for which there exists a real number k such that: f(x+k) = f(x) for ALL x. the number k is called the PERIOD of the function. for trigonometric functions (the standard ones) the period is 2π (2pi).

the vector space P₃ is all polynomials of the form:

{p(x) = a₀+a₁x+a₂x²+a₃x⁴: a₀,a₁,a₂,a₃ in R}.

this vector space has the BASIS {1,x,x²,x³} (four coordinates = four coefficients), and thus has dimension 4 (dimension is the number of elements in ANY basis).

Originally Posted by burton1995

thanks. For 4 how did you choose g(x)?

Deveno didn't choose g(x). It, like, f(x), can be any function satisfying the requirements of the problem:
a) g(0)= 0,
b) g(0)= 1,
c) g(-x)= -g(x)

14) a) I don't get how we go from cos(2x) to the rest

is a well known trig identity. Review your trig.

b) what's periodic?

Oh dear! You should not be taking Linear Algebra if you have not taken at least enough trigonometry to know what "periodic function" means. A function, f, is "periodic with period p" if f(x+p)= f(x) for all x. sin(x) and cos(x) are periodic with period and, so, cos(2x), , and are periodic with period .

d) same

and T/F

how did you figure out it is in the 4th dimension?