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Answer Key for Exam #
- Use elimination on an augmented matrix:
The fourth column has no pivot, so x 4 is a free variable. The corresponding system is x 1 + 3x 4 = 10, x 2 = −2, x 3 − x 4 = − 2
which we solve for the pivot variables: x 1 = 10 − 3 x 4 x 2 = − 2 x 3 = − 2 +x 4 x 4 = x 4
Therefore (^)
x 1 x 2 x 3 x 4
+^ x 4
- We perform the eliminations
A =
= R
A basis for the row space is the pivot rows of R, or of A. A basis for the column space is the pivot columns of A (but not of R). A basis for the nullspace can be found as in problem 1 or by taking the negative of the upper right corner (^) ( 2 3 5 1 − 1 − 2
of R, putting a 3 × 3 identity matrix below it, and taking the three columns of that. So the only basis that requires more work is the left nullspace. To get it we transpose the pivot columns of A and eliminate:
( 1 2 1 1 3 2
Here we can solve the corresponding system, or throw away the 2 × 2 identity on the left, negate the rest, and put a 1 × 1 identity under it. We also have another basis for the column space in the rows of the last matrix above. In conclusion
A row space basis is
and
or
and
A null space basis is
and
and
A column space basis is
(^) and
(^) or
(^) and
A left null space basis is
The factored form of A that displays bases for all four is
A =
- To see which vector to keep we start by computing all the dot products for the three vectors. If
~v 1 =
and^ ~v 2 =
and^ ~v 3 =
then
~v 1 · ~v 2 = 3 + 10 − 1 + 2 = 14, ~v 1 · ~v 3 = 6 + 6 + 4 + 2 = 18, ~v 2 · ~v 3 = 2 + 15 − 4 + 1 = 14, ~v 1 · ~v 1 = 9 + 4 + 1 + 4 = 18, ~v 2 · ~v 2 = 1 + 25 + 1 + 1 = 28, ~v 3 · ~v 3 = 4 + 9 + 16 + 1 = 30
Recall that the projection of ~b onto ~a is
~b · ~a ~a · ~a
~a. If we take ~a = ~v 2 then the ratios will both be 1428 = 12 , so ~v 2
seems like a good one to keep. Then the projection of ~v 1 onto ~v 2 is
~v 1 · ~v 2 ~v 2 · ~v 2
~v 2 =
and therefore
~v 1 =
+^ ~e,
where ~e is the error in the projection. We want to replace ~v 1 by some multiple of ~e. We have
2 ~e =
=^ w~ 1 ,
so we throw away ~v 1 and replace it by w~ 1. Next we do the same thing with ~v 3. The projection of ~v 3 onto ~v 2 is
~v 3 · ~v 2 ~v 2 · ~v 2 ~v 2 =
If we had kept ~v 1 initially and changed the others we would have wound up with the orthonormal basis
and^
and^
and if we had kept ~v 3 at the beginning and changed the others we could have come out with
and^
and^
or with
1 √ 30
and^
and^
- Let A be the matrix with ~v 1 and ~v 2 as columns. Then
AT^ A =
and (^) ( 16 12 12 16
so
P =
and so we find that the projection matrix P onto the subspace S is
P =
We also have that
R = 2P − I =
is the reflection matrix through S. The projection of ~v 3 =
onto S is
P~v 3 =
and the reflection of ~v 3 through S is
R~v 3 =
The projection is the average of the reflection and ~v 3 itself, and this could have been used to avoid one of the last two matrix multiplications.
- If P =
then^ P^ is symmetric and
P 2 =
=^ P.
Any matrix P which satisfies P 2 = P = P T^ is a projection matrix. The trace of P is 201 (1 + 11 + 9 + 19) = 2, so the subspace T that P projects onto is 2-dimensional. Therefore any two rows or columns of P will be a basis for it as long as they are not multiples of each other. We can use the same trick on I − P to get a basis for T ⊥. Since only one of the rows of P is very nice, though, let’s eliminate:
P =
P projects onto its own column space, or its row space since P is symmetric. This gives us a nice basis for T , and a basis for T ⊥^ comes from the null space of P :
A basis for T is
and
and^ A basis for^ T^ ⊥^ is
and
- We would like to find the line y = mx + b which best fits the four points (1, 2), (2, 1), (3, 3) and (4, 2) in the sense of least squares. If the line fit exactly we would have
2 = m + b 1 = 2m + b 3 = 3m + b 2 = 4m + b
or
m b
