# Math Help - Vector

1. Originally Posted by Gamma
Are you talking about the other question on the other Thread? I am really confused as to what you are talking about? Again I cannot stress enough, there is no "THE basis" there is no preferred basis, it is just a basis.

If you need to find a basis for $\mathbb{R}^n$,just pick the standard basis for $\mathbb{R}^n$.
{(1,0,...,0), (0,1,0,...,0),... (0,0,...,1)}
I was talking about this post. What i wanted to understand is how can i derive a standard basis for this, how can i find the rank of every row how can i know which rows have pivot?
Thanks
KC

2. ## Definition

Man, I am really lost with what you are asking now. I think you need to review some of your definitions, it is getting to the point that these posts are becoming incoherent to me.

You do not find a basis for a matrix.

A matrix represents a linear transformation, a basis is for a vector space, not a function. Matrices have ranks, it does not make sense to talk about the rank of a row.

The dual space to $\mathbb{R}^n$ does not have a "standard" basis. It is not canonically isomorphic to $\mathbb{R}^n$. It is isomorphic, but only in terms of a basis. If you give me a basis for $\mathbb{R}^n$, I will give you a basis for $(\mathbb{R}^n)^*$.

I already wrote out the basis for the dual space in a previous post on here titled "bassis(I can't type apparently) for the Dual Space".

Look at that. If you want the basis for the dual space for $\mathbb{R}^n$ subordinant to the standard basis, then $e_1=(1,0,0,...,0)$
$e_2=(0,1,0,...,0)$
.
.
.
$e_n=(0,0,0,...,1)$

and this yields the basis for $(\mathbb{R}^n)^*$
$\epsilon_1(a_1e_1+a_2e_2+...+a_ne_n)=a_1$
$\epsilon_2(a_1e_1+a_2e_2+...+a_ne_n)=a_2$
.
.
.
$\epsilon_n(a_1e_1+a_2e_2+...+a_ne_n)=a_n$

3. Originally Posted by Gamma
This is the point, it depends on the basis, there are an infinite number of possible choices for basis of $\mathbb{R}^n$ each of these yields a different basis for $(\mathbb{R}^n)^*$.

To give you the idea though for any given basis for $\mathbb{R}^n$ call it $\{e_1,e_2,...,e_n\}$, you get n linear functionals in $(\mathbb{R}^n)^*$ that will form the basis, say they are $\{\epsilon^1,\epsilon^2,...,\epsilon^n \}.$

Let the n functionals be defined as follows.

$\epsilon^i(a_1e_1 +a_2e_2 + ... + a_ie_i + ... +a_ne_n)=a_i$

Yeah, there is your basis. If you took the standard basis for $\mathbb{R}^n$ the ith coordinate function would simply pick out the ith coefficient.

could you please explain to me in a little more detail on how i can find the basis for this.
thanks

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