I've modified parts of this quote from "Introduction to Linear Algebra by Strang", for brevity in the questions below. If the quote itself seems illogical or incorrect in any form, please inform me and I will make the necessary edits.

  • The linear combination of one vector in $\mathbb{R^n}$ will always fill out a line ($\mathbb{R^1}$)
  • The linear combination of two vector in $\mathbb{R^n}$ will always fill out a plane ($\mathbb{R^2}$)
  • The linear combination of vectors vector in $\mathbb{R^n}$ will always fill out a three-dimensional space ($\mathbb{R^3}$)

When Strang talks about filling a line, plane, etc. what he is really referring to is actually constructing vector spaces from linear combinations correct?

If so, then am I correct in saying that the linear combination of $m$ vectors where each vector $v \in \mathbb{R^n}$, will create a $m$-dimensional vector space $\mathbb{R^m}$. where $\mathbb{R^m}$ is a subspace of $\mathbb{R^n}$, i.e. $\mathbb{R^m} \subset \mathbb{R^n}$.

Again if I'm correct is the below a correct mathematical translation of what I've just stated above?

$$\text{Given} \ \left(\ \mathbb{R^m} =\ \sum_{i=1,\ \ j=1}^{m} \lambda_j\cdot v_i \ \ \text{where} \ \lambda \in \mathbb{R}, v \in \mathbb{R^n}\right) \implies \mathbb{R^m} \subset \mathbb{R^n}$$

Lastly does this notion of construction vector spaces from linear combinations generalize to arbitrary fields $\mathbb{F}$ and arbitrary vectors spaces $\mathbb{F^n}$? Does the generalization below hold true?

$$\text{Given} \ \left(\ \mathbb{F^m} =\ \sum_{i=1,\ \ j=1}^{m} \lambda_j\cdot v_i \ \ \text{where} \ \lambda \in \mathbb{F}, v \in \mathbb{F^n}\right) \implies \mathbb{F^m} \subset \mathbb{F^n}$$

  • $\begingroup$ If you have $m$ linearly independent vectors in $K^n$ (for any field $K$) then their span is a $m$-dimensional subvector space. Moreover any $m$-dimensional vector space is isomorphic to $K^m$. $\endgroup$ – Hamed Jun 6 '16 at 20:07

This is almost right, but there is just two point that you should check:

The first one is that if you have $m$ vectors it will generate an $m$-dimensional space if and only if the vectors are lineary independent.

The second one is that actually is not $\mathbb R^m\subset\mathbb R^n$, but what you get is $\sum^m_{i=1}\lambda_i v_i=W\subset\mathbb R^n$, and $W\cong\mathbb R^m$, note that is not the same as $W=\mathbb R^m$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.