Determine which system of vectors span C^3

[maNoFchangE]

Homework Statement

Please see the attached picture

Homework Equations

Reduced echelon form of the column matrix

The Attempt at a Solution

I can solve for the first part to find which ones are the bases in $\mathbb{R}^3$ by determining whether in the echelon form, there is a pivot in each column and row. But what about $\mathbb{C}^3$? Since a general vector in this space takes the form $(a+ib,c+id,e+if)^T$, if there is no imaginary number in the vector, which is the case in all those three sets of vectors in the problem, then they cannot span $\mathbb{C}^3$?

 

[PeroK]

Homework Statement

Please see the attached picture
View attachment 97201

Homework Equations

Reduced echelon form of the column matrix

The Attempt at a Solution

I can solve for the first part to find which ones are the bases in $\mathbb{R}^3$ by determining whether in the echelon form, there is a pivot in each column and row. But what about $\mathbb{C}^3$? Since a general vector in this space takes the form $(a+ib,c+id,e+if)^T$, if there is no imaginary number in the vector, which is the case in all those three sets of vectors in the problem, then they cannot span $\mathbb{C}^3$?

What would be an example basis for $\mathbb{C}^3$?

 

[maNoFchangE]

$(1,0,0)^T$, $(i,0,0)^T$, $(0,1,0)^T$, $(0,i,0)^T$, $(0,0,1)^T$, and $(0,0,i)^T$, no?

 

[PeroK]

$(1,0,0)^T$, $(i,0,0)^T$, $(0,1,0)^T$, $(0,i,0)^T$, $(0,0,1)^T$, and $(0,0,i)^T$, no?

Are those linearly independent?

(There is a subtlety here, but I assume you are considering $\mathbb{C}^3$ as a complex vector space. It can, however, also be taken as a real vector space. If that makes sense.)

 

[maNoFchangE]

What would be an example basis for $\mathbb{C}^3$?

Sorry I am confused here, do you mean there are more than one meaning of $\mathbb{C}^3$. To me, it should mean a complex vector space, what else can it be?

 

[PeroK]

Sorry I am confused here, do you mean there are more than one meaning of $\mathbb{C}^3$. To me, it should mean a complex vector space, what else can it be?

Take it to be a complex vector space. Now, are those vectors you gave linearly independent?

 

[maNoFchangE]

On a second thought, they are not, because for instance $(i,0,0)^T=i(1,0,0)^T$.
So, this means the standard basis for $\mathbb{R}^3$: $(1,0,0)^T$, $(0,1,0)^T$, and $(0,0,1)^T$ can also be a basis in $\mathbb{C}^3$?

 

[PeroK]

On a second thought, they are not, because for instance $(i,0,0)^T=i(1,0,0)^T$.
So, this means the standard basis for $\mathbb{R}^3$: $(1,0,0)^T$, $(0,1,0)^T$, and $(0,0,1)^T$ can also be a basis in $\mathbb{C}^3$?

It does indeed!

 

[maNoFchangE]

This means in the original problem, any bases for $\mathbb{R}^3$ are also bases in $\mathbb{C}^3$?

 

[PeroK]

This means in the original problem, any bases for $\mathbb{R}^3$ are also bases in $\mathbb{C}^3$?

Can you prove that? It's not hard.

 

[maNoFchangE]

Can you prove that? It's not hard.

Not every basis of $\mathbb{R}^3$, because $\mathbb{R}^3$ is a subspace of $\mathbb{C}^3$?

 

[PeroK]

Not every basis of $\mathbb{R}^3$, because $\mathbb{R}^3$ is a subspace of $\mathbb{C}^3$?

Can you see why $\mathbb{R}^3$ is not a subspace of $\mathbb{C}^3$?

 

[maNoFchangE]

Can you see why $\mathbb{R}^3$ is not a subspace of $\mathbb{C}^3$?

Why not? $\mathbb{R}^3$ is just $\mathbb{C}^3$ with zero imaginary parts.

 

[PeroK]

Why not? $\mathbb{R}^3$ is just $\mathbb{C}^3$ with zero imaginary parts.

Be careful. $\mathbb{R}^3$ is a real vector space, that is a vector space over the field of real numbers. But, the subset of $\mathbb{C}^3$ represented by vectors of the form $(x_1, x_2, x_3)$ is not a subspace. It's not closed under (complex) scalar multiplication.

 

[maNoFchangE]

Be careful. $\mathbb{R}^3$ is a real vector space, that is a vector space over the field of real numbers. But, the subset of $\mathbb{C}^3$ represented by vectors of the form $(x_1, x_2, x_3)$ is not a subspace. It's not closed under (complex) scalar multiplication.

Ok I get it, $\mathbb{R}^3$ is not a subspace of $\mathbb{C}^3$, but a subset.
Then to answer my question in post #9, any bases of $\mathbb{R}^3$ in matrix form can also be bases in $\mathbb{C}^3$, but the converse is not true.

 

[PeroK]

Ok I get it, $\mathbb{R}^3$ is not a subspace of $\mathbb{C}^3$, but a subset.
Then to answer my question in post #9, any bases of $\mathbb{R}^3$ in matrix form can also be bases in $\mathbb{C}^3$, but the converse is not true.

Yes, although I think this is where you need to be more precise with your terminology. Viewing $\mathbb{R}^3$ as interchangeable with the subset of $\mathbb{C}^3$ consisting of real-valued vectors can lead to problems. Instead, to be precise:

1) Any basis in $\mathbb{R}^3$ when mapped to the equivalent vectors in $\mathbb{C}^3$ forms a basis for $\mathbb{C}^3$.

2) There is not necessarily a direct mapping from a basis in $\mathbb{C}^3$ to vectors in $\mathbb{R}^3$.

Remember also that $\mathbb{R}^3$ is a real vector space (over the field of real scalars) and $\mathbb{C}^3$ is a complex vector space (over the field of complex scalars). This means that you have to be careful not to link the two together too closely. In general, you cannot look at a real vector space as a subspace of a complex vector space, because the field of scalars is different.

 

[maNoFchangE]

Remember also that $\mathbb{R}^3$ is a real vector space (over the field of real scalars) and $\mathbb{C}^3$ is a complex vector space (over the field of complex scalars). This means that you have to be careful not to link the two together too closely. In general, you cannot look at a real vector space as a subspace of a complex vector space, because the field of scalars is different.

I guess that bolded part is the key point.
Alright then, by computing the echelon form and analyzing the existence of pivots in each column and each row in the original problem in post #1, I found that the system of vectors a) and c) form a basis in $\mathbb{R}^3$. Then, basing my argument on

1) Any basis in R3\mathbb{R}^3 when mapped to the equivalent vectors in C3\mathbb{C}^3 forms a basis for C3\mathbb{C}^3.

these systems of vectors are also bases in $\mathbb{C}^3$?

 

[PeroK]

I guess that bolded part is the key point.
Alright then, by computing the echelon form and analyzing the existence of pivots in each column and each row in the original problem in post #1, I found that the system of vectors a) and c) form a basis in $\mathbb{R}^3$. Then, basing my argument on

these systems of vectors are also bases in $\mathbb{C}^3$?

Yes.

 

[maNoFchangE]

Thank you!

 

[HallsofIvy]

Since every imaginary number can be represented as a pair of real numbers, every $C^n$ is equivalent to $R^{2n}$. Of course, then, a basis for $C^n$, having only n vectors, cannot be a basis for $R^{2n}$ which requires 2n vectors.