Linear Algebra

A Basis for the Vector Space of Polynomials of Degree Two or Less and Coordinate Vectors

Problem 150

Show that the set
\[S=\{1, 1-x, 3+4x+x^2\}\] is a basis of the vector space $P_2$ of all polynomials of degree $2$ or less.

Add to solve later

Proof.

We know that the set $B=\{1, x, x^2\}$ is a basis for the vector space $P_2$.
With respect to this basis $B$, the coordinate vectors of vectors in $S$ are
\[ [1]_B=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \quad [1-x]_B=\begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}, \quad [3+4x+x^2]_B=\begin{bmatrix}
3 \\
4 \\
1
\end{bmatrix}.\]

Recall the fact that the set $S$ is a basis for $P_2$ if and only if the coordinate vectors
$T=\{[1]_B, [1-x]_B, [3+4x+x^2]_B\}$ is a basis for $R^3$.

Hence we check that $T$ is a basis for $R^3$. Note that $T$ is a linearly independent set.
In fact, if we have
\[a_1 \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}
+a_2 \begin{bmatrix}
1 \\
-1 \\
0
\end{bmatrix}
+a_3 \begin{bmatrix}
3 \\
4 \\
1
\end{bmatrix}=\mathbf{0},\] then we have
\[\begin{bmatrix}
a_1+a_2+3a_3 \\
-a_2+4a_3 \\
a_3
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}\] and we obtain $a_1=a_2=a_3=0$, and thus $T$ is a linearly independent set.

Since $T$ consists of three linearly independent vectors in $\R^3$ and the dimension of $\R^3$ is $3$, the set $T$ must be a basis for $\R^3$.
Therefore, the set $S$ is a basis for $P_2$.

Add to solve later

More from my site

Find a Basis and Determine the Dimension of a Subspace of All Polynomials of Degree $n$ or Less 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 $U$. Solution. […]
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 […]
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 […]
$Find a Basis for a Subspace of the Vector Space of $2\times 2$ Matrices$ Find a Basis for a Subspace of the Vector Space of $2\times 2$ Matrices Let $V$ be the vector space of all $2\times 2$ matrices, and let the subset $S$ of $V$ be defined by $S=\{A_1, A_2, A_3, A_4\}$, where \begin{align*} A_1=\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}, \quad A_2=\begin{bmatrix} 0 & -1 \\ 1 & 4 […]
$Linear Transformation and a Basis of the Vector Space $\R^3$$ 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 […]
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 […]
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 $p_i(1)=0$ […]
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 […]