Linear Independent Vectors and the Vector Space Spanned By Them
Problem 141
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.
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 […]
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Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set
\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\]
still a spanning set for […]
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 […]
The Subspace of Linear Combinations whose Sums of Coefficients are zero
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
\[W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } […]
Determine Whether Each Set is a Basis for $\R^3$
Determine whether each of the following sets is a basis for $\R^3$.
(a) $S=\left\{\, \begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}, \begin{bmatrix}
2 \\
1 \\
-1
\end{bmatrix}, \begin{bmatrix}
-2 \\
1 \\
4
\end{bmatrix} […]
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*}
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Let $V$ be the following subspace of the $4$-dimensional vector space $\R^4$.
\[V:=\left\{ \quad\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix} \in \R^4
\quad \middle| \quad
x_1-x_2+x_3-x_4=0 \quad\right\}.\]
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Column Rank = Row Rank. (The Rank of a Matrix is the Same as the Rank of its Transpose)
Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$.
Hint.
Recall that the rank of a matrix $A$ is the dimension of the range of $A$.
The range of $A$ is spanned by the column vectors of the matrix […]