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Linear Algebra - Spanning / Matrix Equation

  • #1

Homework Statement



Let B be 4x4 matrix:

[ 1 3 -2 2 ]
[ 0 1 1 -5 ]
[ 1 2 -3 7 ]
[ -2 -8 2 -1 ]

a) Do the columns of B span R^4?
b) Does the equation Bx = y have a solution for each y in R^4?

Sorry for the crappy matrix, its a 4x4, R^4 is the funky double R for 'Reals'

Homework Equations



A) So, started out with row reduction. To find B spans R^4, B must have pivot positions in every row.

I won't go through all the row reduction steps, but I came out to a matrix looking like so:

[ 1 3 -2 2 ]
[ 0 1 1 -5 ]
[ 0 0 0 0 ]
[ 0 0 0 -7 ]

There are only pivot positions in row 1 + 2, so no, B does NOT span R^4.

B) Bx = y does NOT have solutions for each y in R^4. This is due to (what we called in class Theorem 4) with the logically equivalent assumptions that:

1) For each y in R^4, Bx = y has solution
2) Each y (element of R^4) is linear combo of A columns
3) Columns of A span R^4
4) B has pivot in every row

Since B does not span R^4 and does not have pivots in every row, hence forth the statement that Bx = y has solutions for y in R^4 is incorrect.


...Does this look right to anyone?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Looks ok to me.
 

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