Need help in understanding a verbatim of the Replacement theorem in Linear Algebra from the book, Linear Algebra, by Stephen H Friedberg,…

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I will start with enumerating some definitions, which I use in Linear Algebra:

  • The basis for a vector space $V$ is defined as a subset $S$ of $V$ such that $\text{span}(S)=V$ and $S$ is a linearly independent subset of $V.$

  • If $S$ is a finite basis for $V$ then if, $W$ is another basis for $V$ we have, $|V|=|W|.$

  • Finally, if a vector space $V$ have, a finite basis $S$ then, the $V$ is said to have a finite dimension with dimension of $V$ defined to be equal to the number of vectors in $S.$ The finute dimension of $V$ is generally denoted as $\dim (S).$

  • If $V$ is not of a finite -dimension then $V$ is called infinite dimensional.

Next, I state a theorem.

The Replacement Formula in Linear Algebra, says that,

Let $V$ be a vector space and let $G$ be a generating set of $V$ having exactly $n$ elements. If $L$ is a linearly independent subset of $V$ having exactly $m$ elements then, we have, $m\leq n$ and $\exists H\subseteq G$ having exactly $n-m$ elements such that, $L\cup H$ spans $V.$

Now, in the book "Linear Algebra " by Stephen H Friedberg, Insel and Spence,(4th Edition), Chapter-1, Pg- 47 (Section-1.6 titled Bases and Dimension) there was a piece of text written as below:

In the terminology of dimension, the first conclusion in the replacement
theorem states that if $V$ is a finite-dimensional vector space, then no linearly
independent subset of $V$ can contain more than $\dim(V)$ vectors.

But I am doubting this line. This is because of a couple of unclear issues, I am facing. Let's analyze this sentence, part by part:

  • They start the sentence by saying, "In the terminology of dimension, the first conclusion in the replacement
    theorem states that if $V$ is a finite-dimensional vector…" and in here, it's clear that the first conclusion of the replacement theorem refers to the fact:

Let $V$ be a vector space and let $G$ be a generating set of $V$ having exactly $n$ elements. If $L$ is a linearly independent subset of $V$ having exactly $m$ elements then, we have, $m\leq n.$

Moreover, $V$ is a finite-dimensional vector means, that the basis for $V$, which is also a generating set of $V$ say, $S$ is finite. With respect to the theorem we have, $G=S.$

  • Next, they goes on as, "… then no linearly independent subset of $V$ can contain more than $\dim(V)$ vectors." This means, if $L$ is a linearly independent subset of $V$ then, it is finite and has no more than $\dim(V)=|S|$ vectors.

But here's the problem. The replacement theorem says, specifically, if $L$ is a linearly independent set having exactly $m$ elements then, $m\leq n.$

In this case, $n=|S|.$ We can never say, $|L|\leq n.$ This is because, no information regarding the finiteness of $L$ is given. It may happen that, $L$ is an infinite set and then, the given statement is clearly incorrect. The statement is true, only if, $L$ is finite say, has $m(\in\Bbb N)$ number of elements as, this is when, we can correctly apply Replacement Theorem to conclude, $m\leq n.$

But then again, I doubt whether the below assertion is valid:

If $V$ is a vector space that have a finite dimension, then all linearly independent subset of $V$ is finite.

If the above claim is true, then only the statement given in the book, holds true. But otherwise, I feel that the statement is incorrect.

I need some help with this issue.

Best Answer

The statement given in the book is completely correct and justified.

The assertion

If $V$ is a vector space that have a finite dimension, then all linearly independent subset of $V$ is finite.

towards the end of the OP is a valid claim which can be proved.

We give the proof of this claim as follows:

The Replacement Theorem says, that a finite linearly independent subset of $V$ has at most $\dim V$ vectors. Now, an infinite linearly independent set, has more than $\dim V$ vectors.

This implies we can find a subset, say $H$ of the infinite linearly independent set (say, $S'$ ) having more than $\dim V$ vectors. We note that, $H$ will be linearly independent as well. By the Replacement Theorem, the number of vectors in $H$ is less than or equal to $\dim V.$ This is clearly a contradiction.

Thus, such an $H$ does not exist, which in turn implies that no infinite linearly independent subset of V i.e $S'$ existed in the first place.

So, all linearly independent subset of $V$ is finite.

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