# Vector Space of Polynomials and Coordinate Vectors

## Problem 157

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 $B=\{1, x, x^2\}$ of $P_2$, give the coordinate vectors of the vectors in $Q$.
(b) Find a basis of the span $\Span(Q)$ consisting of vectors in $Q$.
(c) For each vector in $Q$ which is not a basis vector you obtained in (b), express the vector as a linear combination of basis vectors.

## Solution.

### Coordinate vectors

(a) With respect to the given basis $B=\{1, x, x^2\}$, the coordinate vectors are
\begin{align*}
&[p_1(x)]_B=\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}, &[p_2(x)]_B=\begin{bmatrix}
1 \\
3 \\
2
\end{bmatrix},\\
& [p_3(x)]_B=\begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}, &[p_4(x)]_B=\begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}.
\end{align*}
For example, to obtain $[p_3(x)]_B$, we express $p_3(x)$ as a linear combination of vectors in $B$ and we get
$p_3(x)=0\cdot 1 +0 \cdot x+2\cdot x^2.$ Note that the order of the sum in the linear combination is the order of basis vectors in $B=\{1, x, x^2\}$. Then we read the coefficients of this linear combination and obtain the coordinate vectors
$[p_3(x)]_B=\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}.$ The rest are the same.

### Find a basis of the span

(b) Let
\begin{align*}
T:&=\{[p_1(x)]_B, [p_2(x)]_B, [p_3(x)]_B, [p_4(x)]_B\}\\
&=\left\{\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
3 \\
2
\end{bmatrix}, \begin{bmatrix}
0 \\
0 \\
2
\end{bmatrix}, \begin{bmatrix}
1 \\
1 \\
2
\end{bmatrix}\right\}.
\end{align*}
Then a basis of $\Span(Q)$ consisting of vectors in $Q$ corresponds to a basis of $\Span(T)$ consisting of vectors in $T$.
Consider the matrix whose columns are the vectors in $T$. We reduce it by elementary row operations as follows.
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 & 1 \\
2 &3 & 0 & 1 \\
1 & 2 & 2 & 2
\end{bmatrix}
\xrightarrow[R_3-R_1]{R_2-2R_1}
\begin{bmatrix}
1 & 1 & 0 & 1 \\
0 &1 & 0 & -1 \\
0 & 1 & 2 & 1
\end{bmatrix}
\xrightarrow[R_3-R_2]{R_1-R_2}\\
\begin{bmatrix}
1 & 0 & 0 & 2 \\
0 &1 & 0 & -1 \\
0 & 0 & 2 & 2
\end{bmatrix}
\xrightarrow{\frac{1}{2}R_3}
\begin{bmatrix}
1 & 0 & 0 & 2 \\
0 &1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form and the first three columns contain the leading 1’s.
Therefore the set consisting of the first three vectors in $T$
$\left\{ \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}\right\}$ is a basis for $\Span(T)$.
Therefore by the coordinate vector correspondence, it follows that the set
$\{p_1(x), p_2(x), p_3(x)\}$ is a basis for $\Span(Q)$ consisting of vectors in $Q$.

### Linear combination of basis vectors

(c) From part (b), the only vector in $Q$ which is not a basis vector is $p_4(x)$.
Thus we want to express $p_4(x)$ as a linear combination of $p_1(x), p_2(x), p_3(x)$.
Consider a linear combination
$a_1p_1(x)+a_2 p_2(x)+a_3p_3(x)+a_4p_4(x)=0 \tag{*}.$ By taking coordinate vectors, we obtain
$a_1\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}+a_2\begin{bmatrix} 1 \\ 3 \\ 2 \end{bmatrix}+a_3\begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}+a_4\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}=0.$

The augmented matrix is
$\left[\begin{array}{rrrr|r} 1 & 1 & 0 & 1 &0\\ 2 &3 & 0 & 1 &0\\ 1 & 2 & 2 & 2 &0 \end{array} \right].$ Note that this is the matrix we considered in part (b) except for the last zero column.
Thus by the same elementary row operations, we obtain the reduced row echelon form of this matrix
$\left[\begin{array}{rrrr|r} 1 & 0 & 0 & 2 &0\\ 0 &1 & 0 & -1 &0 \\ 0 & 0 & 1 & 1&0 \end{array} \right].$ Therefore we obtain the solutions
\begin{align*}
a_1&=-2a_4\\
a_2&=a_4\\
a_3&=-a_4\\
\end{align*}
and $a_4$ is a free variable.
(This is the Gauss-Jordan elimination method.)
Let us take $a_4=1$. Then from the above solutions, we obtain $a_1=-2, a_2=1, a_3=-1$.
We plug these values in the linear combination (*) and obtain
$-2p_1(x)+p_2(x)-p_3(x)+p_4(x)=0.$ Solving this for $p_4(x)$, we obtain the linear combination
$p_4(x)=2p_1(x)-p_2(x)+p_3(x).$

##### Give the Formula for a Linear Transformation from $\R^3$ to $\R^2$
Let $T: \R^3 \to \R^2$ be a linear transformation such that \[T(\mathbf{e}_1)=\begin{bmatrix} 1 \\ 4 \end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix} 2 \\ 5...