# Proof that two vectors with the same basis are equal

• May 5th 2009, 07:18 PM
igodspeed
Proof
Proof that two vectors with the same basis are equal
• May 5th 2009, 08:06 PM
Gamma
Vectors do not have a basis. A vector space has a basis. A vector is represented in a vector space as a linear combination of basis elements uniquely. So if you are asking if two vectors in a vector space share the same coefficients under the same basis, then yes, they are the same.

Alternatively, perhaps you were asking if two vector spaces have the same basis, are they the same? This too is true if the vector spaces are over the same field, because they would then span the same set.
• May 5th 2009, 08:23 PM
igodspeed
is there a proof that i could use to prove this? it cannot be numbers because i need to prove this for a general case.
• May 5th 2009, 08:25 PM
Gamma
What is the question? It does not make sense is what my first post is getting at.

You need to copy down word for word what the problem states if you want help.
• May 5th 2009, 08:42 PM
igodspeed
well words are exactly the one i posted in my first post. Alternatively i can prove that the $\displaystyle Col A$ is the same as the $\displaystyle Row({A^T})$ as a vector space. Proof mean that i need to be able to show this but i cannot use a numbered matrix, but i can use an alphabetical matrix or any other method.
• May 5th 2009, 08:56 PM
Gamma
"is there a way that i can prove that if two vectors have the same basis then it means that they are the same?"
Somehow I do not think so.

I do not see how the above question relates at all to the Col(A) or the Row$\displaystyle (A^T)$, but all the same.

Do you know what $\displaystyle A^T$ is? It means take the columns of A and make them the rows, in that same order.
Col(A) is the span of the columns of A, it is all possible linear combinations of the column vectors of A.
Row(B) is the set of all possible linear combinations of the rows of B.
Do you see why these are the same?
• May 5th 2009, 09:13 PM
igodspeed
it did not have the words "is there a way that i can" rest are his words. He is looking for some kind of a mathematical proof, and i am crying.
• May 5th 2009, 09:32 PM
Gamma
Let $\displaystyle A\in M_{m,n}$
Let $\displaystyle A=[a_1, a_2, ..., a_n]$ where $\displaystyle a_i \in \mathbb{R}^m$ These are are the column vectors that make up A.

Then $\displaystyle Col(A)=span\{a_1, a_2,... a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$

Now consider $\displaystyle A^T$. It has rows $\displaystyle {a_1, a_2, ..., a_n}$ in that order. I dunno how to do it in TeX, so this is as good as it is gonna get, write them in order from top to bottom on your paper and put them in square brackets.

Now consider $\displaystyle Row(A^T)$.
$\displaystyle Row(A^T)=span\{a_1,a_2,...,a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$

There, they are the same and I didn't use a "number" anywhere.
• May 5th 2009, 10:00 PM
igodspeed
Quote:

Originally Posted by Gamma
Let $\displaystyle A\in M_{m,n}$
Let $\displaystyle A=[a_1, a_2, ..., a_n]$ where $\displaystyle a_i \in \mathbb{R}^m$ These are are the column vectors that make up A.

Then $\displaystyle Col(A)=span\{a_1, a_2,... a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$

Now consider $\displaystyle A^T$. It has rows $\displaystyle {a_1, a_2, ..., a_n}$ in that order. I dunno how to do it in TeX, so this is as good as it is gonna get, write them in order from top to bottom on your paper and put them in square brackets.

Now consider $\displaystyle Row(A^T)$.
$\displaystyle Row(A^T)=span\{a_1,a_2,...,a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$

There, they are the same and I didn't use a "number" anywhere.

$\displaystyle A=[a_1, a_2, ..., a_n]$ where $\displaystyle a_i \in \mathbb{R}^m$ These are are the column vectors that make up A.
Then $\displaystyle Col(A)=span\{a_1, a_2,... a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$
$\displaystyle {A^T} = \left[ {\begin{array}{*{20}{c}} {{a_1}} \\ {{a_2}} \\ {{a_3}} \\ {{a_n}} \\ \end{array}} \right]$

$\displaystyle Row(A^T)=span\{a_1,a_2,...,a_n\}=\{r_1a_1 + r_2a_2 + ... + r_na_n | r_i \in \mathbb{R} \}$

did i get it right?

Thanks
• May 5th 2009, 10:07 PM
Gamma
Yeah you copied down what I wrote correctly for sure, but do you actually understand any of it? That seems to be the bigger problem.

I think you just need to review the basic definitions to get familiar with them. Understand what a vector space is. Understand what a basis for a vector space is. Understand what a Linear Transformation is. Understand what dimension is. Understand what rank of a matrix or of a map is. Understand what the span of a set is. Understand what linear independece and dependence is. Understand what a matrix represents.

I stress understand because it is one thing to be able to quote these definitions, but another thing to actually understand and internalize these things. Without a concrete understanding of the jargon of linear algebra, it is very difficult to prove even the simplest question.
• May 6th 2009, 03:33 AM
HallsofIvy
Two vector spaces "are the same", of course, if they contain exactly the same vectors (with, naturally, the same definitions of addition and scalar multiplication).

a vector space, V, [b]is[/v] the set of all linear combinations of its basis vectors. If the vector space U has the same basis, then the linear combinations of those basis vectors is the same so U= V.

What's wrong with that proof?
• May 6th 2009, 02:03 PM
igodspeed
So $\displaystyle row(A^T)$= $\displaystyle Rank(A)$=$\displaystyle Col(A)$

Thanks
• May 6th 2009, 02:09 PM
Gamma
not true.
Row(A)=$\displaystyle Col(A^T)$ as above, these are vector spaces.

but the rank(A) is just an integer.

It is the dimension of the column span of A. ( it turns out the dimension of the column span is the same as the dimension of the row span). The other things are actual vector spaces they are different things.
• May 6th 2009, 02:22 PM
Gamma
Perhaps you meant $\displaystyle dim(Row(A^T))=rank(A)=dim(Col(A))$

That would be true. And the proof is done by definition of rank of A and what I proved above that $\displaystyle Row(A^T)=Col(A)$