Using QR decomposition to find a nontrivial solution to Ax=0

[Random Variable]

Homework Statement

Supposedly the process to solve Ax=0 is to solve Transpose(R).Ry=z (where z is a random vector) and then x=y/(norm-2 of y).

Homework Equations

The Attempt at a Solution

Ay=b for some random vector b

Transpose(A).Ay= Transpose(A).b

Transpose(QR).QRy=Transpose(A).b

Transpose(R).Transpose(Q).QRy = z where z is Transpose(A).b

Transpose(R).Ry=z since Q is orthogonal

Then x (the solution to the homogeneous system) is y/(norm-2 y)? But why?

 

[mighty2000]

it is important to approach problems with a critical and analytical mindset. While the provided solution may work in some cases, it is important to understand the reasoning behind it and not simply accept it as the correct method without further investigation.

Firstly, it is important to clarify that the given solution assumes that A is a square matrix with full rank (i.e. it has linearly independent columns). In this case, the solution to Ax=0 is indeed given by x=y/(norm-2 of y).

To understand why this is the case, let's break down the steps in the solution provided. The first step is to convert the original problem, Ax=0, into the equivalent problem, Ay=b, where b is a random vector. This is achieved by multiplying both sides of the equation by A transpose, as shown in the second line of the solution.

Then, the next step is to use the QR decomposition of A to rewrite the equation as Transpose(QR).QRy=Transpose(A).b. This step utilizes the fact that the transpose of an orthogonal matrix (Q) is also orthogonal, and the transpose of a triangular matrix (R) is also triangular.

The third step involves simplifying the equation by multiplying both sides by the transpose of Q, which is still orthogonal. This results in Transpose(R).Ry=z, where z is a new random vector (Transpose(A).b).

Finally, since R is an upper triangular matrix, the solution to this system can be easily obtained by back substitution. This results in y/(norm-2 of y) as the solution to the original problem, Ax=0.

In summary, the given solution utilizes the properties of orthogonal and triangular matrices to simplify the original problem and obtain the solution. However, it is important to note that this solution may not work for all cases, especially when A is not square or does not have full rank. it is important to always question and understand the methods and assumptions used in solving a problem.