Can the Span of Three Vectors Be Contained in R^2?

[TheWrathOfMath]

For the matrix:

(1 0 1)
(0 1 1)

is is true to claim that:

Row(A) = sp{(1 0 1), (0 1 1)} ⊆ R^n=R^3

Col(A) = sp{(1 0), (0 1) (1 1)} ⊆ R^m=R^2

and that the columns of A span R^m=R^2,

but it's not true to claim that the rows of A span R^n=R^3?

 

[Zermelo]

For the matrix:

(1 0 1)
(0 1 1)

is is true to claim that:

Row(A) = sp{(1 0 1), (0 1 1)} ⊆ R^n=R^3

Col(A) = sp{(1 0), (0 1) (1 1)} ⊆ R^m=R^2

and that the columns of A span R^m=R^2,

but it's not true to claim that the rows of A span R^n=R^3?

Take an arbitrary vector (x, y, z) from R3. Can you express it as a linear combination of your rows?
What is a basis of a vextor space? What is a dimension of a vector space?

 

[TheWrathOfMath]

Take an arbitrary vector (x, y, z) from R3. Can you express it as a linear combination of your rows?
What is a basis of a vextor space? What is a dimension of a vector space?

Two vectors cannot span R^3, can't they?

 

[Zermelo]

Two vectors cannot span R^3, can't they?

You’re right, because the dimension of R3 is 3 => R3 basis has 3 vectors. If 2 linearly independant vectors spanned R3, then the dimension of R3 would be 2!
But feel free to check it manually, (x,y,z) = a (1,0,1) + b (0,1,1), where (x,y,z) is any vector in R3. Try to find values of a and b! You’ll end up prooving such a and b don’t exist. You also need to confirm Col(A) = R2, in the same way as above, or by taking a closer look at the vectors

 

[TheWrathOfMath]

You’re right, because the dimension of R3 is 3 => R3 basis has 3 vectors. If 2 linearly independant vectors spanned R3, then the dimension of R3 would be 2!
But feel free to check it manually, (x,y,z) = a (1,0,1) + b (0,1,1), where (x,y,z) is any vector in R3. Try to find values of a and b! You’ll end up prooving such a and b don’t exist. You also need to confirm Col(A) = R2, in the same way as above, or by taking a closer look at the vectors

And what about this?

Row(A) = sp{(1 0 1), (0 1 1)} ⊆ R^n=R^3

Col(A) = sp{(1 0), (0 1) (1 1)} ⊆ R^m=R^2

This is true, right?
In this case, the linear spaces of the spans are contained in the corresponding linear space, right?

So the rows of A are CONTAINED in R^3, but they are NOT EQUAL to R^3?

 

[Zermelo]

You’re right.
But you can get out a bit more information on Col(A). Can you see something special about those vectors?

 

[TheWrathOfMath]

You’re right.
But you can get out a bit more information on Col(A). Can you see something special about those vectors?

I can throw (1,1), for instance, out, since it can be written as a nontrivial linear combination of (0,1) and (1,0)?

 

[Zermelo]

I can throw (1,1), for instance, out, since it can be written as a nontrivial linear combination of (0,1) and (1,0)?

Yes. Do you now see what the span of (0, 1) and (1, 0) is?

 

[TheWrathOfMath]

Yes. Do you now see what the span of (0, 1) and (1, 0) is?

The span is R^2, basically.

 

[Steven G]

I can throw (1,1), for instance, out, since it can be written as a nontrivial linear combination of (0,1) and (1,0)?

I would not mention the nontrivial part.

 

[TheWrathOfMath]

I would not mention the nontrivial part.

I suppose a more accurate statement would be that a vector in the span can be written as a linear combination of the other vectors in the span, hence that vector can be omitted from the span.