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

Question

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?

Answer 1

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 $$ \newcommand{\CC}{\mathbb{C}} W = \{ f \in V : f(z+1) = 0 \text{ for all } z \in \CC \}. $$

If I'm wrong, please disregard this answer!

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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 $V$.

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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!