Factor the matrix into the form QR where Q is orthogonal

[Dustinsfl]

Factor the matrix into the form QR where Q is orthogonal and R is upper triangular.

\begin{bmatrix}<br /> a &amp; b\\ <br /> c &amp; d<br /> \end{bmatrix}*\begin{bmatrix}<br /> e &amp; f\\ <br /> 0 &amp; g<br /> \end{bmatrix}=\begin{bmatrix}<br /> -1 &amp; 3\\ <br /> 1 &amp; 5<br /> \end{bmatrix}

\begin{bmatrix}<br /> a &amp; c <br /> \end{bmatrix}*\begin{bmatrix}<br /> b\\ <br /> d<br /> \end{bmatrix}=0

ae=-1

af+bg=3

ce=1

cf+dg=5

Skipping some steps but I arrive at:\begin{bmatrix}<br /> 1 &amp; \frac{4}{g}\\ <br /> -1 &amp; \frac{4}{g}<br /> \end{bmatrix}*\begin{bmatrix}<br /> -1 &amp; -1\\ <br /> 0 &amp; g<br /> \end{bmatrix}=\begin{bmatrix}<br /> -1 &amp; 3\\ <br /> 1 &amp; 5<br /> \end{bmatrix}

So as long as g \neq 0 it is all good?

 

[Mark44]

Works for me. You have a matrix that is orthogonal and another that is upper triangular, and they multiply to make the matrix on the right. The moral of the story seems to be that such factorizations aren't unique.

 

[Hurkyl]

"Q is orthogonal" consists of three conditions, not one...

 

[Mark44]

I forgot about the part where the columns have to be unit vectors...

 

[Dustinsfl]

Ok so the column vectors also have to be unit vectors and what is the other stipulation?

 

[Hurkyl]

Ok so the column vectors also have to be unit vectors and what is the other stipulation?

I was counting polynomial equations -- so what you just said counts as 2 conditions.