Question

Looking for answer to part b currently

3a. Given the vector x and the basis B = [b1, b2, b3] of a subpace of R4 determine if x is in the span of B, and if so find the coordinates of x with respect to B and give the coordinate vector [x]B.

x:[1,1,1,-1] b1: [1,0,2,0] b2:[0,1,3,0] b3:[0,0,4,1]

**Part A Answered** (Credits to abtrehearn)

b. Given the basis C = [c1, c2, c3], determine if C is in the span of B. If so write the vector x using the C coordinates. and give the coordinate vector [x]C.

c1:[1,2,-4,-3] c2:[1,3,3,-2] c3:[-1,3,-9,4]

Answer 1

Answer to part A:
Part a:
Vector x is in the span of B if and only if it is a linear combination of the vectors in B. Then there are scalars
x1,x2,x3

such that
x1b1+x2b2+x3b3=x,
or
x1(1,0,2,0)+x2(0,1,3,0)+x3(0,0,4,1)=(1,1,1,−1)

a system of four simultaneous linear equations in unknowns
x1,x2,x3,
which has the unique solution
(x1,x2,x3)=(1,1,−1),

so the vector x can be expressed as a linear combination if the vectors in B, and x is in the span of b. The coordinate vector [x]B is [1,1,-1].

Answer 2

B=1 0 0 1
0 1 0 1
2 3 4 1
0 0 1 -1
C=1 1 -1 1
2 3 3 1
-4 3 -9 1
-3 -2 -4-1
like this im assuming although im not entirely sure i understand what you mean after these are composed

Answer 3

No forget about x for the moment.

Answer 4

Sorry one second.

Answer 5

so do you mean something like this?
c1*(column B1)+c2*(column B2+ c3*(column B3)?

Answer 6

might have to elaborate on that, don't entirely follow at the moment <_<

Answer 7

You know what just setup 3 systems, one for each vector.

Answer 8

If each vector in C can be expressed as a linear combination of B then span(B)=span(C)

Answer 9

would doing row reduction of C then having it show that it has a pivot in r1c1 r2c2 and r3c3 prove span (B) = span(C)?

Answer 10

Do the same thing you did for the vector "x" but then replace x with a vector from C, do it for each in C.

Answer 11

That might be valid, I guess I have to think about it.

Answer 12

1 1 -1 1
2 3 3 1
-4 3 -9 1
-3 -2 -4 -1

Row reduced form of C would be:
1 0 0 0.5
0 1 0 0.25
0 0 1 -0.25
0 0 0 0

unfortunately im still quite lost as to if i use the answer of part a being 1,1,-1 for the answer of c instead of x

Answer 13

Check first vector in C:
\[ \left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&2
\\ 2&3&4&-4\\ 0&0&1&-3\end {array}
\right] \]
Check second vector in C:
\[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&3
\\ 2&3&4&3\\ 0&0&1&-2\end {array}
\right] \]

etc.

Answer 14

I'll see if I can think of a cleaner way of seeing of C is in the span of B, but this will work even if its messy.

Answer 15

Unfortunately i dont know what you mean in the post above yours ;_;

Answer 16

Reduce each one and see if there is a solution.

Answer 17

We are seeing of each vector in C can be expressed as a linear combination of vectors in B.

Answer 18

(1, 2, -4, -3) = 1*(1, 0, 2, 0) + 2*(0, 1, 3, 0) + -3*(0, 0, 4, 1)

So the first element is in the span, etc.

Answer 19

do recall most of the advanced stuff i don't have experience with yet due to my teacher not teaching the class any thing, none of the basics for that matter, im just now barely able to grasp the concept of the basics about linear algebra but all the steps in between causes me to get lost. also i have no clue where you're getting that from. are you multiplying each other then taking the row reduced form of the product of both?

Answer 20

its exactly the same concept we used in the beginning to find out how to express x as a l.c. of B.

Answer 21

this time we want to see if we can express each vector in C using B

Answer 22

while i do understand what you mean by using b in c im not entirely sure of the steps involved to solve this for example (1, 2, -4, -3) = 1*(1, 0, 2, 0) + 2*(0, 1, 3, 0) + -3*(0, 0, 4, 1) where did this come from? while i know (1, 2, -4, -3) is the first column of C and those other three columns are from B where does the multiplication of 1 2 and -3 come from?

Answer 23

Thats the solution to the system.

Answer 24

You need to do the same thing you did with x. Row reduce B augmented with 1 vector from C. Do it for each vector in C.

Answer 25

You will actually find that the last vector in C cant be expressed. So the answer is actually no its not in the span of B.

Answer 26

Matrix B
1 0 0
0 1 0
2 3 4
0 0 1
Matrix C
1 1 -1
2 3 3
-4 3 -9
-3 -2 -4
Ok so i see you used the first column of matrix C as the answer to matrix B but im not entirely sure why you used it as the answer to B can you elaborate if possible?

Answer 27

\[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&2
\\ 2&3&4&-4\\ 0&0&1&-3\end {array}
\right] \]
Reduced:
\[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&2
\\ 0&0&1&-3\\ 0&0&0&0\end {array}
\right] \]

\[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&3
\\ 2&3&4&3\\ 0&0&1&-2\end {array}
\right] \]
Reduced:
\[\left[ \begin {array}{cccc} 1&0&0&1\\ 0&1&0&3
\\ 0&0&1&-2\\ 0&0&0&0\end {array}
\right] \]

\[\left[ \begin {array}{cccc} 1&0&0&-1\\ 0&1&0&3
\\ 2&3&4&-9\\ 0&0&1&4\end {array}
\right] \]
Reduced:
\[\left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0
\\ 0&0&1&0\\ 0&0&0&1\end {array}
\right] \]

Answer 28

You can express the first vector in C as a L.C of B, the second, but not the third.

Answer 29

As you can see the third has no solution (the system is inconsistent)

Answer 30

If C is in the span of B, you must be able to express each element as a L.C of B, do you understand that? The purpose of solving these systems is to see if that's possible. If a system has no solution it means there is no linear combination that will yield the vector.

Answer 31

ahhh that last part cleared that up as to why you did that

Answer 32

The answer then is no, since there are vectors in the span of C that are not in B.

Answer 33

Otherwise you could express the last vector as an L.C of B.

Answer 34

very neat thanks Alchemista, theres one last part to this question and ive got no clue about it im hoping its possible to solve

Answer 35

c. Find the change of coordinates matrix P from B to C.
[x]C = P [x]B
this is the last part of it, any idea?

Answer 36

The question states: "If so", but it is not so.

Answer 37

Wait a second a third part. one second.

Answer 38

its part c the part you just completed was part b :P

Answer 39

Well in any case, it'd be nice if you could move the last part to a new question. All the LaTeX is slowing my browser down to a crawl.

Answer 40

ahh right same here, before i do that though i need to go through this and try to figure out how to sum up part b in one answer