# Linear Algebra - Proof of Subspace $V = \mathcal{F}(\mathbb{C},\mathbb{C})$

Let $$V = \mathcal{F}(\mathbb{C},\mathbb{C})$$, and consider $$W =\left\{ \mathcal{f} \in V: \mathcal{f}{(z+1)}, \forall z \in \mathbb{C} \right\}$$. $$W$$ is a vectorial subspace of $$V$$?

By definition, is needed to show three things:

1) the zero vector (0,0) is in W;
2) $$v,w\in W\implies v+w\in W$$;
3) $$w\in W$$ and $$\lambda \in \mathbb{R}$$ implies that $$\lambda w \in W$$.

Trying to proof 1.

Let $$\mathcal{f}$$ and $$\mathcal{g}$$ $$\in W$$.
Then, $$(\mathcal{f} + \mathcal{g})_{(z+1)} = \mathcal{f}(z+1) + \mathcal{g}(z+1)$$.

Here I got stuck. How can I make the proof through using complex numbers and subspace axioms?

• Can you clarify what you mean by $\mathcal{F}(\mathbb{C},\mathbb{C})$ and "$f \in V : f(z+1)$"? Jan 22, 2019 at 0:28
• @DavidKraemer I don't know. It is how is wrote in my text book. Jan 22, 2019 at 0:32
• My guess for $\mathcal{F}(\mathbb{C},\mathbb{C})$ is the class of all functions $f : \mathbb{C} \to \mathbb{C}$. I think there's a typo in the latter. Jan 22, 2019 at 0:35
• Then your book has a huge mistake: the definition of $\;W\;$ makes no sense at all. Jan 22, 2019 at 0:35
• @Arduin can you share a link to the pdf? Jan 22, 2019 at 0:43

I am going to assume that $$V = \mathcal{F}(\mathbb{C}, \mathbb{C})$$ is the class of all functions $$f : \mathbb{C} \to \mathbb{C}$$ and that the typo in the definition of $$W$$ can be resolved with $$\newcommand{\CC}{\mathbb{C}} W = \{ f \in V : f(z+1) = 0 \text{ for all } z \in \CC \}.$$

To show that $$W$$ is a subspace, we need to show (as you have said):

1. $$0 \in W$$,
2. if $$f, g \in W$$, then $$f+g \in W$$,
3. if $$f \in W$$ and $$c \in \CC$$, then $$c \cdot f \in W$$.

Let's begin with 1. I want to preface by drawing your attention to the fact that $$0$$ is a function $$\CC \to \CC$$ satisfying $$0(z)=0$$ for all $$z \in \CC$$. This is contrary to the usual understanding of $$0$$ as a point or vector. (Of course, the $$0$$ function is a vector, since it's an element of a vector space. But this is confusing to newer students!)

To show that $$0 \in W$$, then, we need to show that $$0(z+1) = 0$$ for all $$z \in \CC$$. Actually, by the very definition of the $$0$$ function we know that $$0(z+1) = 0$$, so we're done here.

Next, step 2. Suppose $$f, g \in W$$ are arbitrary functions. We need to show that the function $$h : \CC \to \CC$$ defined by $$h(z) = f(z) + g(z)$$ is also a member of $$W$$. To this end, we need to show that $$h(z+1) = f(z+1) + g(z+1)$$ for all $$z \in \CC$$. But since $$f, g \in W$$, it follows that \begin{align} h(z+1) &= f(z+1) + g(z+1) \\ &= 0 + 0 \\ &= 0, \end{align} which implies that $$h \in W$$.

Finally, step 3. Let $$f \in W$$ be an arbitrary function and let $$c \in \CC$$ be any scalar. We want to show that the function $$h : \CC \to \CC$$ defined by $$h(z) = c \cdot f(z)$$ is a member of $$W$$. But since $$f \in W$$, it follows that $$f(z) = 0$$ for all $$z \in \CC$$. Then \begin{align} h(z) &= c \cdot f(z) \\ &= c \cdot 0 \\\ &= 0 \end{align} which shows that $$h \in W$$, as needed.

We now can conclude that $$W$$ is indeed a subspace of $$V$$.

Since I'm inferring that the typo is fixed by writing "$$f(z+1) = 0$$" for all $$z$$, you might ask whether the above result would hold if we instead wrote $$f(z+1) = \zeta$$ for some scalar $$\zeta \in \CC$$. (Try to see where the proof fails!) Alternatively, you can explore the situation where $$W$$ is the subclass of functions satisfying $$f(z+1) = f(z)$$ for all $$z \in \CC$$. Follow the same steps from above, and you'll get it!

• Thank you for your answer. I'm going to check if there is a mistake in the question and fix tomorrow. I suppose that is correct your definition and therefore is a valid answer! Jan 22, 2019 at 1:11