Linear algebra EA=R factorization question

In summary, the "easy way" to get the matrix E that puts A into reduced row echelon form is to perform row operations on A, which correspond to multiplying by elementary matrices, until the A part becomes the identity matrix. The product of all the elementary matrices used is the inverse matrix E. This is explained by the fact that every row operation corresponds to an elementary matrix, and applying a sequence of row operations is the same as multiplying by the corresponding elementary matrices in order. This is also why multiplying E and A gives R, the reduced row echelon form of A.
  • #1
FocusedWolf
81
0
ok i know about the thing, i think its called gauss-jordan, where you do elimination on [A I] and you get [I A^-1] or [R E]or something like that.

question is how can you get the E that puts A into reduced row echelon form with "row operations" only...supposedly some easy way ?

on pg 134 Gilbert Strangs book "introduction to linear algebra" he says:

"EA = R
The square matrix E is the product of elementary matrices Eij and Pij and D^-1..." yea that's all he says about doing that, but it was good enough to get it on the exam, and probably the final now lol

on pg 139 he does some work that doesn't look like steps but he gets two different E that can get A unto rref... so what's the secret? :P
 
Physics news on Phys.org
  • #2
I'm a little confused; what exactly is it that you want to know?
 
  • #3
FocusedWolf said:
ok i know about the thing, i think its called gauss-jordan, where you do elimination on [A I] and you get [I A^-1] or [R E]or something like that.

question is how can you get the E that puts A into reduced row echelon form with "row operations" only...supposedly some easy way ?

on pg 134 Gilbert Strangs book "introduction to linear algebra" he says:

"EA = R
The square matrix E is the product of elementary matrices Eij and Pij and D^-1..." yea that's all he says about doing that, but it was good enough to get it on the exam, and probably the final now lol

on pg 139 he does some work that doesn't look like steps but he gets two different E that can get A unto rref... so what's the secret? :P

Well, yes, there is an "easy way" to get E- you DO the row operations that put A into "reduced echelon form". For example, to find the inverse of
[tex]A= \left[\begin{array}{cc}2 & 1 \\ 3 & 2\end{array}\right][/tex]
First set it up as:
[tex]\left[\begin{array}{cccc}2 & 1 & 1 & 0\\ 3 & 2 & 0 & 1\end{array}\right][/tex]
Now do row operations to reduce the "A" part to the identity matrix (ie.e reduced row echelon). Do one column at a time. Since you want a "1" in the upper left, divide the first row by 2. Since you want a "0" in the lower left, multiply the new first row by -3 and add to the second row:
[tex]\left[\begin{array}{cc}1 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{3}{2} & 1\end{array}\right][/tex]
Now you want the second column to be "0" and "1": multiply the second row by 2 and subtract 1/1 times the new second row from the first row:
[tex]\left[\begin{array}{cc}1 & 0 & 2 & -1 \\ 0 & 1 & -3 & 2\end{array}\right][/tex]
The inverse matrix is
[tex]A^{-1}= \left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right][/tex]

The quote from Strang is not saying HOW to do that but explaining WHY it works. Every row operation corresponds to an elementary matrix constructed by applying that row operation to the identity matrix: applying the row operation to any matrix is the same as multiplying the matrix by the corresponding elementary matrix. Applying a sequence of row operations is the same as multiplying by the corresponding elementary matrices in order. The product of all those elementary matrices is the same as applying all the row operations to the identity matrix.

In the above calculation, the sequence of row operations reduces A to the identity matrix so multiplying by the product of their corresponding elementary matrices does the same: that product is the inverse matrix. Since we apply those same operations to the identity matrix as we work, we produce that inverse matrix.
 

1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, vectors, and matrices. It involves the manipulation and analysis of these mathematical objects to solve problems in various fields such as physics, engineering, and computer science.

2. What is the purpose of EA=R factorization?

EA=R factorization, also known as the QR decomposition, is a method used to decompose a matrix into two orthogonal matrices and an upper triangular matrix. This process is useful in solving systems of linear equations, finding eigenvalues and eigenvectors, and performing other operations in linear algebra.

3. How is EA=R factorization performed?

EA=R factorization involves finding the QR factorization of a matrix A, where Q is an orthogonal matrix and R is an upper triangular matrix. This can be done using various methods such as Gram-Schmidt process, Householder reflections, or Givens rotations. The resulting matrices can then be used to solve linear equations or perform other operations.

4. What are the applications of linear algebra in real life?

Linear algebra has numerous applications in various fields such as engineering, physics, economics, and computer science. Some examples include image and signal processing, machine learning, optimization problems, and 3D graphics. It is also used in the development of algorithms and solving complex systems of equations.

5. How does linear algebra relate to other branches of mathematics?

Linear algebra is closely related to other branches of mathematics such as calculus, differential equations, and abstract algebra. It provides a foundation for understanding and solving problems in these areas by providing tools and techniques for manipulating and analyzing linear systems. Many advanced mathematical concepts and theories also rely on the principles of linear algebra.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
2K
  • STEM Academic Advising
Replies
6
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • STEM Academic Advising
Replies
1
Views
1K
Replies
14
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
4K
  • Science and Math Textbooks
Replies
2
Views
3K
  • Poll
  • Science and Math Textbooks
Replies
4
Views
6K
  • Poll
  • Science and Math Textbooks
Replies
1
Views
3K
Back
Top