## DevinBlade 4 years ago Looking for answer to part c c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

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) 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]. 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] *Part B Answered* (Credit to Alchemista, explain further if you would like to) In order to determine if C is in the span of B, Use c1, c2, and c3 in a linear combination with B to determine if its in the span of B. Matrix B = c1 1 0 0 1 0 1 0 2 2 3 4 -4 0 0 1 -3 Reduced: 1 0 0 1 0 1 0 2 0 0 1 -3 0 0 0 0 Matrix B = c2 1 0 0 1 0 1 0 3 2 3 4 3 0 0 1 -2 Reduced: 1 0 0 1 0 1 0 3 0 0 1 -2 0 0 0 0 Matrix B = c3 1 0 0 -1 0 1 0 3 2 3 4 -9 0 0 1 4 Reduced: 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 You can express the first vector in C as a Linear Combination of B, the second, but not the third. As you can see the third has no solution (the system is inconsistent) The answer then is no, since there are vectors in the span of C that are not in B. c. Find the change of coordinates matrix P from B to C. [x]C = P [x]B

2. Alchemista

If you want, you can also perform the reduction on an augmented matrix with all three vectors at once. $\left[ \begin {array}{cccccc} 1&0&0&1&1&-1\\ 0&1&0& 2&3&3\\ 2&3&4&-4&3&-9\\ 0&0&1&-3&- 2&4\end {array} \right]$ Reduced: $\left[ \begin {array}{cccccc} 1&0&0&1&1&0\\ 0&1&0&2 &3&0\\ 0&0&1&-3&-2&0\\ 0&0&0&0&0&1 \end {array} \right]$

i dont think that way would work, if i remember correctly the way you wrote it implies that columns 1 and 2 of C would be part of B making it a 4x5 matrix wouldnt it?

4. Alchemista

It is perfectly valid, you are solving the system for all three vectors simultaneously.

5. Alchemista

As you can see the last vector can't be expressed as a linear combination of the vectors in B.

6. abtrehearn

I am drawing the same conclusion, Alchemista. It's too bad the vector c(3) doesn't work out.

hmmm thats interesting, well if thats the case what are the steps involved to answer part c?

i dont think c was answered o_o unless it cant be answered due to the last row not having a solution

9. abtrehearn

[x]C does not exist.

so because [x]C does not exist, part c: c. Find the change of coordinates matrix P from B to C. [x]C = P[x]B cant be answered then?

11. abtrehearn

It is impossible to find a linear combination of vectors in C that is equal to x, since no linear combination of vec

12. abtrehearn

ors in exists

13. abtrehearn

of vectors in B exists for c(3).

14. Alchemista

I was thinking that perhaps he might have copied down the values for the vectors in C incorrectly.

15. Alchemista

Otherwise I'm not sure.

when i say part c i mean the third part of the question being: c. Find the change of coordinates matrix P from B to C. [x]C = P[x]B unless im just being dumb and im repeating my self do let me know if i am and i just double checked the matrix for C its exactly as the teacher wrote it

17. abtrehearn

No change coordinates from C to C exists.

18. abtrehearn

... from B to C exists.

lol i was like say whaaat for a second there XD