I have 5 vectors $v_1, v_2,...,v_5$ each of them having $5$ components:

$v_1 = \left[\begin{matrix} 5 \\ 4 \\ 3\\ 2 \\ 1 \\\end{matrix}\right] $ $v_2 = \left[\begin{matrix} -1 \\ 2 \\ 0 \\ -2 \\ 1 \\\end{matrix}\right] $ $v_3 = \left[\begin{matrix} 8 \\ 7 \\ 6 \\ 5 \\ 4 \\\end{matrix}\right] $ $v_4 = \left[\begin{matrix} 0 \\ 3 \\ 1\\ -1 \\ 2 \\\end{matrix}\right] $ $v_5 = \left[\begin{matrix} 10 \\ 8 \\ 6\\ 4 \\ 2 \\\end{matrix}\right] $

The question is determine a basis $B = \{b_1, b_2, ...\}$ for the vector space $V = span(v_1, v_2, ... , v_5)$.

What I know:

This is the way I understand the concept of basis: a set with the minimum number of vectors that combined can represent all other vectors in a vector space.

The concept of span is also familiar: all possible linear combinations of some vectors.

I have seen that to find the basis, I have basically to make the matrix created by putting next to each other each of the vectors in $RREF$.

We can observe from the problem that the $v_1$ is the double of $v_5$, and that all components of $v_2$ are smaller exactly one unit respect to the components of $v_4$.


1 .Is $\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\\end{matrix}\right]$ a basis for the vector space $V$, no matter the values of the vectors?

  1. If I use the RREF I can find a basis, what about if I want to find others?
  • $\begingroup$ I think you meant that each of them has $5$ components. This gives us that dim$(V) = 5$. And because $\{v_1,\cdots,v_5\}$ is a span of V, having a cardinality equal to the dimension of $V$, then it is a basis for $V$. $\endgroup$ – user207710 Mar 19 '15 at 20:11
  • $\begingroup$ @Ahmed But the vectors are not linearly independent, because there's at least one that is the multiple of another. I will edit my question to include also the vectors. $\endgroup$ – nbro Mar 19 '15 at 20:12

Q1. The columns of the identity matrix indeed form a basis of $\mathbb{R}^{5}$; the whole $5$-dimensional space. It is in fact the standard basis. However, we are interested in the subspace $V = \text{span}(\lbrace v_{1}, \dots, v_{5}\rbrace)$. And as you mentioned, a basis is a minimal set. $V$ is clearly a subspace of $\mathbb{R}^{5}$. If the span of these five vectors coincides with the entire $\mathbb{R}^{5}$, then indeed your proposed basis works. This happens if the $5$ vectors are linearly independent. If they are not, then we can describe their span using fewer vectors and the columns of the identity matrix are no longer a basis because they are not a minimal set.

Q2 There are infinitely many bases for a subspace of $\mathbb{R}^{n}$.

RREF allows us to identify a (maximal) subset of the vectors that are linearly independent and can hence be used to form a basis. A (nice) property of this approach is that it yields a basis that it consists of vectors among the original set.

Another well known procedure to extract a basis is the Gram-Schmidt process. Starting from a vector of our original set, we end up with an orthogonal (or orthonormal) basis.

Once we have identified a basis $B = \lbrace b_{1}, \dots, b_{k} \rbrace$ for $V$, we can use that to produce many different bases. Let $\mathbf{B} = \left[ b_{1}, \dots, b_{k}\right]$ be the $n \times k$ matrix formed by stacking the basis vectors (here, $n=5$). Let $\mathbf{C}$ be an arbitrary $k \times k$ matrix. Then, each column of the $n \times k$ matrix $$ \widehat{\mathbf{B}} = \mathbf{B}\mathbf{C} $$ is a linear combination of the columns of $\mathbf{B}$ and hence lies in $V$. If $\mathbf{C}$ is full rank (i.e., if its columns are linearly independent), then the columns of $\widehat{\mathbf{B}}$ will also be linearly independent (Note that the columns of $\mathbf{B}$ are by definition linearly independent). Hence, the columns of $\widehat{\mathbf{B}}$ also form a basis of $V$. Choosing different full-rank $k\times k$ matrices $\mathbf{C}$ will yields new bases for $V$.

A additional remark: Interestingly, if we know how many vectors among $v_{1}, \dots, v_{5}$ are linearly independent (say $k$, here $1\le k \le 5$), then we could also take an equal number of random linear combinations of all vectors (say with coefficients selected i.i.d. according to the normal distribution). The $k$ created vectors will also be linearly independent (and hence, form a basis for $V$) with probability $1$.

  • $\begingroup$ Thanks, great answer! So, if I understood correctly, if I find first a basis using the RREF, I can find other basis, if we know the number of linearly independent vectors. Could you please show a simple concrete example on how to do it? $\endgroup$ – nbro Mar 19 '15 at 21:00
  • $\begingroup$ The RREF will allow you to identify which a subset of your original set that contains only linearly independent vectors. These lin. independent vectors form a basis. According to your original post, you already know how to do that. As for the Gram-Schmidt procedure, even wikipedia has an example. And you can find many others online. $\endgroup$ – megas Mar 19 '15 at 21:05
  • $\begingroup$ Yes, but the thing I did not understand is the part of finding other basis from my found basis using RREF... $\endgroup$ – nbro Mar 19 '15 at 21:08
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    $\begingroup$ I did not describe such a way in the answer. RREF is just one way to get a basis. Then I suggested a couple of more different ways to get a basis. But in any case, once you find a basis using RREF (or any other method), then you can used that to create a new basis as follows: lets say that the basis contains $k$ vectors $b_{1}, \dots, b_{k}$. Then, you can take $k$ linear combinations of these $k$ vectors. But note these combinations must be themselves linearly independent. I will edit my answer to describe that. $\endgroup$ – megas Mar 19 '15 at 21:13

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