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Answer Key for Exam #
- Use elimination on an augmented matrix:
The corresponding system is x 1 + 6x 4 + 5x 5 = 6, x 2 − 2 x 4 − x 5 = −1, x 3 + 2x 4 = 0
which we solve for the pivot variables x 1 , x 2 and x 3 :
x 1 = 6 − 6 x 4 − 5 x 5 x 2 = − 1 +2x 4 +x 5 x 3 = − 2 x 4 x 4 = x 4 x 5 = x 5
Therefore
~x =
x 1 x 2 x 3 x 4 x 5
- 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 (^) ( − 1 2 4 3
of R, putting a 2 × 2 identity matrix below it, and taking the two 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 2 5 1 1 3 3
Here we can solve the corresponding system, or throw away the 2 × 2 identity on the left, negate the rest, and put a 2 × 2 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
A column space basis is
and
or
and
A left null space basis is
and
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 = 1 + 2 + 1 + 10 = 14, ~v 1 · ~v 3 = 1 + 4 + 1 + 8 = 14, ~v 2 · ~v 3 = 1 + 8 + 1 + 20 = 30, ~v 1 · ~v 1 = 1 + 1 + 1 + 4 = 7, ~v 2 · ~v 2 = 1 + 4 + 1 + 25 = 31, ~v 3 · ~v 3 = 1 + 16 + 1 + 16 = 34
Recall that the projection of ~b onto ~a is
~b · ~a ~a · ~a ~a. If we take ~a = ~v 1 then both ratios will be 147 = 2, so ~v 1
seems like a good one to keep. Then the projection of ~v 2 onto ~v 1 is
~v 2 · ~v 1 ~v 1 · ~v 1
~v 1 = 2~v 1 =
and therefore
~v 2 =
= 2~v 1 +^ ~e^ =
+^ ~e,
where ~e is the error in the projection. We want to replace ~v 2 by some multiple of ~e. We have
~e =
= − w~ 2 , where w~ 2 =
are all perpendicular to each other, and we only have to fix the lengths. The dot product of the first vector with itself is 42, and we did the other two earlier, so we finally get that an orthonormal basis for the subspace of R^4 spanned by ~v 1 , ~v 2 and ~v 3 is
and^
and^
If we had projected w~ 2 onto w~ 3 instead then the answer would have been
and^
and^
There are many other possible answers. A really cheap one, which no one noticed (and I didn’t either until I was almost finished grading them), is
and
and
- Let A be the matrix with ~v 1 and ~v 2 as columns. Then
AT^ A =
and (^) ( 12 8 8 12
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 should be a linear combination of ~v 1 and ~v 2 , so the fact that it came out to ~v 1 exactly, while a little surprising, is consistent. The projection is always 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 for which P 2 = P = P T^ is a projection matrix. The trace of P is 121 (5 + 2 + 5 + 5 + 2 + 5) = 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—in other words, any two of the first three columns or rows should work. But since we also have to find a basis for T ⊥, which must be 4-dimensional, let’s eliminate. The last three rows obviously drop out, and we have
P =
