# MATH 33AH Lecture Notes - Lecture 6: Orthogonal Matrix, Invertible Matrix, BecquerelExam

by OC2716258

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**preview**shows pages 1-2. to view the full**8 pages of the document.**MATH 33A LECTURE 3

2ND MIDTERM SOLUTIONS

FALL 2016

Copyright c

2016 UC Regents/Dimitri Shlyakhtenko.

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MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 2016 2

Problem 1. (True/False, 1 pt each) Mark your answers by ﬁlling in the appropriate box

next to each question.

(i: T F ) Any basis for R3consists of 3 orthonormal vectors. (False): e.g.

1

0

0

,

1

1

0

,

1

1

1

(ii: T F ) If Ais a 4 ×4 orthogonal matrix, Ae1=e2,Ae2=e3,Ae3=e1, then Ae4=e4or

Ae4=−e4(True): if w=Ae4then wmust be perpendicular to the other columns

of A, i.e., to e2,e3and e1so be a multiple of e4. But since its length must be 1,

w=±e4

(iii: T F ) There are no vectors v,win R7so that v·w=7 and kvk=kwk=2. (True):

The Cauchy-Schwartz inequality states that |v•w| ≤ kvkkwkwhich would mean

7≤2·2 which cannot be true.

(iv: T F )(Atx)•y=x•(Ay)for any vectors x,yand any matrix A(here •denotes the

dot product) (True): if Ahas entries aij,xhas entries xiand yhas entries yithen

Atx•y=∑ij yjaij xi=∑ij aij xiyjwhile x•(Ay) = ∑ij xiaijyj=∑ij xiyjaij.

(v: T F ) A square orthogonal matrix is invertible. (True) Since AtA=I,Acannot have any

kernel, so by rank nullity has full rank, so is invertible.

(vi: T F ) If AtA=Ithen Ais an orthogonal matrix. (True) if Ahas columns v1, . . . , vnthen

the i,j-th entry of AtAis exactly vi•vj; so Ais orthogonal iff vi•vjis zero for i6=j

and 0 if i=ji.e., AtA=I.

(vii: T F ) Every basis for the plane x+y+z=0 consists of two vectors. (True) the dimen-

sion of the plane is 2.

(viii: T F ) The columns of an orthogonal matrix are orthonormal. (True) this is the deﬁnition

of an orthogonal matrix.

(ix: T F ) The determinant of an invertible matrix is nonzero. (True) 1 =det I=det AA−1=

(det A)·(det A−1)so det A6=0.

(x: T F ) For any matrix A, the kernel of Ais perpendicular to the image of At. (True) see

book.

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MATH 33A LECTURE 3 2ND MIDTERM SOLUTIONS FALL 2016 3

Problem 2. (10 pts) Let Pbe the subspace of R4consisting of vectors

x

y

z

w

satisfying

x+y=−wand x+z=−w.

(a) Find a basis for P.

Note that by deﬁnition Pconsists of those vectors which satisfy

x+y+w=0, x+z+w=0

which means that

P=ker A

with

A=1101

1011.

Row-reducing

A∼1 1 0 1

0−110∼1 1 0 1

0 1 −1 0 ∼1 0 1 1

0 1 −1 0

The last two columns correspond to free variables so that we have:

x

y

z

w

=a

−1

1

1

0

+b

−1

0

0

1

so that

−1

1

1

0

and

−1

0

0

1

form a basis. Note: other answers are possible; the two

vectors must lie in Pand be linearly independent.

(b) Find the dimension of P.

Since there are two basis vectors, the dimensino is 2.

(c) Find an orthonormal basis for P.

Using Gram-Schmidt we get:

w1=

−1

1

1

0

,u1=1

√3

−1

1

1

0

w2=v2−(v2•u1)u1=

−1

0

0

1

−1

3

−1

1

1

0

=

−2/3

−1/3

−1/3

1

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