# Same vector space for arbitrary independent vectors?

• I
Gold Member
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.

## Answers and Replies

micromass
Staff Emeritus
Homework Helper
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?

I see no reason to conclude ##V## and ##S## are the same.

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?

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

kelvin490
jbunniii
Homework Helper
Gold Member
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?
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.

kelvin490
Gold Member
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.

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.

jbunniii
Homework Helper
Gold Member
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.
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##.

kelvin490