QR factorization of a n x 1 matrix

[abajaj2280]

Homework Statement

Consider the vector a as an n × 1 matrix.

A) Write out its reduced QR factorization, showing the matrices $\hat{Q}$ and $\hat{R}$ explicitly.

B) What is the solution to the linear least squares problem ax ≃ b where b is a given n-vector.

Homework Equations

I was using the equation from 1.1 (http://www.math.ucla.edu/~yanovsky/Teaching/Math151B/handouts/GramSchmidt.pdf) to help me solve this problem.

The Attempt at a Solution

I haven't taken linear algebra for about 2 years and this is kind of hazy. I'm really confused here, and I really don't know where to start. I know that I'm supposed to come to this website with some sort of progress, but I'm really confused here.

 

[middleCmusic]

So, the first step in the Gram-Schmidt process is to think of the matrix as being a row of column vectors. Since your matrix is n x 1, it's like having a matrix of only one column vector. Thus $A = [\mathbf{a_1}]$ in this case. So, going by the pdf you provided, let $\mathbf{u_1} = \mathbf{a_1}$ and then
$\mathbf{e_1} = \frac{\mathbf{u_1}}{\left\| \mathbf{u_1} \right\|}$ $= \frac{\mathbf{a_1}}{\sqrt{(a_{11})^2+(a_{21})^2+...+(a_{n1}^2)}}$.

In this case, $Q = [\frac{ \mathbf{a_1} }{\left\| \mathbf{a_1} \right\|}]$ and R is the 1x1 matrix
$[ \mathbf{a_1} \bullet (\frac{ \mathbf{a_1} }{\left\| \mathbf{a_1} \right\|}) ]$ = $[ \frac{ \mathbf{a_1} \bullet \mathbf{a_1} }{ \left\| \mathbf{a_1} \right\| } ]$ = $[\frac{(\left\| \mathbf{a_1} \right\|)^2}{\left\| \mathbf{a_1} \right\|}] = [\left\|\mathbf{a_1}\right\|]$.

Then, $QR = [ \frac{\mathbf{a_1}}{\left\|\mathbf{a_1}\right\|} \left\|\mathbf{a_1}\right\| ] = [\mathbf{a_1}]$.

In this case, the result isn't very interesting, because it's an nx1 matrix. But I think that's also to (supposedly) make it easier for you. I can see how in this case it made it even more confusing.

 

[abajaj2280]

Thank you so much for your help. I have an exam in this class in one week, so I will be referring back to this when studying.