Find dimension and a basis of $W = \{f(x) \in P_n[x] \,| \,f(1) = 0\}$

Question

Given that $W = \{f(x) \in P_n[x] \,| \,f(1) = 0\}$ is a subspace of $P_n[x]$. Find its dimension and a basis.

My attempts: I know that dimension of $P_n[x] = n + 1$

here constraints is that is $f(1) = 0$. that is only one constraints..

so dimension of $P_n[x] = n $ and basis will be $1,x,x^2,\ldots,x^{n-1}$.

is its true/false

Any solution/hints will be appreciated

Answer 1

Hint:

$W$ is a subspace of $P_n[x]$ so $\dim W \le \dim P_n[x] = n+1$. However, clearly $W \ne P_n[x]$ so actually $\dim W \le n$.

Now check that the set $$\{x-1, x^2-1, \ldots, x^n-1\} \subseteq W$$

is linearly independent and conclude that it is a basis for $W$.

Answer 2

A polynomial in $W$ is in the general form

$$a_nx^n+a_{n-1}x^{n-1}+...+a_1x-a_n-a_{n-1}-a_1$$

therefore a basis is given by

$$\{x^n-1,x^{n-1}-1,...,x^2-1,x-1\}$$

Answer 3

It is true that there is only one constraint. However, one constraint is not resolved by removing one element of a given basis, and saying that whatever is left spans the constrained space.

In fact, in your case, none of $1,x,x^2...,x^n$ even belong to $W$, let alone span it, because they don't satisfy $f(1) = 0$! (They satisfy $f(1) = 1$).

Consequently, any basis that you produce for the new space, must not contain any of these elements!

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There is a better way of thinking here. We know that $1,x,x^2,...,x^n$ is a basis for $P_n$. It is not very difficult to see that $1,(x+1),(x+1)^2,...,(x+1)^n$ is also a basis for $P_n$. If you cannot see why, you can ask me to show this.

Now, if you take any linear combination of these which avoids $1$ (i.e. the constant term is zero), then it is a multiple of $x+1$.

Consequently, the new basis can be shortened to a basis of $W$, given by $(x+1),...,(x+1)^n$. It will have dimension $n$.

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Moral(s) of the story :

• When you have a basis with you, adding even a single constraint can make a basis of the new space differ radically from the larger space's basis. In particular, one cannot delete elements from any given basis, to get a basis of a smaller space.

• However, there exists a choice of basis of the larger space, from which you can delete elements to get a basis of the smaller space. Often, this is written the other way, but here we constructed a basis of the smaller space by deletion, hence I am telling you this. In case of polynomials and the sort of question given to you, this answer will be helpful.

Answer 4

Another basis, more natural from my point of view, results from the isomorphism which can be deduced from the fact that $f(1)=0$ if and only if $f(x)$ is divisible by $x-1$: \begin{align} P_{n-1}(x)&\longrightarrow W,\\ g(x)&\longmapsto f(x)=(x-1)g(x). \end{align} This isomorphism implies $\dim W=n$, and we can use as a basis for $W$ $$\bigl\{x-1, x^2-x,\dots, x^n-x^{n-1}\bigr\}.$$

Answer 5

$(x-1)$ will be a factor of each element of the subspace (by the factor theorem). In $P_n(x)$ there are $n-1$ linearly independent vectors of degree $\le n$ satisfying $f(1)=0$. Namely, $\{x-1,x(x-1),x^2(x-1),\dots,x^{n-1}(x-1)\}$. These span an $n$-dimensional subspace. $1$ is not in the subspace, showing we have a subspace of dimension $n$...