# Is $Y$ necessarily a subspace of $V^n$?

The question is given below:

My guess for the answer is yes.

My trial for justification is:

I know that for $$Y$$ to be a subspace of $$V^n$$ it must satisfy that $$\forall r \in F$$ and $$\forall u,v \in Y$$, we must have that $$ru + v \in Y$$. And we are sure that this is true because all vectors in $$Y$$ are linearly dependent.

Am I correct?

• Does it contain $0$ vector? – Prof.Shanku Sep 11 at 15:36
• @Prof.Shanku I think the zero vector by itself is linearly dependent .... correct? as 0 = 1.0 =0 so $a=1$ a nonzero coefficient – Smart Sep 11 at 15:41
• yes I know this information. @Prof.Shanku – Smart Sep 11 at 15:43
• @Prof.Shanku See below for the answer : it very well contains the zero vector, but is still not a subspace. – астон вілла олоф мэллбэрг Sep 11 at 15:48

$$\newcommand{\o}{\mathbf 0}$$(Note : $$\dim V = \infty$$ if $$V$$ is infinite dimensional)

Hint 1 : If $$2 \leq n \leq \dim V$$, then let $$\{e_1,...e_n\}$$ be a linearly independent subset of $$V$$. Consider $$(\o,\o , ... , e_i, ...,\o) = f_i \in V^n$$, where $$\o$$ denotes the zero vector.

• Are each of the $$f_i \in Y$$? Why/Why not?

• Is $$\sum_{i=1}^n f_i \in Y$$? Why/Why not?

• Can you conclude if $$Y$$ is a subspace or not from the above?

Hint 2 : If $$n > \dim V$$, then why is $$Y=V^n$$?

I think you have not understood the question properly.

So $$V$$ is a vector space. You want me to assume it is finite dimensional, so I will assume that. We call the elements of $$V$$ as vectors.

Now, what is $$V^n$$? Note that $$V^n$$, is the set of all $$n$$-tuples of vectors in $$V$$. That is, if say $$v_1,v_2,...,v_n$$ are vectors in $$V$$, then the element $$(v_1,v_2,...,v_n)$$ is an element of $$V^n$$.

For example, if $$\o$$ denotes the zero vector in $$V$$, then $$(\o,\o,...,\o)$$ is an element of $$V^n$$, where all the $$n$$ entries are $$\o$$.

Note that $$V^n$$ is a vector space, under component wise addition and scalar multiplication. Also, there is no relation between $$n$$ and the dimension of $$V$$ for now.

What is $$Y$$? $$Y$$ is a subset of $$V^n$$ which is defined as follows : take an element of $$V^n$$. As I mentioned before, it has $$n$$ entries, which are all vectors, say $$v_1,...,v_n$$ (they need not be distinct, remember that).

Now, take the set $${v_1,...,v_n}$$. This is either linearly independent or linearly dependent in $$V$$. If it is linearly dependent in $$V$$, then $$(v_1,...,v_n) \in Y$$. Otherwise it does not belong in $$Y$$.

Let us take an example. Is $$(\o,\o,...,\o) \in Y$$? Yes, because the elements are linearly independent : see, we need a non-zero linear combination of $$\o,\o,...,\o$$ to be zero. Non-zero means at least one of the coefficients must be non-zero. But this can be done easily , because : $$1\o + 0\o + ... + 0\o = \o$$ which is a non-zero linear combination of the elements of the tuple and equals zero. Therefore, $$(\o,\o,...,\o) \in Y$$.

Consider the element $$(\o,v_2,...,v_n) \in V^n$$. I claim it is in $$Y$$. Why? Because : $$1 \o + 0v_2+...+0v_n = \o$$ is a non-zero linear combination of the entries which equals zero. Therefore , $$(\o,v_2,...,v_n) \in Y$$ for all choices of $$v_2,...,v_n$$.

From here, you should be able to see the following lemma :

Let $$(v_1,v_2,...,v_n) \in V^n$$. If at least one of the $$v_i$$ is the zero vector ,then $$(v_1,v_2,...,v_n) \in Y$$.

Why? Well, think what I did when the zero was in the first position. What non-zero linear combination would I take if the zero was in the $$i$$th position, for example?

With the lemma behind us, we now need to recall a common result :

Let $$W$$ be a subspace of dimension $$m$$. Then, any set of $$m+1$$ or more elements in $$W$$ is linearly dependent.

Now, let $$n > \dim V$$. Then, any element of $$V^n$$ contains at least $$\dim V + 1$$ entries, so must be linearly dependent. That is, $$Y = V^n$$, because I just showed that every element of $$V^n$$ must be in $$Y$$. So of course, in this case, $$Y$$ is a subspace.

Finally, what happens when $$2 \leq n \leq \dim V$$? Well, there we have the following result :

If $$W$$ is a vector space and $$p \leq \dim W$$ then there exists a subset $$\{v_1,...,v_p\}$$ of $$W$$ which is linearly independent.

With this I claim $$Y$$ is not a subspace. Why? We will show that $$Y$$ is not closed under addition : recall that closed under addition means (or can shown to be equivalent to) that for every $$w_1,...,w_n \in Y$$, we have that $$w_1+...+w_n \in Y$$. We will find vectors $$w_1,...,w_n$$ such that each one is in $$Y$$ but their sum is not in $$Y$$.

What we do for this is use the common result : since $$n \leq \dim V$$, we get a set $$\{e_1,...,e_n\} \subset V$$ which is linearly independent. This set has nothing to do with the standard basis vectors in $$\mathbb R^n$$, which are also often labelled $$e_i$$.

Now, we create the elements $$f_i\in V^n$$ ,for $$1 \leq i \leq n$$, as follows : $$f_i$$ needs to have $$n$$ vector entries. Let us take all entries except the $$i$$th entry as $$\o$$, and the $$i$$th entry as $$e_i$$. In other words, $$f_i = (\o,\o,...,\o,\underbrace{e_i}_{\text{ith entry}},\o,...,\o)$$. Convince yourself that $$f_i$$ is actually an element of $$V^n$$, and then from the lemma, an element of $$Y$$.

Now, what is $$\sum_{i=1}^n f_i$$? The addition is component wise, so we add up the first components of the $$f_i$$, then the second components of the $$f_i$$, and the corresponding answers become the components of $$\sum_{i=1}^n f_i$$, which we call $$f \in Y^n$$.

For example, the first entries of the $$f_i$$ are all $$\o$$, except that of $$f_1$$ whose first entry is $$e_1$$. The sum of all these is $$e_1+\o+\o+...+\o = e_1$$, so the first entry of $$f$$ is $$e_1$$.

Similarly, the second entries of the $$f_i$$ are all $$\o$$, expect for $$f_2$$ whose first entry is $$e_2$$. The sum of all these is $$\o+e_2+...+\o = e_2$$. Thus the second entry of $$f$$ is $$e_2$$.

Now, do the rest and convince yourself that $$f = (e_1,e_2,...,e_n)$$.

But then , when you take the entries of $$f$$ as a set , you get $$\{e_1,...,e_n\}$$ which is linearly independent by assumption. Therefore, $$f \notin Y$$.

Which shows that $$Y$$ is not a subspace.

(Note : Have you seen where I used $$n > 1$$ above? It was crucial).

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Sep 13 at 13:36
• So by saying this " If$W$ is a vector space and $p≤dimW$ then there exists a subset${v1,...,vp}$ of $W$ which is linearly independent." you are saying that for every number less than or equal $dim W$ there exists a linearly independent set in $W$ ? do you have a source for this theorem, may be a reference? – Intuition Sep 14 at 19:10
• (Note: Have you seen where I used $n>1$ above? It was crucial) .... I think any single non zero vector is linearly independent this is why this case is excluded from the problem ....... am I correct? – hopefully Sep 14 at 23:26
• yes you are right @hopefully – астон вілла олоф мэллбэрг Sep 15 at 9:53
• Could you please look at this question ?math.stackexchange.com/questions/3358228/… – Smart Sep 16 at 13:21