Verifying $\beta$ as a Basis of $\Bbb{R}^2$ & Finding $|v|_{\beta-}$

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In summary, "basis" is a set of vectors in a vector space with three properties: the vectors span the space, they are independent, and there are a certain number of vectors in the set. Without the exact definition being used, it is not possible to determine if a given set of vectors is a basis. However, it is possible to find a vector that is not expressible as a linear combination of the vectors in the given set.
  • #1
karush
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verify that
$\beta =\left\{
\left[\begin{array}{c}0 \\ 2 \end{array}\right],
\left[\begin{array}{c}3 \\ 1\end{array}\right]\right\}$
is a basis for $\Bbb{R}^2$
for $v=\left[\begin{array}{c}6\\ 8
\end{array}\right]$
find $|v|_{\beta-}$

ok $x_2=2$ and $x_1=3$
not sure how to answer the rest
 
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  • #2
What is the definition of basis?
 
  • #3
A basis for a vector space of dimension n has three properties
a) The vectors span the space.
b) The vectors are independent.
c) There are n vectors in the set.
And any two of those imply the third!

Since there are 3 two statement subsets from those 3 statements, there are 3 diffferent ways to prove this is a basis:
1) There are 2 vectors, (c), so it is sufficient to prove (a), that the two vectors span $R^2$. That, in turn, means that we must show that for any vector (x, y) in $R^2$ there exist numbers a and b such that a(0, 2)+ b(3, 1)= (x, y). That is equivalent to the two equations 0a+ 3b= 3b= x and 2a+ b= y. From the first equation b= x/3 and then the second becomes 2a+ x/3= y so a= y/2- x/6. The fact that those solutions proves this is a basis.

2) There are 2 vectors, (c), so it is sufficient to prove (b), that the two vectors are independent. These two vectors are independent if a(0, 2)+ b(3, 1)= (0, 0) implies a= b= 0. That is easy to prove. We have the two equations 3b= 0 and 2a+ b= 0. The first gives b= 0 immediately and then the second equation is 2a+ 0= 0 so a= 0. That again proves that this is a basis.

3) (a) and (b), that the vector span the space and are independent is actually the definition of "basis". The two proofs in (2) and (3) prove that this is a basis.

The last exercise is to write the vector (6, 8) as a linear combination of (0, 2) and (3, 1). That is the same as in (1) but instead of using (x, y), use (6, 8): we want to find a and b such that a(0, 2)+ b(3, 1)= (6, 8). That gives the two equations 3b= 6 and 2a+ b= 8. From 3b= 6, b= 2. So the second equation becoms 2a+ 2= 8. 2a= 6, a= 3. (6, 8)= 3(0, 2)+ 2(3, 1).
 
  • #4
Evgeny.Makarov said:
What is the definition of basis?

not sure but the images they show in the book it looks like a plane of which all the vectors that in basis lay on.

kinda maybe...
 
  • #5
karush said:
the images they show in the book it looks like a plane of which all the vectors that in basis lay on.
Did not understand this phrase.

Trying to solve a problem about a concept without knowing the precise definition of that concept is the silliest thing you can do. Fortunately, Country Boy provided three equivalent definitions (there are more). You should select the one used in your textbook.
 
  • #6
Country Boy said:
The last exercise is to write the vector (6, 8) as a linear combination of (0, 2) and (3, 1). That is the same as in (1) but instead of using (x, y), use (6, 8): we want to find a and b such that a(0, 2)+ b(3, 1)= (6, 8). That gives the two equations 3b= 6 and 2a+ b= 8. From 3b= 6, b= 2. So the second equation becoms 2a+ 2= 8. 2a= 6, a= 3. (6, 8)= 3(0, 2)+ 2(3, 1).

by combination do you mean this?

$\beta =
\left[\begin{array}{cc|c}0 & 3&6 \\ 2 & 1&8\end{array}\right]
\xrightarrow{R_1/3}
\left[\begin{array}{cc|c}0 & 1&2 \\ 2 & 1&8\end{array}\right]
\xrightarrow{R_3-R_1}
\left[\begin{array}{cc|c}0 & 1&2 \\ 2 & 0&6\end{array}\right]
\xrightarrow{R_3/2}
\left[\begin{array}{cc|c}0 & 1&2 \\ 1 & 0&3\end{array}\right]$
so $x_1=3\quad x_2=2$
 
  • #7
karush said:
by combination do you mean this?
A linear combination of vectors $v_1,\ldots,v_n$ is $x_1v_1+\dots+x_nv_n$ where $x_1,\ldots,x_n$ are numbers (scalars, elements of the field). You correctly found $x_1$ and $x_2$ so that the linear combination $\displaystyle x_1\begin{pmatrix}0\\2\end{pmatrix}+x_2\begin{pmatrix}3\\1\end{pmatrix}$ equals $\begin{pmatrix}6\\8\end{pmatrix}$.
 
  • #8
ok I have another if I may.:cool:

Determine if
$$\beta =
\left\{\left[
\begin{array}{r}-1\\2\\0\end{array}\right],
\left[
\begin{array}{r}3\\-1\\1\end{array}
\right]\right\}$$
is a basis for $\Bbb{R}^3$ok by observation this does not have a third vector and there is no $v=$So not sure if really anything can be done with it.
 
  • #9
I still think that it does not make sense to move forward without the definition you are supposed to use. Proofs that something is or is not a basis depend on the definition.

But you can find a vector that is not represented as a linear combination of the vectors is $\beta$. For example, if we look at the first two components, then $(2, 1)$ is represented as a linear combination in a unique way, namely, as a sum of the two vectors. This means that the only vector of the form $(2, 1, z)$ is expressible through $\beta$ is $(2, 1, 1)$. So, for example, $(2, 1, 2)$ is not expressible.
 

FAQ: Verifying $\beta$ as a Basis of $\Bbb{R}^2$ & Finding $|v|_{\beta-}$

1. What is the definition of a basis in linear algebra?

A basis is a set of linearly independent vectors that span a vector space. This means that any vector in the vector space can be written as a unique linear combination of the basis vectors.

2. How do you verify if a set of vectors is a basis of a vector space?

To verify if a set of vectors is a basis of a vector space, we need to check two conditions:

  • The vectors are linearly independent, meaning that no vector in the set can be written as a linear combination of the other vectors.
  • The vectors span the entire vector space, meaning that every vector in the vector space can be written as a linear combination of the basis vectors.

3. What is the process for verifying $\beta$ as a basis of $\Bbb{R}^2$?

To verify $\beta$ as a basis of $\Bbb{R}^2$, we need to check if the two vectors in $\beta$ are linearly independent and if they span $\Bbb{R}^2$. This can be done by setting up a system of equations and solving for the coefficients of the linear combination. If the system has a unique solution, then the vectors are linearly independent and span $\Bbb{R}^2$, thus making $\beta$ a basis of $\Bbb{R}^2$.

4. How do you find the norm of a vector $v$ with respect to a given basis $\beta$?

The norm of a vector $v$ with respect to a given basis $\beta$ is calculated by taking the square root of the sum of the squares of the coefficients of the linear combination of $v$ with respect to $\beta$. This can be represented as $|v|_{\beta-} = \sqrt{a^2 + b^2}$, where $a$ and $b$ are the coefficients of the linear combination of $v$ with respect to $\beta$.

5. Can there be multiple bases for a vector space?

Yes, there can be multiple bases for a vector space. In fact, any set of linearly independent vectors that span the vector space can be considered a basis. However, the number of vectors in a basis for a vector space is always the same, known as the dimension of the vector space.

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