# Question about my approach to linear algebra proof, Axler Ch.2A #14

I need some help verifying my approach because the answers that I found online use a different approach.

Problem statement:

Prove that a vector space V is infinite-dimensional if and only if there is a sequence of vectors $$v_1, v_2, ...$$ in $$V$$ such that $$v_1,..,v_m$$ is linearly independent for every positive integer $$m$$.

Forward direction.

Premise: Suppose $$V$$ is infinite-dimensional. We want to show that a linearly independent list of vectors $$v_1,..,v_m$$ exists for every positive integer $$m$$.

Suppose no linearly independent list exists for some list of $$m$$ vectors. This implies that the spanning list is less than $$m$$. To see this, consider that a list of vectors will always be dependent if and only if the size of the list is greater than a linearly independent list that also spans the vector space.

So, there is some list size $$ that spans V. This contradicts the original premise that V is infinite dimensional. Therefore, a linearly independent list must exist for every positive integer $$m$$.

Reverse direction.

Premise: Suppose a linearly independent list exists for some $$v_1,..,v_m$$ for every positive integer $$m$$. We want to show that $$V$$ is infinite-dimensional.

Suppose $$V$$ is finite dimensional. Then, suppose some arbitrary list of vectors of size $$m-1$$ spans this finite-dimensional $$V$$. But, $$v_1,..,v_m$$ must be a linearly independent list also exists in $$V$$ according to our premise. A spanning list cannot be less than the length of a linearly independent list. So, $$V$$ must be infinite dimensional.

Therefore, a vector space $$V$$ is infinite dimensional if and only if there is a sequence of vectors $$v_1, v_2, ...$$ in $$V$$ such that $$v_1,..,v_m$$ is linearly independent for every positive integer $$m$$.

Thanks.

• In your proof of the reverse direction, you might want to assume the finite dimension of $V$ is $n.$ No list of more than $n$ vectors can be linearly independent in $V$, which contradicts your premise. – Chris Leary Jul 24 '19 at 16:37
• The above comment by Chris (+1) follows my own general critique (see answer below) that you need to instantiate the objects you use: Suppose $V$ is finite dimensional, then there is a positive integer $n$ and a list of vectors $\{v_1, ..., v_n\}$ such that $Span\{v_1, ..., v_n\}=V$. You see, now we have instantiated an explicit integer $n$ and explicit vectors $v_1,...,v_n$ that we can talk about. – Michael Jul 24 '19 at 16:42

Premise: Suppose 𝑉 is infinite-dimensional.
We want to show that a linearly independent
list of vectors 𝑣_1,..,v_𝑚 exists for every
positive integer 𝑚.


Good and clear premise.

Proof by contradiction.
Suppose no linearly independent list exists for some list of 𝑚 vectors.


No, the contradiction should not start by assuming you have a list of $$m$$ vectors. It should start by assuming you have a positive integer $$m$$ for which no linearly independent list of $$m$$ vectors exists.

 This implies that the spanning list is less than 𝑚


What do you mean by "the spanning list"? What spanning list? Spanning list of what?

To see this, consider that a list of
vectors will always be dependent if and
only if the size of the list is greater
than a linearly independent list that
also spans the vector space.


The "if and only if" is incorrect. You are trying to state the general fact that if (but not only if) a collection of vectors is larger than a collection of vectors that span the space, then the first collection cannot be linearly independent. However, while that is a general fact, you have not set up your proof in a way that clearly makes use of this fact. What list/lists are you talking about? You have not fixed any list, you only make vague claims about nonexistence of a list. You need to instantiate the objects you use with phrases such as "let xyz be a pdq with property abc" or "There exists an object abc with property xyz."

So, there is some list size m
that spans V.


How do you conclude this?

This contradicts
the original premise that V is infinite
dimensional. Therefore, a linearly
independent list must exist for every
positive integer 𝑚.
To see this, consider that a list of
vectors will always be dependent if
and only if the size of the list is
greater than a linearly independent
list that also spans the vector space.

• While the premise is clear, a higher-level critique wonders if the premise is sufficient for the forward direction: It seems the question asks for a single infinite sequence $\{v_n\}_{n=1}^{\infty}$ rather than a finite set of vectors that exists for each $m$ (which might use completely different vectors for each $m$, i.e., $m=2$ list is $\{v_1,v_2\}$; $m=3$ list is $\{w_1,w_2,w_3\}$, ...). – Michael Jul 24 '19 at 16:46
• It may be better to change your premise/approach: Try recursively building your list $\{v_n\}_{n=1}^{\infty}$: Let $v_1$ be any nonzero vector in $V$. Now form $v_2$. In general you start with $\{v_1, ..., v_k\}$ and then find another $v_{k+1}$. – Michael Jul 24 '19 at 16:53
• For the forward direction, I want to state the fact that an arbitrary collection of vectors will always be dependent if and only if that arbitrary collection of vectors has a size greater than another collection of vectors that is both linearly independent and spans the vector space $V$. Since that linearly independent set spans $V$, it contradicts the premise that $V$ is infinite-dimensional. Point taken about instantiation! – Richard K Yu Jul 24 '19 at 16:58
• The "if and only if" regarding the size of your list is incorrect: The vectors $\{(0,0,1), (0,0,2)\}$ are linearly dependent but $\{(0,0,1), (0,1,0), (1,0,0)\}$ are linearly independent. You want to use if rather than if and only if. – Michael Jul 24 '19 at 17:00
• So I mean to say that any list of vectors with length greater than 3 in $R^3$ will necessarily be linearly dependent (Any list in the set of all lists of vectors greater than 3 in $R^3$ is linearly dependent). In your example, while a list of vectors length 2 can be linearly dependent, it is not always linearly dependent (i.e. there is a linearly independent list of length 2 as well). Is the if and only if true in that sense? – Richard K Yu Jul 24 '19 at 17:47