# Vector spaces, Spans and Matrix Determinants

1. Dec 13, 2007

### ND3G

I think I have something mixed up so if someone can please point out my error.

1. the set of all linear combinations is called a span.

2. If a family of vectors is linearly independent none of them can be written as a linear combination of finitely many other vectors in the collection.

3. If the determinant of a matrix is not equal to zero the vectors are linearly independent.

Therefore, if the determinant of the matrix does not equal zero the vectors can not be written as a linear combination, hence there is no span.

Last edited: Dec 13, 2007
2. Dec 13, 2007

### Coin

The only problem I see here is that a "span" is something that applies to a set of vectors, or a space. Then all of a sudden you're talking about matrices. It does not make sense in any way I can think of to talk about the "span" of a matrix-- the words are just not meaningful.

Now maybe the idea is, you want to take the basis set of vectors for some space, and create a matrix whose columns consist of those basis vectors. If this is the case then yes, the determinant would be nonzero (this is actually one way of testing whether a set of vectors form a basis), and yes, it will not be possible to represent any of the basis vectors as a linear combination of the others (since this is part of what a basis means). The span of the basis vectors meanwhile will be the entire space.

3. Dec 13, 2007

As Coin pointed out, the term "span" applies to a set of vectors, so you'll have to be a bit more precise. So, you could say that the set of all linear combinations of a set of vectors forms the span of that very set (or linear shell, as I have been tought).

4. Dec 13, 2007

### ND3G

Ok, to clear things up a little

Given vectors X1, X2..., X3 in R^n, a vector in the form X = t1 X1 + t2 X2 + ... tk Xk

1. the set of such linear combinations is called a span of the Xi and denoted by span{X1, X2, ...Xk}

2. If a family of vectors is linearly independent none of them can be written as a linear combination of finitely many other vectors in the collection.

3. If the determinant of a matrix is not equal to zero the vectors are linearly independent.

Therefore, if the determinant of the matrix does not equal zero the vectors can not be written as a linear combination, hence there is no span.

Now my text gives me a solution where a matrix whose columns consist of basis vectors has a determinant of -42. It also states that the vectors span R^4.

Now, if the non zero determinant means that no linear combination can be written, and with no linear combination there is no span as it is a combination of the linear combinations. So, either the vectors span R^4 or the determinant is non-zero. I can't see how it can be both.

The vectors are: {[1 3 -1 0]T, [-2 1 0 0]T, [0 2 1 -1]T, [3 6 -3 -2]T}

5. Dec 13, 2007

### HallsofIvy

Staff Emeritus
No, "non zero determinant" means that none of the vectors in the set can be written as a linear combination of the others. It surely does not mean "no linear combination can be written". A linear combination of a set of vectors is just a sum of numbers times those vectors. That can always be done.

6. Dec 13, 2007

### ND3G

Yeah, I am starting to see that.

I also found another theorem which counters what I posted before.

An nxn matrix is invertible if and only if
1) the rows are linearly independent (otherwise det = 0)*
2) the columns are linearly independent (otherwise det = 0)*
3) rows of A span R^n

So in order for the rows of A to span R^n the det must be non-zero, not the other way around.

*a matrix is not invertible if the determinant = 0*

Last edited: Dec 13, 2007
7. Dec 13, 2007