What are the subspaces of a matrix and how do bases relate to them?

[mmmboh]

Hi, I have an assignment due and I have done most of the questions there are just a couple things I have left, if someone can help that would be amazing :)

1. In this problem we suppose that F is a field, A is an m by n matrix over
F and that W is a subspace of Fm.
(a) Show that U = {v$$\in$$Fⁿ:Av$$\in$$W} is a subspace of Fⁿ.
(b) Now suppose that m = n and A is invertible, and that B = {v1, v2,...vk} is a basis for W. Show that {A^-1v₁, ... A^-1v_n} is a basis for U.

2. Let P5(x) = R5[x] be the real vector space of polynomials of degree at
most 5. Find a basis for each of P5(x) $$\cap$$ U, P5(x) $$\cap$$ W, and P5(x) $$\cap$$ V. V is the real vector space of all functions from R to R, U is the set of even functions, W is the set of odd functions.

1. Alright I have already done part a, but I can't figure out part b, can anyone help please?

2. I am lost for this question help please.

Thanks :)

 

[jambaugh]

I'm retyping your post using tex so I can better read it:

1. In this problem we suppose that F is a field, A is an m by n matrix over
F and that W is a subspace of $$F^m$$.
(a) Show that $$U = \{v \in F^n:Av\in W\}$$ is a subspace of $$F^n$$.
(b) Now suppose that m = n and A is invertible, and that
$$B = \{v_1, v_2, \ldots,v_k\}$$ is a basis for W. Show that
$$\{A^{-1}v_1,A^{-1}v_2, \cdots A^{-1}v_n\}$$ is a basis for U.

2. Let $$P_5(x) = \mathbb{R}_5[x]$$
be the real vector space of polynomials of degree at most 5.
Find a basis for each of $$P_(x) \cap U$$,$$P_5(x) \cap W$$, and $$P_5(x) \cap V$$.
V is the real vector space of all functions from $$\mathbb{R} \to \mathbb{R}$$, U is the set of even functions, W is the set of odd functions.

Recall that a function f(x) is even if f(-x) = f(x). It is odd if f(-x) = - f(x). The main reason for referring to these properties as "even" and "odd" is that even and odd degree power functions are respectively "even" and "odd" in this sense.

So you have the space of fifth degree polynomials which is a space of functions.
a.) Can you identify a basis for the whole space?
b.) Can you identify which polynomials are even functions, or odd functions respectively?

Look at the definition of a polynomial (a linear combination of powers of x) and the definition of a basis of a vector space.

Here's another potentially helpful bit of info: ANY function can be written as a sum of even and odd parts:

$$f(x) = f_{ev}(x) + f_{od}(x)$$
where
$$f_{ev}(x) = \frac{1}{2}(f(x)+f(-x))$$
and
$$f_{od}(x) = \frac{1}{2}(f(x)-f(-x))$$

 

[mmmboh]

I know a basis is a linearly independent set of vectors that can represent any vector in a vector space.
For P₅(x)$$\cap$$U(which is the set of all even vectors)
So would x⁴+x² do it? I am confused about how to do this.
And for P₅(x)$$\cap$$W would it then be x⁵+x³+x?
and for P₅(x)$$\cap$$V, would it be x⁵+x⁴+x³+x²+x?
Polynomials in vector spaces confuse me :S

And For number one I got the answer for part A)...but part B I am a little bit confused about how to prove it, if you can give another hint that would be great, I'll try until I get it though!

Thanks a lot for the help.

 

[jambaugh]

I know a basis is a linearly independent set of vectors that can represent any vector in a vector space.

Represent How? (That's the key point of the hint.)

Polynomials in vector spaces confuse me :S

They will until they don't and then you'll better understand what vectors really are. They aren't arrows. They are any mathematical element of some set of things you can add and scale, i.e. take linear combinations of. So for example the set of all functions on the real line is a vector space since you can add them and get functions and you can multiply them by constants and you get functions.

The polynomials you listed are single elements and not sets of elements (sets of polynomials) so they won't be a basis. A basis for a space of polynomials should look something like:
Basis = { 2x+3, x^8 - 2x + 4, 1}

Another hint. Compare the most general quadratic polynomial: P=ax^2 + bx + c with the most general vector in 3-space: V=ai + bj + ck.

 

[mmmboh]

Ok so a basis for the intersection with all even functions would be (1,x²,x⁴), and a basis for all odd functions is (1,x,x³,x⁵) and a basis for the real vector space from all functions R to R is (1,x,x²,x³,x⁴,x⁵).
Is this correct?

 

[jambaugh]

Ok so a basis for the intersection with all even functions would be (1,x²,x⁴), and a basis for all odd functions is (1,x,x³,x⁵) and a basis for the real vector space from all functions R to R is (1,x,x²,x³,x⁴,x⁵).
Is this correct?

Almost. Careful there $$1 = x^0$$ is even.

 

[mmmboh]

Ah Thank's a lot!

For the first question part B, I know all the vectors in $$\beta$$ are independent because $$\beta$$ is a basis for W, so I know the inverse of all these vectors are also independent. And I know that A maps the vectors in U onto W, so the inverse of A will map the vectors in W, or $$\beta$$, onto U...but how should I write this out?

 

[jambaugh]

Ah Thank's a lot!

For the first question part B, I know all the vectors in $$\beta$$ are independent because $$\beta$$ is a basis for W, so I know the inverse of all these vectors are also independent. And I know that A maps the vectors in U onto W, so the inverse of A will map the vectors in W, or $$\beta$$, onto U...but how should I write this out?

To write it out formally you should start with something like

"Let x be an arbitrary element of U..."
and end with:
"... thus x =..."
expressing x as a linear combination of the asserted basis of U.

 

[mmmboh]

Let x be an arbitrary element of U, Ax lies in W, and can be expressed as C1V1+...+CnVn. Thus (A^-1)x lies in U?
:S

 

[mmmboh]

If I show that A^-1v₁,...,A^-1v_n is independent because $$\alpha$$₁A^-1v₁+...+$$\alpha$$_nA^-1v_n= 0 only when all the $$\alpha$$'s are zero, would this answer the question?

The assignment is due tomorrow

 

[jambaugh]

Let x be an arbitrary element of U, Ax lies in W, and can be expressed as C1V1+...+CnVn. Thus (A^-1)x lies in U?
:S

Note A inverse times x but A inverse times A x which you've just written as ...

 

[mmmboh]

I think I got it, thanks!