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 $P_2$.

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 $B=\{1, […]

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 $\R^2$.
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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
[…]

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Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
\[V:=\left\{ \quad\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \R^4
\quad \middle| \quad
x_1-x_2+x_3-x_4=0 \quad\right\}.\]
Find a basis of the subspace $V$ […]

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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 […]

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 […]