Dtyjlui

Just a question. Let's say that you are given four vectors: ${v_1=[1,0,0] v_2=[0,2,0], v_3=[0,0,3] v_4=[1,1,1].}$. And with the help of the determinant of a matrix containing these vectors in column form, I have to show that they span $R^3$ but do not form a basis for $R^3.$

Obviously, there is another way to attack this problem. By having it in 3x4 matrix form just as mentioned above and reducing it to reduced row echelon form gives you that the v4 vector does not have a pivot, then therefore it can be thrown out of the set. Now you are left with three vectors in your set, and therefore they DO span R3 and also form a basis since they are linearly independent unlike before when you actually had 4 vectors and v4 was dependent on the previous vectors. Is this way of thinking right? I know that I have my book here, but it is useless.

In order for a set of vectors to span (Lets say R3) and form a basis for R3, they must be linearly independent and also the amount of vectors must be equivalent to the n-dimension. Is this right thinking?

That is my first question, and my second question is whether the general theorem of nxn matrices that say that if det is nonzero, then the vector set span Rn and are linearly independent. Well, basically if this theorem also can be applied to mxn matrices as far as the determinant goes when determining spanning and linearly independence? .. If not, how do I do i show that my vectors do span r3 but dont form a basis in r3 with determinant when having an mxn matrix which my vectors create in column form?...

$$v_1 + frac 12 v_2 + frac 13v_3 = v_4$$

That is, we can write, e.g., $v_4$ as a linear combination of $v_1, v_2, v_3$.

So the four given vectors are not linearly independent, and therefore cannot, all of them, be a basis for $mathbb R^3$. You also discovered this fact. (And yes, you're correct, four vectors will not form a basis of a 3-dimension vector space.) But the four vectors, because they inclue $v_1, v_2, v_3$, which spans $mathbb R^3$, certainly spans $mathbb R^3$. (Note Theorem 2 below.)

You can show that $v_1, v_2, v_3$ span all of $mathbb R^3$, and are linearly independent, by showing the determinant of the $3times 3$ matrix they form, when listed as columns in a matrix to form a $3times 3$ matrix, is not equal to $0$.

So while all four vectors span $mathbb R^3$, we also know that since $v_4$ is in the span of ${v_1, v_2, v_3}$, all four do not form a basis of $mathbb R^3$. Instead, a basis for $mathbb R^3$ is given by ${v_1, v_2, v_3}$.

Further notes:

You may find this entry on Wikipedia helpful to help disambiguate "spanning set" of a space from the "basis" of a space.

Theorem 1: The subspace spanned by a non-empty subset $S$ of a vector space $V$ is the set of all linear combinations of vectors in $S.$ [The span of $operatorname{Span}{v_1, v_2, v_3, v_4} = operatorname{Span}{v_1, v_2, v_3}$]. (Brackets mine).
This theorem is so well known that at times it is referred to as the definition of span of a set.

Theorem 2: Every spanning set S of a vector space V must contain at least as many elements as any linearly independent set of vectors from $V.$ [Bold-face mine.]

Theorem 3: Let V be a finite-dimensional vector space. Any set of vectors that spans V can be reduced to a basis for V by discarding vectors if necessary (i.e. if there are linearly dependent vectors in the set). [In our case, we can form the basis of $mathbb R^3$ by discarding $v_4$.](Brackets mine.)

This also indicates that a basis is a minimal spanning set when $V$ is finite-dimensional.

See also this example given in wikipedia, to help illustrate the theorems above.

I think you understand this material well.

The natural way to determine whether these vectors span is, as you say, to calculate the reduced row echelon form. They can't form a basis since there are four of them in a three dimensional space.

You can use determinants to show that they span, but it's not pretty. You look at the four possible $3 times 3$ matrices formed by omitting each of the four vectors in turn. When you find one with a nonzero determinant you have a basis, so a spanning set.