Linear Algebra - Spanning / Matrix Equation

In summary, the conversation discusses a 4x4 matrix B and whether its columns span R^4 and if the equation Bx = y has a solution for each y in R^4. Through row reduction, it is determined that B does not span R^4 and the equation does not have solutions for each y in R^4. This is supported by Theorem 4, which states that for Bx = y to have solutions, B must have pivot positions in every row and its columns must span R^4. Since B does not meet these criteria, the statement that Bx = y has solutions for y in R^4 is incorrect.
  • #1
twiztidmxcn
43
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
  • #2
Looks ok to me.
 

1. What is the concept of spanning in linear algebra?

Spanning in linear algebra refers to the set of all possible linear combinations of a given set of vectors. In other words, it is the set of all possible ways that a vector can be written as a combination of the given set of vectors. This concept is important in understanding the dimension of a vector space and determining if a set of vectors is linearly independent or not.

2. How do you determine if a set of vectors spans a vector space?

To determine if a set of vectors spans a vector space, you can use the row reduction method or Gaussian elimination to solve the matrix equation Ax = b. If the resulting matrix is consistent (has a solution), then the set of vectors spans the vector space. Another method is to check if the vectors in the set are linearly independent, and if they are, then they automatically span the vector space.

3. What is a matrix equation in linear algebra?

A matrix equation in linear algebra represents a system of linear equations in the form of a matrix. It is typically written as Ax = b, where A is a coefficient matrix, x is a vector of unknown variables, and b is a vector of constants. Solving this equation involves finding the values of x that satisfy the equation, which can be done using various methods such as row reduction and Gaussian elimination.

4. Can a set of vectors span more than one vector space?

No, a set of vectors can only span one vector space at a time. This is because a vector space is a unique mathematical structure with specific properties, and a set of vectors can only span a vector space if it satisfies those properties. If a set of vectors spans more than one vector space, then it means that those vector spaces are equivalent, and the set of vectors is a basis for both of them.

5. What is the relationship between span and linear independence?

Span and linear independence are closely related concepts in linear algebra. A set of vectors is linearly independent if none of the vectors can be written as a linear combination of the others. On the other hand, the span of a set of vectors is the set of all possible linear combinations of those vectors. Therefore, a set of linearly independent vectors will automatically span a vector space, while a set of dependent vectors will not span a vector space unless some of the vectors are removed from the set.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
786
  • Calculus and Beyond Homework Help
Replies
14
Views
581
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
558
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
928
Back
Top