## SUGGESTIONS FOR HOMNEWORK #7

Homework suggestions:

4.1
16. If you let the entries of the fourth vector be unknowns, the fact that it is orthogonal to the other three gives a system of three equations in four unknowns. This should give you a line of solutions. You need a vector of length one on this line.

28. The vectors spanning the subspace are already orthogonal. You can convert them to an orthonormal set by dividing each by its length. (This is always true when you have mutually orthogonal nonzero vectors. This is all you wind up doing when you go through the Gram-Schmidt process, because all the projections you compute are zero.)

4.2 <
In [4]+[18] and [8]+[22] the work of doing the Gram-Schmidt orthonormalization should also give QR factorization. Q is just made up of the vectors w_i that give the orthonormal basis. r_ij for i < j is v_j . w_i (dot product) which you need to find anyway in computing the next of the w's. The diagonal entries r_kk are the same as the lengths you need no divide by when you get w_kk.
[ You take v_k - (its projection on the space spanned by the preceding w's). To get w_k you divide that difference by its length, and that length is r_kk. ]

34. First find a basis for the kernel. Then convert to an orthonormal basis by the Gram-Schmidt process.

4.3
12. The condition that

a
b
c

be perpendicular to the other two columns gives two linear equations in 3 unknowns. Solving gives a line of solutions. You need

a
b
c

to be a unit vector on this line.

26. Note that if Q has orthonormal columns (as in the QR factorization), then Q^T Q = 1_m where Q is n by m, Q^T is the transpose, and 1_m is the m by m identity matrix. The reason is that the dot product of the i th row of Q^T and the j th column of Q is the same as the dot product of the i th column of Q and the j th column of Q. This will be 1 if i=j (columns have length one) and 0 if i is different from j (distinct columns are orthogonal). In fact Q^T Q = 1_m is necessary and sufficicent for the columns of Q to be an orthonormal set. We discussed this in class mainly for the case where Q is square (then it gives the condition for Q to be an orthogonal matrix), but it is true when m < n as well, with the same explanation.

Take the equation M = QR and multiply both sides on the left by Q^T.