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

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 $U$.

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 $n$.

Linear Dependent/Independent Vectors of Polynomials
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ […]

Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis
Let $P_3$ be the vector space over $\R$ of all degree three or less polynomial with real number coefficient.
Let $W$ be the following subset of $P_3$.
\[W=\{p(x) \in P_3 \mid p'(-1)=0 \text{ and } p^{\prime\prime}(1)=0\}.\]
Here $p'(x)$ is the first derivative of $p(x)$ and […]

Dimension of the Sum of Two Subspaces
Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
\[\dim(U+V) \leq \dim(U)+\dim(V).\]
Definition (The sum of subspaces).
Recall that the sum of subspaces $U$ and $V$ is
\[U+V=\{\mathbf{x}+\mathbf{y} \mid […]

Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]

Vector Space of Polynomials and Coordinate Vectors
Let $P_2$ be the vector space of all polynomials of degree two or less.
Consider the subset in $P_2$
\[Q=\{ p_1(x), p_2(x), p_3(x), p_4(x)\},\]
where
\begin{align*}
&p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\
&p_3(x)=2x^2, &p_4(x)=2x^2+x+1.
\end{align*}
(a) Use the basis […]

The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero
Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$.
Then prove that $V$ is a subspace of $\R^n$.
Proof.
To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]

Linear Transformation and a Basis of the Vector Space $\R^3$
Let $T$ be a linear transformation from the vector space $\R^3$ to $\R^3$.
Suppose that $k=3$ is the smallest positive integer such that $T^k=\mathbf{0}$ (the zero linear transformation) and suppose that we have $\mathbf{x}\in \R^3$ such that $T^2\mathbf{x}\neq \mathbf{0}$.
Show […]