Linear Algebra - Spanning / Matrix Equation

Click For Summary
SUMMARY

The discussion centers on the properties of a 4x4 matrix B and its ability to span R^4. Through row reduction, it was determined that matrix B does not have pivot positions in every row, specifically only in the first two rows. Consequently, the columns of B do not span R^4, and the equation Bx = y does not have a solution for every y in R^4. This conclusion aligns with Theorem 4 regarding the relationship between spanning sets and the existence of solutions in linear equations.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly spanning sets
  • Familiarity with matrix row reduction techniques
  • Knowledge of pivot positions in matrices
  • Comprehension of linear combinations and their relation to vector spaces
NEXT STEPS
  • Study the implications of pivot positions in matrices for spanning sets
  • Learn about Theorem 4 in linear algebra and its applications
  • Explore row reduction methods in greater detail, including Gaussian elimination
  • Investigate the relationship between linear combinations and the solution sets of linear equations
USEFUL FOR

Students and educators in linear algebra, mathematicians focusing on vector spaces, and anyone seeking to deepen their understanding of matrix equations and spanning sets.

twiztidmxcn
Messages
43
Reaction score
0

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?
 
Physics news on Phys.org
Looks ok to me.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K