determine if v1= [ 2 1 0] v2=[ -1 1 3] v3=[ 0 -1 6] spans the vector space of rows with three real entries which has dimension 3.
so I wanted to make sure I did this correct. First I created a matrix with v1,v2,v3 as the columns (so for example column 1= 2
1
0
then I created an augmented matrix with that by tacking on the column [0;0;0]
I reduced it to the rref and found that it created the identity matrix so a1=0 a2=0 and a3=0. So does this mean it spans V , where V has the dimension 3?
Yes, your method is correct.
By creating a matrix with v1, v2, and v3 as columns and performing row operations to reduce it to reduced row-echelon form (rref), you can determine if v1, v2, and v3 span the vector space of rows with three real entries.
If the rref of the augmented matrix yields the identity matrix, it means that the only solution to the system of equations is the trivial solution (a1 = 0, a2 = 0, a3 = 0). This implies that v1, v2, and v3 are linearly independent and span the vector space.
Since the dimension of the vector space is 3 and you have three linearly independent vectors, the vectors v1, v2, and v3 span the vector space.
To determine if the vectors v1=[2 1 0], v2=[-1 1 3], and v3=[0 -1 6] span the vector space of rows with three real entries, we need to check if these vectors form a basis for the vector space.
To do this, you correctly created a matrix with v1, v2, and v3 as the columns. Let's call this matrix A.
A = [2 -1 0
1 1 -1
0 3 6]
Next, you augmented this matrix with a column of zeros to create an augmented matrix. Let's call this augmented matrix [A | 0].
[A | 0] = [2 -1 0 | 0
1 1 -1 | 0
0 3 6 | 0]
You reduced this augmented matrix to row-echelon form or reduced row-echelon form (rref) and found that it transformed into the identity matrix.
If your resulting matrix looks like this:
[1 0 0 | 0
0 1 0 | 0
0 0 1 | 0]
Then, it means that the augmented matrix reduced to the identity matrix. This indicates that the system of linear equations [A | 0] has only the trivial solution, where a1 = 0, a2 = 0, and a3 = 0.
Since the only solution to the system is the trivial solution, it means that the vectors v1, v2, and v3 are linearly independent. In other words, they do not span the entire vector space.
Therefore, the vectors v1, v2, and v3 do not form a basis for the vector space of rows with three real entries. Consequently, they do not span the vector space of dimension 3.