Exercise 3.3.4. Given
expand the expression , compute its partial derivatives with respect to
and
, and set them to zero. Compare the resulting equations to
to confirm that you obtain the same normal equations in both cases (i.e., using calculus vs. geometry). Then find the projection
and explain why
.
Answer: We have
so that
Taking the partial derivative of this expression with respect to we have
and setting it to zero we have
where and
represent the least squares solution.
Taking the partial derivative of this expression with respect to we have
and setting it to zero we have
where again and
represent the least squares solution.
We then have a system of two equations
that is equivalent to the matrix equation
We now compare this to the normal equations . We have
and
so that the normal equation in matrix form is
Note that this equation is the same as the previous equation calculated from taking the derivatives, except with both sides divided by 2.
We can now solve for . We have
so that
We can multiply by
to obtain the projection
of
onto the column space of
:
We have because
is already in the column space of
; in particular,
is equal to the first column of
plus 3 times the second column of
:
UPDATE: Fixed a typo in the equations involving and
. Thanks go to Matt for finding this error.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition
and the accompanying free online course, and Dr Strang’s other books
.
You have a slight misprint with A transposed about half way through the page. The top row should be 1 0 1 not the 1 0 0 that you have. I see you have it wrong twice but the answer is still correct. Probably just copied it wrong.
Fixed. Thanks for catching this!