Question: Is the zero vector a linear combination of any nonempty set of vectors ? Justify. Is the set of all differential real valued functions defined on R a subspace of c(R)? Justify your answer.

Solution:
First part:
Yes,it is
$$\vec{0} = 0.\vec{v1} + 0.\vec{v2} + ...+0.\vec{vn}$$
Moreever, an empty sum, that is the sum of no vectors, is usually define to be zero, and with that defination 0 is the linear combination of set of vectors, empty or not.
Second Part:
To show differential function is subspace of real valued function define on R a subspace of c(R). Let $$\mathcal{D}(R)$$ denote the set of all differentiable real valued function over R. If $$f : \mathcal {R} \Rightarrow \mathcal{R}$$ is differential then f is continious and so $$\mathcal{D}(R)$$ is a subset of $$\mathcal{C}(R)$$\

We can show that this subset is subspace by showing that this is closed under addition and scalar multiplication, we can proceed as follows.
If f and g are in $$\mathcal{D}(R)$$. Then
$$(f + g)(x)$$ = $$f(x) + g(x)$$ is define for any $$x \in R$$
$$V_{ss1}$$ : Zero is a constant value.
We know that differentiation of constant is zero.
Thus, f(x) = 0
$$\forall x\in R$$. so we have $$f \in \mathcal{c}(R)$$ .
$$V_{ss2}$$ :Sum of two differential function is also differential.
We also have differential function is Continuous.
Let $$f(x) \in R\ and\ f(y) \in R$$
Then, $$f(x) + f(y) \in R \Rightarrow f(x) + f(y) \in \mathcal{C}(R)$$
$$V_{ss3}$$: Scalar multiplication to any differential function is also differential. This implies also Continuous
i.e $$\forall c \in R\ and\ f(x) \in R$$
$$\mathcal{C}f(x) \in (R) \Rightarrow \mathcal{C}f(x) \in \mathcal {C}(R)$$.

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