Proving/verifying dimension and basis

Question

I'm coming from a computer science background and am currently trying to formalize my linear algebra knowledge by going through Linear Algebra Done Right. I have an intuitive grasp on most of the basic concepts but I struggle with trying to prove things rigorously. Here's an example:

Let $U=\{p\in \mathcal{P}_4(F): p(2)=p(5)\}$. Find a basis of U. Extend the basis to a basis of $\mathcal{P}_4(F)$. Find a subspace $W$ of $\mathcal{P}_4(F)$ such that $\mathcal{P}_4(F)=U\oplus W$.

It was pretty straightforward for me to find a basis $\{x^4-203x,x^3-39x,x^2-7x,1\}$, but I have no idea how to really prove that it's in fact a basis. One technique I know is to show that a set is (1) linearly independent, and (2) of the same size as the dimension of the vector space. Condition 1 can be proved by stating that each vector contains a polynomial of a different degree, and so none can be constructed as a linear combination of the others. However, how would I prove condition 2? Specifically, how do I prove that the vector space is of dimension 4? Intuitively it makes sense that

1. $\mathcal{P}_4(F)$ has dimension 5
2. $U$ excludes some members from that space so it has to have fewer dimensions
3. The condition $p(2)=p(5)$ places a single constraint on U, so it makes sense that it has one less dimension.

However, I have no idea how to formalize this.

Help with this specific question would be great, but what I'm really curious about is if there are standard ways to find the dimension of spaces with definitions like this, or if there are ways to verify a basis without relying on the dimension of the vector space.

Answer 1

Let's start by finding a basis of $$ U=\{p\in P_4:p(2)=p(5)\} $$ To do so, suppose $p\in P_4$ is given by $$ p(t)=a_0+a_1t+a_2t^2+a_3t^3+a_4t^4 $$ Then $p\in U$ if and only if $$ a_{0} + 2 \, a_{1} + 4 \, a_{2} + 8 \, a_{3} + 16 \, a_{4} = a_{0} + 5 \, a_{1} + 25 \, a_{2} + 125 \, a_{3} + 625 \, a_{4} $$ That is, $p\in U$ if and only if $$ 3 \, a_{1} + 21 \, a_{2} + 117 \, a_{3} + 609 \, a_{4}=0 $$ Dividing by $3$ gives $p\in U$ if and only if $$ a_{1} + 7 \, a_{2} + 39 \, a_{3} + 203 \, a_{4}=0\tag{1} $$ Now, (1) implies that $p\in U$ if and only if \begin{align*} p(t) &= a_0-(7 \, a_{2} + 39 \, a_{3} + 203 \, a_{4})t+a_2t^2+a_3t^3+a_4t^4 \\ &= a_0+a_2(-7\,t+t^2)+a_3(-39\,t+t^3)+a_4(-203\,t+t^4) \end{align*} This proves that $$ U=\DeclareMathOperator{Span}{Span}\Span\{ 1,-7\,t+t^2,-39\,t+t^3,-203\,t+t^4 \}\tag{2} $$ To prove that the polynomials listed in (2) are linearly independent, suppose $$ \lambda_1\cdot 1+\lambda_2\cdot(-7\,t+t^2)+\lambda_3\cdot(-39\,t+t^3)+\lambda_4\cdot(-203\,t+t^4)=0 $$ Then $$ \lambda_{4} t^{4} + \lambda_{3} t^{3} + \lambda_{2} t^{2} - 7 \, \lambda_{2} t - 39 \, \lambda_{3} t - 203 \, \lambda_{4} t + \lambda_{1}=0\tag{3} $$ In particular, the coefficients of $t^0$, $t^2$, $t^3$, and $t^4$ in (3) must vanish. Hence $\lambda_1=\lambda_2=\lambda_3=\lambda_4=0$ so our polynomials are linearly independent.

This shows that the polynomials listed in (2) is a basis of $U$.

Finally, to find a subspace $W$ of $P_4$ satisfying $P_4=U\oplus W$ we need only find $\dim P_4-\dim U=5-4=1$ linearly independent polynomials in $P_4$ not in $U$. Of course $p(t)=t$ does the trick so we can take $W=\Span\{t\}$.

Another way to verify that the polynomials in (2) are linearly independent is to check their Wronskian. In this case the Wronskian is $$ W=\det \left[\begin{array}{rrrr} 1 & t^{2} - 7 \, t & t^{3} - 39 \, t & t^{4} - 203 \, t \\ 0 & 2 \, t - 7 & 3 \, t^{2} - 39 & 4 \, t^{3} - 203 \\ 0 & 2 & 6 \, t & 12 \, t^{2} \\ 0 & 0 & 6 & 24 \, t \end{array}\right] = 48 \, t^{3} - 504 \, t^{2} + 1872 \, t - 2436 \not\equiv 0 $$ This also proves that the polynomials are linearly independent (assuming we are working over $\Bbb R$).