How Do You Determine the Basis, Dimension, and Rank of Vector Sets?

[_Bd_]

Homework Statement

for the set of vectors:
v_1 = 1, -2, 0, 0, 3
v_2 = 2, -5, -3, -2, 6
v_3 = 0, 5, 15, 10, 0
v_4 = 2, 6, 18, 8, 6
(a) find a basis for the set of vectors and state the dimension of the space spanned by these vectors, what is the rank of this matrix?
(b) construct a matrix whose rows correspond to the vectors v_1 thru v_4 and find the basis for the kernel of that matrix and state the dimension of this null space. what is the rank of this matrix?

Homework Equations

allowed to use calculator

The Attempt at a Solution

According to the book instructions I already solved part A
by taking the RREF of the matrix:
[1, -2, 0, 0, 3]
[2, -5, -3, -2, 6]
[0, 5, 15, 10, 0]
[2, 6, 18, 8, 6]
and noticing that the last row is just zeros, therefore its a rank 3 matrix with dimension n-r which would be 2. (Im not sure if this is 2 or 5. . .since it has 5 variables??)
making a system of eq. with the rref form I get
that Ax = 0 then x=
x_1 = 2t - 3s
x_2 = t
x_3 = -t
x_4 = t
x_5 = s

therefore the basis is:

t[2, 1 ,-1, 1, 0] + s[-3, 0, 0, 0, 1] <-- as column vectors

then for part b. . .when I started to do it I realized it was the exact same process? cause my original matrix was made of the rows and the kernel is the nullspace which is 2 (from what I understood from the dif. chapters in the book its the same as the dimension stated above)
so nullity = n - rank = 2
and then the rank. . .would be 3 again?
what trips me off is that its the exact same process twice? or am I confusing some things here?

 

[HallsofIvy]

Homework Statement

for the set of vectors:
v_1 = 1, -2, 0, 0, 3
v_2 = 2, -5, -3, -2, 6
v_3 = 0, 5, 15, 10, 0
v_4 = 2, 6, 18, 8, 6
(a) find a basis for the set of vectors and state the dimension of the space spanned by these vectors, what is the rank of this matrix?

The rank of what matrix? There is no matrix given here. If the problem means the matrix having those vectors as rows or columns, that should have been said.

(b) construct a matrix whose rows correspond to the vectors v_1 thru v_4 and find the basis for the kernel of that matrix and state the dimension of this null space. what is the rank of this matrix?

Homework Equations

allowed to use calculator

The Attempt at a Solution

According to the book instructions I already solved part A
by taking the RREF of the matrix:
[1, -2, 0, 0, 3]
[2, -5, -3, -2, 6]
[0, 5, 15, 10, 0]
[2, 6, 18, 8, 6]
and noticing that the last row is just zeros, therefore its a rank 3 matrix with dimension n-r which would be 2. (Im not sure if this is 2 or 5. . .since it has 5 variables??)

Yes, the rank is three. Now what dimension are you talking about? A matrix does not have a "dimension". Assuming you are talking about the matrix having these vectors as columns it is a linear transformation from $R^4$ to $R^5$. Since its rank is 3, it nullspace has dimension 5- 3= 2. The "dimension spanned by the vectors", since you have shown that 3 of them are independent, is 3.

making a system of eq. with the rref form I get
that Ax = 0 then x=
x_1 = 2t - 3s
x_2 = t
x_3 = -t
x_4 = t
x_5 = s

therefore the basis is:

t[2, 1 ,-1, 1, 0] + s[-3, 0, 0, 0, 1] <-- as column vectors

Yes, so you have shown that the null space has dimension 2. But there is nothing said about "null space" in part (a).

then for part b. . .when I started to do it I realized it was the exact same process? cause my original matrix was made of the rows and the kernel is the nullspace which is 2 (from what I understood from the dif. chapters in the book its the same as the dimension stated above)
so nullity = n - rank = 2
and then the rank. . .would be 3 again?
what trips me off is that its the exact same process twice? or am I confusing some things here?

If you use those same vectors as rows rather than columns, then you have a linear transformation from \(\displaystyle R^5\) to \(\displaystyle r^4\). You will now find that 3 of the rows are independent so that the rank is 3 and the dimension of the null space is 4- 3= 1.

 

[_Bd_]

(A) The rank of what matrix? There is no matrix given here. If the problem means the matrix having those vectors as rows or columns, that should have been said.
======

(B) Yes, the rank is three. Now what dimension are you talking about? A matrix does not have a "dimension". Assuming you are talking about the matrix having these vectors as columns it is a linear transformation from $R^4$ to $R^5$. Since its rank is 3, it nullspace has dimension 5- 3= 2. The "dimension spanned by the vectors", since you have shown that 3 of them are independent, is 3.

=====
(C)
Yes, so you have shown that the null space has dimension 2. But there is nothing said about "null space" in part (a).

(D)
If you use those same vectors as rows rather than columns, then you have a linear transformation from \(\displaystyle R^5\) to \(\displaystyle r^4\). You will now find that 3 of the rows are independent so that the rank is 3 and the dimension of the null space is 4- 3= 1.

(A) I really don't know what matrix If I had to guess it would be the matrix formed by the vectors in space (so that's how I started the problem) and well from what I saw in the book, and as you saw in my example I wrote the vectors in row form not in column form to begin with. . . and went on from there to the rref, dimension of 2 and finding the basis of the matrix formed by those vectors. . .

(B) again I must mention or I don't understand but I wrote it in row form not in column form. . . so I am not too sure as to where(or why) that linear transformation took place

(C)
Ok I found another theorem on the book about the vector space instead of the solution space, got that cleared up now. . .

(D)
again I don't understand the switching of rows to columns for the first part, (part B ask specifically about row vectors which would be the process I already did isn't it?)

= thanks for the help BTW =