Use the definition of a vector space V to prove the following: a. \(0 \mathbf{u}=\mathbf{0}\) for every \(\mathbf{u} \in V\). b. \(-\mathbf{u}=(-1) \mathbf{u}\) for every \(\mathbf{u} \in V\). (Hint: The distributive property 7 is all important.)
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Textbook Solutions for Linear Algebra: A Geometric Approach
Question
Decide whether each of the following is a subspace of \(\mathcal{C}^{0}(\mathbb{R})\). If so, provide a basis and determine its dimension. If not, give a reason.
a. \(\{f: f(1)=2\}\)
b. \(\left\{f \in \mathcal{P}_{2}: \int_{0}^{1} f(t) d t=0\right\}\)
c. \(\left\{f \in \mathcal{C}^{1}(\mathbb{R}): f^{\prime}(t)+2 f(t)=0 \text { for all } t\right\}\)
d. \(\left\{f \in \mathcal{P}_{4}: f(t)-t f^{\prime}(t)=0 \text { for all } t\right\}\)
e. \(\left\{f \in \mathcal{P}_{4}: f(t)-t f^{\prime}(t)=1 \text { for all } t\right\}\)
f. \(\left\{f \in \mathcal{C}^{2}(\mathbb{R}): f^{\prime \prime}(t)+f(t)=0 \text { for all } t\right\}\)
g. \(\left\{f \in \mathcal{C}^{2}(\mathbb{R}): f^{\prime \prime}(t)-f^{\prime}(t)-6 f(t)=0 \text { for all } t\right\}\)
h. \(\left\{f \in \mathcal{C}^{1}(\mathbb{R}): f(t)=\int_{0}^{t} f(s) d s \text { for all } t\right\}\)
Solution
Step 1 of 8
(a) Let us denote the given set by \(A\) i.e. \(A=\{f: f(1)=2\}\)
It is known that a non-empty subset \(W\) of a vector space \(V\) is a subspace of \(V\) if:
\(\ast\ 0\in W\)
\(\ast u,v\in W\Rightarrow(u+v)\in W\)
\(\ast \text { For } u \in W,(c u) \in W\) for any scalar \(c\)
From the given definition of \(A\), it is clear that the zero function \(f(x)=0\) doesn’t belong to \(A\) as it doesn’t satisfy the \(f(1)=2\). Therefore, \(A\) is not a subspace.
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