Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less

Linear Algebra Problems and Solutions

Problem 137

Let $P_n(\R)$ be the vector space over $\R$ consisting of all degree $n$ or less real coefficient polynomials. Let
\[U=\{ p(x) \in P_n(\R) \mid p(1)=0\}\] be a subspace of $P_n(\R)$.

Find a basis for $U$ and determine the dimension of $U$.

 
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Solution.

Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be an element in $U$.
Since $p(1)=0$, we have $a_n+a_{n-1}+\cdots+a_1+a_0=0$. Thus we have
\begin{align*}
p(x)&=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x-(a_n+a_{n-1}+\cdots+a_1)\\
&=a_n(x^n-1)+a_{n-1}(x^{n-1}-1)+\cdots a_1(x-1).
\end{align*}


Let us define
\[q_n(x)=x^n-1, q_{n-1}(x)=x^{n-1}-1, \dots, q_1(x)=x-1 \in U.\] Then the above computation shows that we have
\[p(x)=a_nq_n(x)+a_{n-1}q_{n-1}(x)+\cdots a_1q_1(x).\] In other words, any element $p(x)\in U$ is a linear combination of $q_1(x), \dots, q_n(x)$.
Hence $B:=\{q_1(x), \dots, q_n(x)\}$ is a spanning set for $U$.


We show that $B$ is a linearly independent set.
Consider a linear combination
\[c_1q_1(x)+\cdots+c_n q_n(x)=\theta(x):=0x^{n}+0x^{n-1}+\cdots+0x+0.\]

Then we have
\begin{align*}
c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x-(c_1+c_2+\cdots+c_n)=0.
\end{align*}
Comparing the coefficients of both sides, we obtain
\[c_1=c_2=\cdots=c_n=0\] and thus $q_1(x), \dots, q_n(x)$ are linearly independent, and hence $B=\{q_1(x), \dots, q_n(x)\}$ is a basis for the subspace $U$. Since the dimension is the number of vectors in a basis, the dimension of $U$ is $n$.


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