1
$\begingroup$

Problem:

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 $v_4$.

Questions:

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?
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.