Linear algebra, find a basis for the quotient space

[Karl Karlsson]

Homework Statement

Let V = C[x] be the vector space of all polynomials in x with complex coefficients and let $W = \{p(x) ∈ V: p (1) = p (−1) = 0\}$.

Determine a basis for V/W

Relevant Equations

V = C[x]
$W = \{p(x) ∈ V: p (1) = p (−1) = 0\}$.

Let V = C[x] be the vector space of all polynomials in x with complex coefficients and let $W = \{p(x) ∈ V: p (1) = p (−1) = 0\}$.

Determine a basis for V/W

The solution of this problem that i found did the following:

Why do they choose the basis to be {1+W, x + W} at the end? I mean since
$dim(ker(L)) + dim (im(L)) = dim(V), dim(ker(L)) = dim(W)$ and then $dim(im(L))=2 = dim(V) - dim(W)$ that means $dim(W) \rightarrow \infty$ because there is an infinite number of linearly independent polynomials that satisfy p(1)=p(-1)=0. Can't I just choose W to be a basis for V/W?

Thanks in advance!

 

[Office_Shredder]

I think there's some confusion. W in the vector space V/W is a single vector, and is the zero vector. Recall V/W is the space of sets of the form x+W for some x, where is x-y is in W then x+W = y + W.

This is what they mean when they write 1+W and x+W.

If you really want the whole set W to be your sweet of basis (which doesn't even work, since they don't even span the space in question) you would have many linearly dependent vectors and you wouldn't have a basis.

As you observed some of the vector spaces have infinite dimensions, so your formula for the dimension of the image and kernel don't actually make sense. I can't read what is written in the solution but it looks like they take a different approach to show V/W has 2 dimensions

 

[fresh_42]

$p(1)=0$ means $(x-1)\,|\,p(x)$ and $p(-1)=0$ gives $(x+1)\,|\,p(x)$, so we have $(x^2-1)\,|\,p(x)$. Now $V/W = \mathbb{C}[x]/(x^2-1)\cdot\mathbb{C}[x]$, which means we can reduce any polynomial $p(x)$ to one of degree at most $1$ by writing $q(x)=f(x)\cdot (x^2-1)+ g(x)\, , \,\deg g \leq 1.$ Thus we have $\pi(p)=g$ if $\pi : V\longrightarrow V/W$ is the canonical projection. Since $g(x)$ is of degree zero or one, we can use $[ g]=1$ and $[g]=x$ as representatives of the equivalence classes. You cannot use dimension formulas, as $\infty$ isn't a number.

 

[pasmith]

Alternatively, since x^{2k} - 1 and x^{2k+1} - x vanish at x = \pm 1 and are non-zero if k \geq 1 one can write <br /> \begin{align*}<br /> \sum_{n \geq 0} a_nx^n &amp;= a_0 + a_1 x + \sum_{k \geq 1} a_{2k}x^{2k} + \sum_{k \geq 1} a_{2k+1} x^{2k+1} \\<br /> &amp;= \left(a_0 + \sum_{k \geq 1} a_{2k}\right) + x\left( a_1 + \sum_{k \geq 1} a_{2k+1}\right)<br /> + \sum_{k \geq 1} a_{2k} (x^{2k} - 1) + \sum_{k \geq 1} a_{2k+1} (x^{2k + 1} - x)\end{align*} which is a polynomial of degree at most 1 plus a polynomial in W.

 

[Karl Karlsson]

I think there's some confusion. W in the vector space V/W is a single vector, and is the zero vector. Recall V/W is the space of sets of the form x+W for some x, where is x-y is in W then x+W = y + W.

This is what they mean when they write 1+W and x+W.

If you really want the whole set W to be your sweet of basis (which doesn't even work, since they don't even span the space in question) you would have many linearly dependent vectors and you wouldn't have a basis.

As you observed some of the vector spaces have infinite dimensions, so your formula for the dimension of the image and kernel don't actually make sense. I can't read what is written in the solution but it looks like they take a different approach to show V/W has 2 dimensions

Alternatively, since x^{2k} - 1 and x^{2k+1} - x vanish at x = \pm 1 and are non-zero if k \geq 1 one can write <br /> \begin{align*}<br /> \sum_{n \geq 0} a_nx^n &amp;= a_0 + a_1 x + \sum_{k \geq 1} a_{2k}x^{2k} + \sum_{k \geq 1} a_{2k+1} x^{2k+1} \\<br /> &amp;= \left(a_0 + \sum_{k \geq 1} a_{2k}\right) + x\left( a_1 + \sum_{k \geq 1} a_{2k+1}\right)<br /> + \sum_{k \geq 1} a_{2k} (x^{2k} - 1) + \sum_{k \geq 1} a_{2k+1} (x^{2k + 1} - x)\end{align*} which is a polynomial of degree at most 1 plus a polynomial in W.

Hi Office_Shredder! Thanks that is just the type of solution i was looking for :)

 

[Karl Karlsson]

$p(1)=0$ means $(x-1)\,|\,p(x)$ and $p(-1)=0$ gives $(x+1)\,|\,p(x)$, so we have $(x^2-1)\,|\,p(x)$. Now $V/W = \mathbb{C}[x]/(x^2-1)\cdot\mathbb{C}[x]$, which means we can reduce any polynomial $p(x)$ to one of degree at most $1$ by writing $q(x)=f(x)\cdot (x^2-1)+ g(x)\, , \,\deg g \leq 1.$ Thus we have $\pi(p)=g$ if $\pi : V\longrightarrow V/W$ is the canonical projection. Since $g(x)$ is of degree zero or one, we can use $[ g]=1$ and $[g]=x$ as representatives of the equivalence classes. You cannot use dimension formulas, as $\infty$ isn't a number.

Hi fresh_42! Thanks for another great solution that was easy to follow!

 

[Karl Karlsson]

Hi just for clarification. Does the basis $\{1+W, x + W\}$ for V/W refer to all multiples of x+W and 1+W meaning for example 2x + 7+ W is also within the basis above?

 

[pasmith]

Hi just for clarification. Does the basis $\{1+W, x + W\}$ for V/W refer to all multiples of x+W and 1+W meaning for example 2x + 7+ W is also within the basis above?

2x + 7 + W is not a basis vector. It is within the space spanned by the basis, since 2x + 7 + W = 2(x + W) + 7(1 + W) since if p \in W then 9p \in W.

 

[fresh_42]

Hi just for clarification. Does the basis $\{1+W, x + W\}$ for V/W refer to all multiples of x+W and 1+W meaning for example 2x + 7+ W is also within the basis above?

All linear combinations: $\mathbb{C}\cdot (1+W) + \mathbb{C}\cdot (x+W)$.

 

[Karl Karlsson]

The answer $\{1+W,x+W\}$ to this problem {W+1,W+x} spans all of V, right? If this problem instead would have been the exact same but V was the vector space of all polynomials of degree n (n+1 vectors). Wouldn't the solution have been exactly the same? And then $dim(V/W) = dim(V)-dim(W) = 2$, but clearly the basis $V/W = \{1+W,x+W\}$ has more then just two linearly independent vectors since W contains n-1 linearly independent vectors, right?

 

[fresh_42]

The answer $\{1+W,x+W\}$ to this problem {W+1,W+x} spans all of V, right?

No. It spans $V/W$ not $V=\mathbb{C}[x].$

If this problem instead would have been the exact same but V was the vector space of all polynomials of degree n (n+1 vectors). Wouldn't the solution have been exactly the same?

Yes, the resulting quotient would be the same, i.e. an isomorphic copy to be exact, namely isomorphic to all linear polynomials.

And then $dim(V/W) = dim(V)-dim(W) = 2$, but clearly the basis $V/W = \{1+W,x+W\}$ has more then just two linearly independent vectors since W contains n-1 linearly independent vectors, right?

No. $W$ is the zero vector in $V/W$. There is no resolution of $W$ into elements of $V$ anymore.

Test the concept with $V=\mathbb{Z}$ and $W= 12\mathbb{Z}$ instead. It isn't a vector space, only a ring and an ideal, but it helps to understand quotient building. What are the generators of $V/W = \mathbb{Z}/12\cdot\mathbb{Z}$ and how can you represent them most conveniently?