# I Same vector space for arbitrary independent vectors?

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1. May 28, 2016

### kelvin490

If we use n linearly independent vectors x1,x2...xn to form a vector space V and use another set of n linearly independent vectors y1,y2...yn to form a vector space S, is it necessary that V and S are the same? Why?

If we have a vector space Q that the dimension is n, can we say that any set of n linearly independent vectors k1,k2...kn can form a basis of Q? Why?

Suppose only real numbers are involved.

2. May 28, 2016

### micromass

Staff Emeritus
I see no reason to conclude $V$ and $S$ are the same.

Yes. This theorem should be proven in any standard linear algebra book.

3. May 28, 2016

### jbunniii

Consider the simplest case, where $n=1$. Then $x_1$ and $y_1$ each span a one-dimensional subspace. If $x_1$ and $y_1$ are linearly independent, then these subspaces are distinct; indeed, they intersect trivially.

4. May 28, 2016

### kelvin490

Thanks. It's seems there is a theorem that a vector space Q that the dimension is n can have any set of n linearly independent vectors k1,k2...kn as its basis. How can we ensure that k1,k2...kn doesn't form a vector space other than Q but not Q?

I have this question because two different set of n linearly independent vectors can form two different vector space with dimension n. It seems hard to ensure k1,k2...kn must form Q.

5. May 28, 2016

### jbunniii

Let $R$ denote the vector space spanned by $k_1,k_2,\ldots,k_n$. Since each $k_j$ is in $Q$ and $Q$ is a vector space, every linear combination of the $k_j$'s is in $Q$, hence $R \subseteq Q$.

Now suppose that $q$ is an arbitrary element of $Q$. Since $k_1,k_2,\ldots,k_n$ is a basis for $Q$, we can write $q = a_1 k_1 + a_2 k_2 + \cdots + a_n k_n$ for some scalars $a_1,a_2,\ldots a_n$. Therefore $q$ is contained in the subspace spanned by $k_1,k_2,\ldots,k_n$, which is $R$. This shows that $Q \subseteq R$.

Since we have shown both containments $R \subseteq Q$ and $Q \subseteq R$, we conclude that $Q = R$.