0

Say you have a matrix named $A$, and it contains vectors $v_n$. Now if the matrix is short and wide, meaning that $n>m$, does that mean that the $m-n$ vectors are redundant in making a basis representing $R^m$?

All I'm trying to ask is, if this was confusing, do the number of vectors and elements correlate to the $R$ being spanned? If you want a basis for $R^2$, you need $2$ linearly independent vectors with $2$ elements each. If you added one more vector to this set it would be redundant for a basis.

Likewise, for $R^3$ a linearly independent $3\times3$ for the basis and for $R^n$ a linearly independent $n\times n$?

I'm still learning this and don't know what to call this particular phenomenon and I wanted to know if this was true?

Bernard
  • 175,478
Asker123
  • 432

1 Answers1

2

For $\mathbb{R}^n$, you need $n$ vectors of size $n$ to span the space. Thus if you have a set $V$ of $n$ vectors that span the space $\mathbb{R}^n$, then any other vector $u$ can be written as a linear combination of vectors from $V$.

Note that the above works only if $V$ spans $R^n$. Just because you have any $n$ vectors in $V$, then there's no guarantee they would span the space (for example take all $v\in V$ to be $v=0$).

So for your matrix: there is a guarantee that at least $n-m$ of your vectors are redundant. However there's no guarantee that your matrix $M$ will span all of $R^n$.

Alex R.
  • 32,771