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

[karush]

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

 

[Evgeny.Makarov]

What is the definition of basis?

 

[HOI]

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).

 

[karush]

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...

 

[Evgeny.Makarov]

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.

 

[karush]

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 $x_1=3\quad x_2=2$

 

[Evgeny.Makarov]

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 $\begin{pmatrix}6\\8\end{pmatrix}$.

 

[karush]

ok I have another if I may.

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.

 

[Evgeny.Makarov]

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.