Let $V$ be a vector space over a field $K$. Let $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$ be linearly independent vectors in $V$. Let $U$ be the subspace of $V$ spanned by these vectors, that is, $U=\Span \{\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n\}$.
Let $\mathbf{u}_{n+1}\in V$. Show that $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly independent if and only if $\mathbf{u}_{n+1} \not \in U$.

$(\implies)$
Suppose that the vectors $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly independent. If $\mathbf{u}_{n+1}\in U$, then $\mathbf{u}_{n+1}$ is a linear combination of $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$.
Thus, we have
\[\mathbf{u}_{n+1}=c_1\mathbf{u}_1+c_2\mathbf{u}_2+\cdots+c_n \mathbf{u}_n\]
for some scalars $c_1, c_2, \dots, c_n \in K$.

However, this implies that we have a nontrivial linear combination
\[c_1\mathbf{u}_1+c_2\mathbf{u}_2+\cdots+c_n \mathbf{u}_n-\mathbf{u}_{n+1}=\mathbf{0}.\]
This contradicts that $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly independent. Hence $\mathbf{u}_{n+1} \not \in U$.

$(\impliedby)$ Suppose now that $\mathbf{u}_{n+1} \not \in U$.
If the vectors $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly dependent, then there exists $c_1, c_2 \dots, c_n, c_{n+1}\in K$ such that
not all of them are zero and
\[c_1\mathbf{u}_1+c_2\mathbf{u}_2+\cdots+c_n \mathbf{u}_n+c_{n+1}\mathbf{u}_{n+1}=\mathbf{0}.\]

We claim that $c_{n+1} \neq 0$. If $c_{n+1}=0$, then we have
\[c_1\mathbf{u}_1+c_2\mathbf{u}_2+\cdots+c_n \mathbf{u}_n=\mathbf{0}\]
and since $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$ are linearly independent, we must have $c_1=c_2=\cdots=c_n=0$. This means that all $c_i$ are zero but this contradicts our choice of $c_i$. Thus $c_{n+1} \neq 0$.

Then we have
\[\mathbf{u}_{n+1}=\frac{-c_1}{c_{n+1}}\mathbf{u}_1+\frac{-c_2}{c_{n+1}}\mathbf{u}_2+\cdots+\frac{-c_n}{c_{n+1}}\mathbf{u}_n.\]
(Note: we needed to check $c_{n+1} \neq 0$ to divide by it.)

This implies that $\mathbf{u}_{n+1}$ is a linear combination of vectors $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n$, and thus $\mathbf{u}_{n+1} \in U$, a contradiction.
Therefore, the vectors $\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n, \mathbf{u}_{n+1}$ are linearly independent.

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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 $P_3$.
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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.
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Let $V$ be a vector space over a scalar field $K$.
Let $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ be vectors in $V$ and consider the subset
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-1
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2 \\
1 \\
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Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
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\begin{align*}
&p_1(x)=x^2+2x+1, &p_2(x)=2x^2+3x+1, \\
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Hint.
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