# Determine which system of vectors span C^3

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1. Mar 11, 2016

### maNoFchangE

1. The problem statement, all variables and given/known data

2. Relevant equations
Reduced echelon form of the column matrix

3. 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 $\mathbb{C}^3$?

2. Mar 11, 2016

### PeroK

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

3. Mar 11, 2016

### 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?

4. Mar 11, 2016

### PeroK

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.)

5. Mar 11, 2016

### maNoFchangE

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?

6. Mar 11, 2016

### PeroK

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

7. Mar 11, 2016

### 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 $\mathbb{C}^3$?

8. Mar 11, 2016

### PeroK

It does indeed!

9. Mar 11, 2016

### maNoFchangE

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

10. Mar 11, 2016

### PeroK

Can you prove that? It's not hard.

11. Mar 11, 2016

### maNoFchangE

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

12. Mar 11, 2016

### PeroK

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

13. Mar 11, 2016

### maNoFchangE

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

14. Mar 11, 2016

### PeroK

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.

15. Mar 12, 2016

### maNoFchangE

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.

16. Mar 12, 2016

### PeroK

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.

17. Mar 12, 2016

### maNoFchangE

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 $\mathbb{C}^3$?

18. Mar 12, 2016

### PeroK

Yes.

19. Mar 12, 2016

### maNoFchangE

Thank you!

20. Mar 12, 2016

### HallsofIvy

Staff Emeritus
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.