MATH 2321 SMU Trail Theorem Orthogonal Complement & Projections Questions

about orthogonal matrix , orthogonal complement, and orthogonal projections, please with clear writing with steps

MATH2321.2 Special Workshop/Quiz #2
April 3, 2020
Please do your work on lined paper and number both the pages and the
lines. Send them to me via e-mail as best you can by 11:59pm. ITSS says that
they cannot increase the size limit on incoming e-mails so you may have to send in
batches.
I will be available via email for help most of the day.
1
2
A. [8 points] Let W be the subspace of R3 with the orthogonal basis {(−1) , (−2)}. Use
4
−1
the method outlined in the Section “Projection matrices” on page 3-4 of the Lecture notes
for April 2 to find the projection matrix from R3 onto W.
B. [8 points] Let u and v be vectors in an inner product space. Prove that
1
1
〈𝑢, 𝑣〉 = ‖𝑢 + 𝑣‖2 − ‖𝑢 − 𝑣‖2
4
4
C. [8 points] This problem relates directly with the proof of Schur’s Theorem. I am
assuming that you have the theorem in front of you and have read over the commentary
−3 1 −1
in the Notes for March 31 (part 1). I will take the matrix A = (−7 5 −1) and tell
−6 6 −2
you that one of its eigenvalues is 4. You are to find a unitary (orthogonal in this case)
triangulation T for A following the method of the proof.
1. Compute a unit eigenvector u corresponding to 4.
2. Extend to an orthonormal basis for R3 with the first vector in the basis and form the
matrix U.
3. Multiply UTAU to find the 2 by 2 matrix B.
4. Find an eigenvalue for B and repeat the above steps to produce the matrix W
5. From this construct the matrix V, and obtain the product UV.
6. Finally write down the matrix T.
D. [8 points]
A “nilpotent” matrix N has the property that for some positive integer k,
k
N = 0. Using Schur’s theorem show that given a nilpotent matrix N, there exists a
unitary matrix U such that U*NU = T where T is a triangular matrix with 0’s along the
main diagonal.
E. [18 points]
Find the Error
1. Trial Theorem. 1 = 2.
Let u be any vector such that ||𝐮|| = √2. Choose a vector v ≠ u such that
Proof:
u∙v = 2. Then we have:
u∙u = 2
2 u∙u = 4
2 u∙u = u∙u + u∙v
2 u∙u −  u∙v = u∙u + u∙v −  u∙v
2 u∙u −  u∙v = u∙u − u∙v
2u∙(u − v) = u∙(u − v) (observe that u − v ≠ 0)
2u = v
2=1
QED
2. Trial Theorem. 1 = √5.
Let u be any vector such that ||𝐮|| = 1. Choose a vector v ≠ u such that
Proof:
u∙v = 3 and ||𝐯|| = √5. Then we have:
2
||𝐮 − 𝐯|| = (u − v)∙(u − v) = u∙u −  u∙v + v∙v = 1×1 − 2×3 + √5 × √5 = 0
Hence ‖𝐮 − 𝐯‖ = 0 and by the properties of the norm, u − v = 0. But then u = v and thus
1 = ||𝐮|| = ||𝐯|| = √5.
QED
𝑥
3
3. Let’s consider the subspace W of R W = {(𝑦) |𝑥 − 𝑦 − 𝑧 = 0}. This is a
𝑧
1
plane through the origin, so it is a subspace and a normal vector is (−1). It
−1
can also be considered to be the nullspace of the matrix ( 1 −1 −1), and we can
obtain a basis for W in the usual way (the matrix is already in echelon form with
𝑡+𝑠
one pivot and 2 free variables). So, a typical vector in the null space is ( 𝑠 )
𝑡
1
1
and hence a basis for W is {(0) , (1)}.
1
0
1
Now we obtain the “orthogonal decomposition” of v = (2) with respect to W as
3
follows: first project v onto W to obtain vW (the component of v in the subspace
W). Then h = v − vW will be the component of v that is orthogonal to W.
Calculating, we obtain:
1 1
(2)∙(0) 1
vW = 31 11 (0)
(0)∙(0) 1
1 1
+
1 1
(2)∙(1) 1
3 0
1 1 (1)
(1)∙(1) 0
0 0
1
1
7
3
1
= 2 (0) + 2 (1) = 2 (3).
1
0
4
4
1
7
−5
1
1
Then h = v − vW = (2) − 2 (3) = 2 ( 1 )
3
4
2
1
1
Now h is orthogonal to all the vectors in W so in particular h∙ (0) and h∙ (1) should both be 0.
1
0
Compute these dot products and explain what went wrong.