Solution

Part a is the computation of the null space of . While you have the general theory, this is best done by inspection. We notice that can be free variables and then a basis for the null space and hence the plane is . Note that I chose the values of and for the free variables to avoid fractions!

(b) The desired matrix is:

For part c, we go thru the usual calculations P =

> A1:=evalm(transpose(A)&*A);

The matrix is and the projection matrix then is:

For part d, note that we can only project vectors, not points! (You can project the vector from the origin to the point, though!) Thus we want the projection of the vector or . The answer is:

P =

For part e, by projection of the unit circle, we can only mean the projection of the various vectors from origin to the points of the unit circle. The vectors can be easily described as and we can find the projected vector as: P =

If we call this vector v the square of the length of v is

=

The length of the major axis will be twice the maximum value of the function

= for from 0 to

Now we have entered something not formally studied in this course, but something that you should know from previous experience. The expression is the square of the length of the projected vector and is a function of . We are looking for its maximum value. You can draw on your calculus skills to do this.

> h1:=diff(h,theta);

We take the derivative of h with respect to

and find the critical numbers by setting this

expression equal to 0.

The expression vanishes when =

How was this done? Think about it. (hint: make it an equation in tan( ) ) Anyway, subtitute in h to check the max/min.

The values of h are 1 and which says that the major axis has length _____ and minor axis has length ____.