Find if 3 vectors form a basis in space R^4

[Deimantas]

Homework Statement

Is the system of vectors a1=(1,-1,0,1), a2=(2,3,-1,0), a3=(4,1,-1,4) linearly independent? Do these vectors form a basis in the vector space R^4? State why.

Homework Equations

The Attempt at a Solution

I have done the first part of the exercise. I have found that the system of vectors is linearly independent and the rank is 3. I think that if there are 3 linearly independent vectors, and the rank is 3, then the vectors will form a basis in vector space R^3. What about R^4? How to prove whether it does form the basis?

 

[Arian.D]

Try to see if they could span R⁴. The answer is no, they won't span it. You need 4 vectors for a basis in R⁴ because the dimension is 4, but that's a theorem, try to figure out why there must be a vector that is not a linear combination of these 3 vectors. (Looking at the system of linear equations will give you some insight into it).

 

[Mark44]

Homework Statement

Is the system of vectors a1=(1,-1,0,1), a2=(2,3,-1,0), a3=(4,1,-1,4) linearly independent? Do these vectors form a basis in the vector space R^4? State why.

Homework Equations

The Attempt at a Solution

I have done the first part of the exercise. I have found that the system of vectors is linearly independent and the rank is 3. I think that if there are 3 linearly independent vectors, and the rank is 3, then the vectors will form a basis in vector space R^3.

No, they can't possibly form a basis for R³. They are vectors in R⁴. They form a basis for a three-dimension subspace of R⁴, but that is very different from R³.

What about R^4? How to prove whether it does form the basis?

The vectors can't form a basis for R⁴ since there aren't enough of them.

 

[DryRun]

I never quite understood the R^n or R^m concept. I tried but gave up. What do you mean when you say "since there aren't enough of them". Is R^n representative of the total number of rows or columns or unique number of independent vectors?

 

[Mark44]

Vectors in R or R¹ have one component (a single real number). Any basis for this vector space contains one vector.
Vectors in R² have two components (e.g., <1, 3>). Any basis for this vector space contains two vectors.
Vectors in R³ have three components (e.g., <1, 3, -2>). Any basis for this vector space contains three vectors.
And so on...

In the problem in this thread we're given three vectors in R⁴. Being vectors in R⁴, they have four components, so they don't belong to, for example, R³ or R² or R⁵. A basis for R⁴ must be a linearly independent set of four vectors.

 

[DryRun]

OK, as an example, if a 4x5 matrix has only 2 linearly independent vectors, then these 4 row vectors form a basis in R^2, corresponding to the number of pivot rows or pivot columns only. Correct?

 

[Mark44]

OK, as an example, if a 4x5 matrix has only 2 linearly independent vectors, then these 4 row vectors form a basis in R^2, corresponding to the number of pivot rows or pivot columns only. Correct?

No.

A 4 x 5 matrix is a map from R⁵ to R⁴. The rows are vectors in R⁵, and have absolutely nothing to do with R².

 

[DryRun]

No.

A 4 x 5 matrix is a map from R⁵ to R⁴. The rows are vectors in R⁵, and have absolutely nothing to do with R².

I'm sorry, but i don't understand. What about the effect of the number of independent linear vectors on R^n or R^m?

Is there a good book or paper that i could refer to? I just can't wrap my head around those R^n or R^m numbers. But i do know the fundamental theorem of linear algebra part 1, which deals with dimensions.

EDIT: OK, i think i understand. The R^{value} is always equal to the number of components in the row/s corresponding to the linearly independent vector/s. So, the rule is always count the number of components in the rows and not in the independent columns. Correct?

 

[Mark44]

I'm sorry, but i don't understand. What about the effect of the number of independent linear vectors on R^n or R^m?

Now I don't understand what you're asking.

Let's go back to your example of a 4 x 5 matrix.

Suppose you ended up with these vectors:
<1, 0, 1, 1, 1>
<0, 1, 1, 1, 1>
You asked whether these could be a basis for R². They couldn't possibly be, because these two vectors belong to R⁵ (they have 5 components). A basis for R² would have to consist of vectors in R², such as <1, 1> and <0, 1>.

The two vectors in R⁵ above are a basis for a two-dimensional subspace of R⁵. This subspace looks like a plane, but it's a plane embedded in 5-dimensional space. The fact that this subspace of R⁵ is a plane has nothing to do with the R² vector space.

Is there a good book or paper that i could refer to? I just can't wrap my head around those R^n or R^m numbers. But i do know the fundamental theorem of linear algebra part 1, which deals with dimensions.

It's just a matter of realizing that n and m in the symbols Rⁿ and R^m refer to vector spaces of dimension n and m respectively. n and m are just placeholders for some integers.

You aren't going to be able to come up with a mental image of 5-dimensional space (or 4- or 6- or n-) but that's OK - you don't need to. You can visualize a 1-dimensional subspace as a line in some blob that represents the higher-dimension space. A 2-dimensional subspace is just a plane in the higher-dimension space.

What is it about Rⁿ that you're having trouble with?

 

[Mark44]

You made your edit after I had already started a response.

EDIT: OK, i think i understand. The R^{value} is always equal to the number of components in the row/s corresponding to the linearly independent vector/s.

That's more complicated than it needs to be. The "value" is equal to the number of components in a vector in R^{value}.

So, the rule is always count the number of components in the rows and not in the independent columns. Correct?

In your example of a 4 x 5 matrix that row-reduces to a matrix with two non-zero rows, Rank(A) = 2, Kernel(A) = 3, and n = 5, where n is the dimension of the domain, R⁵.

 

[DryRun]

In your example of a 4 x 5 matrix that row-reduces to a matrix with two non-zero rows, Rank(A) = 2, Kernel(A) = 3, and n = 5, where n is the dimension of the domain, R⁵.

That's the best way that i can understand and remember it. It's just the rank-nullity theorem.