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.