### FINAL EXAM INFORMATION

作业math | 代做Algorithm – 该题目是一个常规的Algorithm的练习题目代写, 是有一定代表意义的math/Algorithm等代写方向

```
math 207 Section 5 (WWW) Spring 2022
```

#### FINAL EXAM INFORMATION

```
Time:The final exam is open for 72 hours starting Sunday midnight, May 8 and closing at Wednes-
```

```
day midnight, May 11. You have 2 hours 10 minutes alloted to take exam.
```

```
Content:Everything we covered this semester, except for 7.4 (singular value decomposition).
```

```
Details: No external help is allowed. No use of calculators, no use of internet, no use of other
```

```
people, no use of electronic devices, nothing outside your head.
```

#### FINAL EXAM: REVIEW OF TOPICS AND SAMPLE PROBLEMS

- Topic 1:Linear systems of equations. You should be able to:
- Row reduce a matrix to row echelon form (REF) and/or reduced row echelon form (RREF).
- Solve a system of linear equations and write the solution set in parametric vector form.
- Determine whether a given system of linear equations has 0, 1, or infinitely many solutions.

```
Sample problems:Find the solution set of the equationAx=b, and give your answer in para-
```

```
metric form:
```

#### 1)A=

#### 1 2 1 0

#### 0 0 1 1

#### 1 1 0 1

```
, b=
```

#### 1

#### 0

#### 7

#### 2)A=

#### "

#### 1 2 1 1

#### 2 4 1 1

```
, b=
```

#### "

#### 0

#### 0

- Topic 2:Linear independence, spanning, and basis. You should be able:
- Determine if a given sequence of column vectors is linear independent, spansR

```
m
, and/or forms
```

```
a basis forR
```

```
m
.
```

- Determine if a vector lies in the span of some other given vectors.
- Understand that a sequence of exactlynvectors inR

```
n
is linearly independent if and only if it
```

```
spansR
n
, if and only if it forms a basis forR
n
.
```

```
Sample problem:For what value(s) of the parameterhis{v 1 ,v 2 ,v 3 }a basis ofR
```

```
3
?
```

```
v 1 :=
```

#### 1

#### 1

#### 2

```
, v 2 :=
```

#### 1

#### 1

#### 1

```
, v 3 :=
```

```
h
```

#### 3

#### 0

#### .

```
1
```

- Topic 3:Matrices and linear transformations. You should be able to:
- Find the standard matrix of a linear transformation.
- Given a matrix, find bases for its null space, column space, and row space.
- Given a linear transformation, find bases for its range and kernel.
- Determine if a linear transformation is one-to-one and/or onto.
- Compute the rank of a matrix or linear transformation.
- Apply the Rank Theorem (what is also called Rank-Nullity).

```
Sample problems:Consider the linear transformationT:R
```

```
4
R
```

```
3
given by
```

#### T

```
x 1
```

```
x 2
```

```
x 3
```

```
x 4
```

#### =

```
x 1 + 2x 2 + 4x 3 + 2x 4
```

```
2 x 1 + 4x 2 +x 3 3 x 4
```

```
3 x 1 + 6x 2 2 x 3 8 x 4
```

#### .

```
1) Find a basis for the kernel ofT.
```

```
2) Find a basis for the image ofT.
```

```
3) Are there any two different vectorsx=yinR
4
for whichT(x) =T(y)?
```

```
4) Are there any vectorsbinR
```

```
3
for which the equationT(x) =bhas no solution?
```

```
Are the following statements true or false?
```

- There is a 511 matrix with 4 linearly independent columns.
- There is a 47 matrixAfor which dim NulA= 5.
- There is a 47 matrixAfor which dim RowA= 2.
- There is a 512 matrix with rank 7.
- Topic 4:Coordinates. You should be able to:

- Find coordinates for a vector with respect to a given basis.
- Use coordinates to determine if a given sequence of vectors is linearly independent, spans, and/or

```
forms a basis.
```

- Find the matrix of a linear transformation with respect to a given basis or bases.
- Use the matrix of a linear transformation to find the image of a vector, or to say if the linear

```
transformation is one-to-one and/or onto.
```

- Change of coordinates, matrices of linear transformations in different coordinates.

```
a. Find the matrix of change of coordinates from different bases.
```

```
b. Calculate the matrix of a linear transformation in a given basis.
```

```
c. Use the change of coordinates matrix to calculate the matrix of a linear transformation
```

```
in another basis.
```

```
Sample problem 1:Use coordinate vectors to test whether the polynomials spanP 2 :
```

```
1 3 t+ 5t
```

```
2
,3 + 5t 7 t
```

```
2
,4 + 5t 6 t
```

```
2
, 1 t
```

```
2
.
```

```
Sample problem 2:Consider the standard basisE={e 1 ,e 2 ,e 3 }ofR
```

```
3
, where as usual,
```

```
e 1 =
```

#### 1

#### 0

#### 0

```
, e 2 =
```

#### 0

#### 1

#### 0

```
, e 3 =
```

#### 0

#### 0

#### 1

#### .

```
Also consider the basisB={b 1 ,b 2 ,b 3 }, where
```

```
b 1 =
```

#### 2

#### 0

#### 0

```
,b 2 =
```

#### 1

#### 1

#### 0

```
,b 3 =
```

#### 1

#### 1

#### 1

#### .

```
1) Compute the coordinates [x]Eand [x]Bof the vectorx=
```

#### 2

#### 2

#### 2

```
with respect to the basesE
```

```
andB, respectively.
```

```
2) Find the change of coordinates matrixPBEwhich transforms coordinate vectors in the basis
```

```
Eto coordinate vectors in the basisB. For the vectorxabove, verify that [x]B=PBE[x]E.
```

```
3) Consider the linear transformationT :R
```

```
3
R
```

```
3
for whichT(e 1 ) =e 2 ,T(e 2 ) =e 3 and
```

```
T(e 3 ) = 0. Find the matrix [T]ErepresentingTin the standard basisE.
```

```
4) Find the matrix [T]Bthat representsTin the basisB.
```

- Topic 5:Eigenvalues, eigenvectors, diagonalization. You should be able to:
- Find the eigenvalues of a matrix.
- Find bases for the eigenspaces of a matrix.
- Determine if a matrix can or cannot be diagonalized.
- Diagonalize a matrix, where possible.

```
Sample problem:Diagonalize, if possible:
```

#### A=

#### 1 1 1

#### 2 2 2

#### 1 1 1

#### .

- Topic 6:Orthogonality, projections, GramSchmidt process. You should be able to:
- Use dot products to write a vector as a linear combination of orthogonal basis vectors.
- Find the orthogonal projection of a vector in a subspace.
- Find the closest approximation to a vector in a given subspace, and/or find the distance from

```
a vector to a subspace.
```

- Convert a basis for a subspace into an orthogonal or orthonormal basis for that subspace.

```
Sample problems:
```

```
1) Find an orthonormal basis of the column space of the following matrix:
```

#### A=

#### 1 1 0

#### 1 0 1

#### 0 1 1

#### 1 0 1

#### .

```
(Notice that the columns ofAare linearly independent.)
```

```
2) Consider the subspaceWofR
```

```
3
spanned by
```

```
v 1 =
```

#### 1

#### 0

#### 1

```
v 2 =
```

#### 1

#### 1

#### 1

#### .

```
Find the vectorxinWwhich minimizes the distance from the vectorb=
```

#### 1

#### 2

#### 3

```
toW, and
```

```
compute the distance frombtoW.
```

- Topic 7:Least squares solutions of linear systems, linear regression.
- Find the least-squares solution to a linear system of equations.
- Find the equation of the least-squares line that best fits a collection of points inR

```
2
.
```

```
Sample problem:Find the equationy= 0 + 1 xof the least-squares line that best fits the data:
```

#### ( 1 ,0),(0,1),(1,2),(2,4).

- Topic 8:Orthogonal diagonalization of a symmetric matrix. You should be able to:
- Understand that a matrix is symmetric if and only if it is orthogonally diagonalizable, if and

```
only if there is an orthonormal basis consisting of eigenvectors for that matrix
```

- Given a symmetric matrixA, find an orthogonal matrixPand a diagonal matrixDfor which

#### A=PDP

```
.
```

```
Sample problem:Find an orthogonal matrixPand a diagonal matrixDfor whichA=PDP
```

```
:
```

#### A=

#### 3 1 5

#### 1 1 1

#### 5 1 3

#### .

- Topic 9:Quadratic forms and constrained optimization. You should be able to:
- Find the symmetric matrix of a quadratic form.
- Find an orthogonal change of variables that eliminates cross-terms in a quadratic form.
- Determine if a given quadratic form is positive definite, positive semidefinite, negative definite,

```
negative semidefinite, or indefinite.
```

- Given a quadratic formQ, use linear algebra to find the greatest and least values ofQ(x) subject

```
to the constraintx= 1.
```

- Given a quadratic formQ, use linear algebra to find a unit vectorxthat maximizes or minimizes

```
Q(x) on the unit sphere.
```

```
Sample problem:Consider the quadratic formQ(x) = 4x
```

```
2
1 ^4 x^1 x^2 + 7x
```

```
2
2.
```

```
1) Find a unit vectorxthat maximizes the value ofQ(x) subject to the constraintx= 1.
```

```
2) ClassifyQas positive definite, positive semidefinite, negative definite, negative semidefinite, or
```

```
indefinite.
```

#### SOLUTIONS

- Topic 1:Find the solution set of the equationAx=b, and give your answer in parametric form:

#### 1)A=

#### 1 2 1 0

#### 0 0 1 1

#### 1 1 0 1

```
, b=
```

#### 1

#### 0

#### 7

#### 2)A=

#### "

#### 1 2 1 1

#### 2 4 1 1

```
, b=
```

#### "

#### 0

#### 0

#### SOLUTION:

```
1) Row reduce the augmented matrix:
```

```
```

#### 1 2 1 0 1

#### 0 0 1 1 0

#### 1 1 0 1 7

```
R 3 =R 3 R 1
```

#### 1 2 1 0 1

#### 0 0 1 1 0

#### 0 1 1 1 6

```
R 2 R 3
```

#### 1 2 1 0 1

#### 0 1 1 1 6

#### 0 0 1 1 0

```
R 2 =R 2 +R 3
```

#### 1 2 1 0 1

#### 0 1 0 2 6

#### 0 0 1 1 0

```
R 1 =R 1 R 3
```

#### 1 2 0 1 1

#### 0 1 0 2 6

#### 0 0 1 1 0

```
R 1 =R 1 +2R 2
```

#### 1 0 0 3 13

#### 0 1 0 2 6

#### 0 0 1 1 0

```
R 2 =R 2
```

#### 1 0 0 3 13

#### 0 1 0 2 6

#### 0 0 1 1 0

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
```

```
x 1 + 3x 4 = 13
```

```
x 2 2 x 4 = 6
```

```
x 3 + x 4 = 0
```

```
x 4 = free
```

```
x 1 = 13 3 x 4
```

```
x 2 =6 + 2x 4
```

```
x 3 =x 4
```

```
x 4 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

```
x 4
```

#### =

```
13 3 x 4
```

```
6 + 2x 4
```

```
x 4
```

```
x 4
```

#### =

#### 13

#### 6

#### 0

#### 0

```
+x 4
```

#### 3

#### 2

#### 1

#### 1

```
, x 4 free.
```

```
2) Row reduce the augmented matrix:
```

```
"
```

```
1 2 1 1 0
```

#### 2 4 1 1 0

```
R 2 =R 2 2 R 1
```

#### "

#### 1 2 1 1 0

#### 0 0 1 3 0

```
R 1 =R 1 +R 2
```

#### "

#### 1 2 0 2 0

#### 0 0 1 3 0

```
R 2 =R 2
```

#### "

#### 1 2 0 2 0

#### 0 0 1 3 0

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
x 1 + 2x 2 + 2x 4 = 0
```

```
x 2 = free
```

```
x 3 3 x 4 = 0
```

```
x 4 = free
```

```
x 1 = 2 x 2 2 x 4
```

```
x 2 = free
```

```
x 3 = 3x 4
```

```
x 4 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

```
x 4
```

#### =

```
2 x 2 2 x 4
```

```
x 2
```

```
3 x 4
```

```
x 4
```

```
=x 2
```

#### 2

#### 1

#### 0

#### 0

```
+x 4
```

#### 2

#### 0

#### 3

#### 1

```
, x 2 andx 4 free.
```

- Topic 2:For what value(s) of the parameterhis{v 1 ,v 2 ,v 3 }a basis ofR

```
3
?
```

```
v 1 :=
```

#### 1

#### 1

#### 2

```
, v 2 :=
```

#### 1

#### 1

#### 1

```
, v 3 :=
```

```
h
```

#### 3

#### 0

#### .

#### SOLUTION:

```
Row the reduce the matrixA=
```

```
h
```

```
v 1 v 2 v 3
```

```
i
```

```
to REF:
```

```
1 1 h
```

#### 1 1 3

#### 2 1 0

```
R 2 =R 2 R 1
```

```
1 1 h
```

```
0 0 3h
```

#### 2 1 0

```
R 3 =R 3 +2R 1
```

```
1 1 h
```

```
0 0 3h
```

```
0 3 2 h
```

```
R 2 R 3
```

```
1 1 h
```

```
0 3 2 h
```

```
0 0 3h
```

```
The vectors{v 1 ,v 2 ,v 3 }form a basis forR
```

```
3
if and only if the square matrixA=
```

```
h
```

```
v 1 v 2 v 3
```

```
i
```

```
has pivots all down its diagonal. That happens if and only ifh= 3.
```

- Topic 3:Consider the linear transformationT:R

```
4
R
```

```
3
given by
```

#### T

```
x 1
```

```
x 2
```

```
x 3
```

```
x 4
```

#### =

```
x 1 + 2x 2 + 4x 3 + 2x 4
```

```
2 x 1 + 4x 2 +x 3 3 x 4
```

```
3 x 1 + 6x 2 2 x 3 8 x 4
```

#### .

```
1) Find a basis for the kernel ofT.
```

```
2) Find a basis for the image ofT.
```

```
3) Are there any two different vectorsx=yinR
4
for whichT(x) =T(y)?
```

```
4) Are there any vectorsbinR
```

```
3
for which the equationT(x) =bhas no solution?
```

#### SOLUTION:

```
1) Find the standard matrix forT. Denotinge 1 ,...,e 4 for the standard basis vectors ofR
```

```
4
, the
```

```
standard matrix is
```

#### A=

```
h
```

```
T(e 1 ) T(e 2 ) T(e 3 ) T(e 4 )
```

```
i
```

```
=
```

#### 1 2 4 2

#### 2 4 1 3

#### 3 6 2 8

#### .

```
The kernel ofTis the same as the null space ofA, so we need to find a basis for NulA. Row
```

```
reduce:
```

```
h
```

```
A 0
```

```
i
```

```
=
```

#### 1 2 4 2 0

#### 2 4 1 3 0

#### 3 6 2 8 0

```
R 2 =R 2 2 R 1
```

#### 1 2 4 2 0

#### 0 0 7 7 0

#### 3 6 2 8 0

R 3 =R 3 3 R 1

#### 1 2 4 2 0

#### 0 0 7 7 0

#### 0 0 14 14 0

```
R 3 =R 3 2 R 2
```

#### 1 2 4 2 0

#### 0 0 7 7 0

#### 0 0 0 0 0

```
R 2 =
```

```
1
7
R 2
```

#### 1 2 4 2 0

#### 0 0 1 1 0

#### 0 0 0 0 0

```
R 1 =R 1 4 R 2
```

#### 1 2 0 2 0

#### 0 0 1 1 0

#### 0 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
```

```
x 1 + 2x 2 2 x 4 = 0
```

```
x 2 = free
```

```
x 3 + x 4 = 0
```

```
x 4 = free
```

```
x 1 = 2 x 2 + 2x 4
```

```
x 2 = free
```

```
x 3 =x 4
```

```
x 4 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

```
x 4
```

#### =

```
2 x 2 + 2x 4
```

```
x 2
```

```
x 4
```

```
x 4
```

```
=x 2
```

#### 2

#### 1

#### 0

#### 0

```
+x 4
```

#### 2

#### 0

#### 1

#### 1

```
, x 2 andx 4 free.
```

```
The kernel is two-dimensional, and a basis is given by
```

```
```

#### 2

#### 1

#### 0

#### 0

#### ,

#### 2

#### 0

#### 1

#### 1

#### .

```
2) The image ofTis the same as the column space of its standard matrixA, so we just need to
```

```
find a basis for ColA. One basis is given by the pivot columns ofA. From our work above, we
```

```
know the first and third columns contain pivots. Our answer is
```

```
```

#### 1

#### 2

#### 3

#### ,

#### 4

#### 1

#### 2

#### .

```
3) This is just asking ifTis one-to-one. The kernel ofTis bigger than{ 0 }, soTis not one-to-one.
```

```
The answer is yes, there are different vectorsx=yinR
```

```
4
for whichT(x) =T(y).
```

```
4) This is just asking ifTis onto. The image ofTis two-dimensional, so it is not all ofR
```

```
3
```

. The

```
answer is yes, there are vectorsbinR
```

```
3
for which the equationT(x) =bhas no solution.
```

```
True / False questions: See class notes.
```

- Topic 4:Use coordinate vectors to test whether the polynomials spanP 2 :

```
1 3 t+ 5t
```

```
2
,3 + 5t 7 t
```

```
2
,4 + 5t 6 t
```

```
2
, 1 t
```

```
2
.
```

#### SOLUTION:

```
Find coordinate vectors for the polynomials with respect to the standard basisB={ 1 ,t,t
```

```
2
}.
```

```
(You could use a different basis if you want, but that would probably be harder.) Since we are using
```

```
the standard basis, the coordinates are just the coefficients:
```

```
h
```

```
1 3 t+ 5t
2
c
```

```
i
```

```
B
```

#### =

#### 1

#### 3

#### 5

#### ,

```
h
```

```
3 + 5t 7 t
2
c
```

```
i
```

```
B
```

#### =

#### 3

#### 5

#### 7

#### ,

```
h
```

```
4 + 5t 6 t
```

```
2
c
```

```
i
```

```
B
```

#### =

#### 4

#### 5

#### 6

#### ,

```
h
```

```
1 t
```

```
2
c
```

```
i
```

```
B
```

#### =

#### 1

#### 0

#### 1

#### .

```
Instead of asking whether the polynomials spanP 2 , we can just ask if their coordinates spanR
```

```
3
.
```

```
Put the columns in a matrix and row reduce to echelon form:
```

#### 1 3 4 1

#### 3 5 5 0

#### 5 7 6 1

```
R 2 =R 2 +3R 1
```

#### 1 3 4 1

#### 0 4 7 3

#### 5 7 6 1

```
R 3 =R 3 5 R 1
```

#### 1 3 4 1

#### 0 4 7 3

#### 0 8 14 6

```
R 3 =R 3 +2R 2
```

#### 1 3 4 1

#### 0 4 7 3

#### 0 0 0 0

#### .

```
There is not a pivot in every row, so the coordinate vectors do not spanR
3
```

. That means the answer

```
is no, these polynomials do not spanP 2.
```

- Topic 5:Diagonalize, if possible:

#### A=

#### 1 1 1

#### 2 2 2

#### 1 1 1

#### .

#### SOLUTION:

```
Our goal is to find a basis forR
3
consisting of eigenvectors forA.
```

```
The first step is to find the eigenvalues. Compute the characteristic polynomial and set it equal
```

```
to zero:
```

det(AI) =

```
1 1 1
```

```
2 2 2
```

```
1 1 1
```

```
= (1)
```

```
2 2
```

```
1 1
```

#### 2 2

```
1 1
```

#### +

```
2 2
```

#### 1 1

```
= (1)
```

```
h
```

```
(2)( 1 )2(1)
```

```
i
```

```
```

```
h
```

```
2( 1 )2(1)
```

```
i
```

```
+
```

```
h
```

```
2(1)(2)(1)
```

```
i
```

```
= (1)(
```

```
2
)( 2 ) + ()
```

```
=
```

```
3
+ 2
```

```
2
```

```
=
```

```
2
(2).
```

```
The eigenvalues are= 0 (with algebraic multiplicity 2) and= 2 (with algebraic multiplicity 1).
```

```
The next step is to find bases for each of the eigenspaces. We start with the 0-eigenspace,
```

```
Nul(A 0 I) = Nul(A). It consists of solutions to the homogeneous equationAx= 0 , so we row
```

```
reduce an augmented matrix:
```

```
h
```

```
A 0
```

```
i
```

```
=
```

#### 1 1 1 0

#### 2 2 2 0

#### 1 1 1 0

```
R 2 =R 2 2 R 1
```

#### 1 1 1 0

#### 0 0 0 0

#### 1 1 1 0

```
R 3 =R 3 +R 1
```

#### 1 1 1 0

#### 0 0 0 0

#### 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
x 1 +x 2 +x 3 = 0, x 1 =x 2 x 3.
```

```
Writexas a vector in parametric vector form, using the free variables as parameters:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

#### =

```
x 2 x 3
```

```
x 2
```

```
x 3
```

```
=x 2
```

#### 1

#### 1

#### 0

```
+x 3
```

#### 1

#### 0

#### 1

A basis for the 0-eigenspace is given by:

```
```

#### 1

#### 1

#### 0

#### ,

#### 1

#### 0

#### 1

#### .

```
Repeat to find a basis for the 2-eigenspace, Nul(A 2 I). It consists of solutions to the homogeneous
```

equation (A 2 I)x= 0 , so we row reduce an augmented matrix:

```
h
```

```
A 2 I 0
```

```
i
```

```
=
```

#### 1 1 1 0

#### 2 0 2 0

#### 1 1 3 0

```
R 2 =R 2 +2R 1
```

#### 1 1 1 0

#### 0 2 4 0

#### 1 1 3 0

```
R 3 =R 3 R 1
```

#### 1 1 1 0

#### 0 2 4 0

#### 0 2 4 0

```
R 3 =R 3 +R 2
```

#### 1 1 1 0

#### 0 2 4 0

#### 0 0 0 0

```
R 2 =
```

```
1
2
R 2
```

#### 1 1 1 0

#### 0 1 2 0

#### 0 0 0 0

```
R 1 =R 1 R 2
```

#### 1 0 1 0

#### 0 1 2 0

#### 0 0 0 0

```
R 1 =R 1
```

#### 1 0 1 0

#### 0 1 2 0

#### 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
(
x 1 + x 3 = 0
```

```
x 2 + 2x 3 = 0
```

#### (

```
x 1 =x 3
```

```
x 2 = 2 x 3
```

```
Writexas a vector in parametric vector form, using the free variables as parameters:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

#### =

```
x 3
```

```
2 x 3
```

```
x 3
```

```
=x 3
```

#### 1

#### 2

#### 1

#### .

A basis for the 2-eigenspace is given by

```
```

#### 1

#### 2

#### 1

#### .

```
Finally, put all the eigenspace bases together and see if we have as many vectors as the size ofA.
```

In this case, 2 + 1 = 3, and we have a basis of eigenvectors. To writeA=PDP

```
1
withPinvertible
```

andDdiagonal, use the basis of eigenvectors for the columns ofP, and put their corresponding

eigenvalues on the diagonal ofD:

#### P=

#### 1 1 1

#### 1 0 2

#### 0 1 1

#### , D=

#### 0 0 0

#### 0 0 0

#### 0 0 2

#### .

- Topic 6:

```
1) Find an orthonormal basis of the column space of the following matrix:
```

#### A=

#### 1 1 0

#### 1 0 1

#### 0 1 1

#### 1 0 1

#### .

```
(Notice that the columns ofAare linearly independent.)
```

```
2) Consider the subspaceWofR
3
spanned by
```

```
v 1 =
```

#### 1

#### 0

#### 1

```
v 2 =
```

#### 1

#### 1

#### 1

#### .

```
Find the vectorxinWwhich minimizes the distance from the vectorb=
```

#### 1

#### 2

#### 3

```
toW, and
```

```
compute the distance frombtoW.
```

#### SOLUTION:

```
1) Letv 1 ,v 2 ,v 3 be the columns ofA:
```

```
v 1 =
```

#### 1

#### 1

#### 0

#### 1

```
,v 2 =
```

#### 1

#### 0

#### 1

#### 0

```
,v 3 =
```

#### 0

#### 1

#### 1

#### 1

#### .

```
We want an orthonormal basis forW := ColA, and a regular (not even orthogonal) basis is
```

```
given byv 1 ,v 2 ,v 3. Use the GramSchmidt Algorithm to find an orthogonal basis. In every
```

```
step, we find an orthogonal projection and then rip it out.
```

```
First set
```

```
u 1 =v 1 =
```

#### 1

#### 1

#### 0

#### 1

#### .

```
Letp 2 be the orthogonal projection ofv 2 onto span{u 1 }:
```

```
p 2 =
```

```
v 2 u 1
```

```
u 1 u 1
```

```
u 1 =
```

#### 1

#### 3

```
u 1 =
```

```
1
3
```

```
```

```
1
3
```

```
0
```

```
1
3
```

#### .

Then define

```
u 2 :=v 2 p 2 =
```

#### 1

#### 0

#### 1

#### 0

```
1
3
```

```
```

```
1
3
```

```
0
```

```
1
3
```

#### =

#### 2 3 1 3 1

```
1
3
```

#### .

Nowu 1 andu 2 form an orthogonal basis for span{v 1 ,v 2 }. We can get a nicer basis by rescaling

u 2 to clear the fractions:

```
u
```

```
2 := 3u^2 =
```

#### 2

#### 1

#### 3

#### 1

#### .

Nowu 1 andu 2 also form an orthogonal basis for span{v 1 ,v 2 }.

Go again. Letp 3 be the orthogonal projection ofv 3 onto span{v 1 ,v 2 }:

```
p 3 =
```

```
v 3 u 1
```

```
u 1 u 1
```

```
u 1 +
```

```
v 3 u
2
```

```
u
```

```
2
u
```

```
2
```

```
u
```

```
2 =
```

#### 0

#### 3

```
u 1 +
```

#### 3

#### 15

```
u
```

```
2 =
```

#### 1

#### 5

#### 2

#### 1

#### 3

#### 1

#### .

Then define

```
u 3 :=v 3 p 3 =
```

#### 0

#### 1

#### 1

#### 1

#### 1

#### 5

#### 2

#### 1

#### 3

#### 1

#### =

#### 1

#### 5

#### 0

#### 5

#### 5

#### 5

#### 2

#### 1

#### 3

#### 1

#### =

#### 1

#### 5

#### 2

#### 4

#### 2

#### 6

#### .

Nowu 1 ,u 2 , andu 3 form an orthogonal basis forW = ColA. We can get a nicer-looking

orthogonal basis by rescalingu 3 to clear the fractions:

```
u
```

```
3 :=
```

#### 5

#### 2

```
u 3 =
```

#### 1

#### 2

#### 1

#### 3

#### .

Finally, normalize the orthogonal basis to get an orthonormal basis

```
n
1
u 1
u 1 ,
```

```
1
u 2
u
```

```
2 ,
```

```
1
u 3
u
```

```
3
```

```
o
```

```
.
```

We have

```
u 1 =
```

```
3 , u
```

```
2 =
```

```
15 , u
```

```
3 =
```

#### 15 ,

so our answer is

```
```

#### 1

#### 3

#### 1

#### 1

#### 0

#### 1

#### ,

#### 1

#### 15

#### 2

#### 1

#### 3

#### 1

#### ,

#### 1

#### 15

#### 1

#### 2

#### 1

#### 3

#### .

- The closest vector tobinWis given by its orthogonal projection

```
b. In order to find that, we
```

```
first need an orthogonal basis forW. We can see thatv 1 andv 2 are linearly independent, so
```

```
they form a basis forW. Use the GramSchmidt algorithm. Start by setting
```

```
u 1 :=v 1 =
```

#### 1

#### 0

#### 1

#### .

```
Letp 2 be the orthogonal projection ofv 2 onto span{u 1 }:
```

```
p 2 =
```

```
v 2 u 1
```

```
u 1 u 1
```

```
u 1 =
```

#### 2

#### 2

```
u 1 =
```

#### 1

#### 0

#### 1

#### .

```
Then define
```

```
u 2 =v 2 p 2 =
```

#### 1

#### 1

#### 1

#### 1

#### 0

#### 1

#### =

#### 0

#### 1

#### 0

#### .

```
Nowu 1 andu 2 form an orthogonal basis forW, and we can use them to find our projection:
```

```
b=
```

```
bu 1
```

```
u 1 u 1
```

```
u 1 +
```

```
bu 2
```

```
u 2 u 2
```

```
u 2 =
```

#### 2

#### 2

```
u 1 +
```

#### 2

#### 1

```
u 2 =
```

#### 1

#### 0

#### 1

#### + 2

#### 0

#### 1

#### 0

#### =

#### 1

#### 2

#### 1

#### .

```
This is the closest approximation tobinW.
```

```
Then the distance frombtoWis
b
b
```

. We have

```
bb=
```

#### 1

#### 2

#### 3

#### 1

#### 2

#### 1

#### =

#### 2

#### 0

#### 2

#### ,

```
so the distance is
```

```
b
b
=
```

```
p
2
2
+ 2
2
= 2
```

#### 2.

- Topic 7:Find the equationy= 0 + 1 xof the least-squares line that best fits the data:

#### ( 1 ,0),(0,1),(1,2),(2,4).

#### SOLUTION:

```
We wish we could find 0 and 1 so that all of our data points lie on the liney= 0 + 1 x. In
```

```
other words, we wish we could solve the following system of equations:
```

```
0 + 1 (1) = 0
```

```
0 + 1 (0) = 1
```

```
0 + 1 (1) = 2
```

```
0 + 1 (2) = 4.
```

```
The system is inconsistent because there is no line that passes through all four data points. Short
```

```
of solving the system, we find a least-squares solution. The corresponding matrix equation is
```

```
1 1
```

#### 1 0

#### 1 1

#### 1 2

#### "

```
0
```

```
1
```

#### =

#### 0

#### 1

#### 2

#### 4

#### ,

```
orX=y, where
```

#### X=

#### 1 1

#### 1 0

#### 1 1

#### 1 2

```
, y=
```

#### 0

#### 1

#### 2

#### 4

#### .

```
Solve the normal equationsX
```

```
X=X
```

```
y. The coefficient matrix is
```

#### X

```
X=
```

#### "

#### 1 1 1 1

#### 1 0 1 2

#### 1 1

#### 1 0

#### 1 1

#### 1 2

#### =

#### "

#### 4 2

#### 2 6

#### ,

```
and the augmented column is
```

#### X

```
y=
```

#### "

#### 1 1 1 1

#### 1 0 1 2

#### 0

#### 1

#### 2

#### 4

#### =

#### "

#### 7

#### 10

#### .

```
Row reduce the augmented matrix:
```

```
h
```

```
X
```

```
X X
```

```
y
```

```
i
```

```
=
```

#### "

#### 4 2 7

#### 2 6 10

```
R 1 R 2
```

#### "

#### 2 6 10

#### 4 2 7

```
R 1 =
```

```
1
2
R 1
```

#### "

#### 1 3 5

#### 4 2 7

```
R 2 =R 2 4 R 1
```

#### "

#### 1 3 5

#### 0 10 13

```
R 2 =
```

```
1
10
R 2
```

#### "

#### 1 3 5

#### 0 1

```
13
10
```

```
R 1 =R 1 3 R 2
```

#### "

#### 1 0

```
11
10
```

```
0 1
```

```
13
10
```

#### .

```
The least-squares solution is 0 =
```

```
11
10
, 1 =
```

```
13
10
, so the least-squares line of best fit isy=
```

```
11
10
```

#### +

```
13
10
x.
```

- Topic 8:Find an orthogonal matrixPand a diagonal matrixDfor whichA=PDP :

#### A=

#### 3 1 5

#### 1 1 1

#### 5 1 3

#### .

#### SOLUTION:

```
First, find the eigenvalues ofAby computing its characteristic polynomial and setting it equal to
```

```
zero:
```

```
det(AI) =
```

```
3 1 5
```

```
1 1 1
```

```
5 1 3
```

#### .

```
We do a cofactor expansion across the top:
```

```
det(AI) = ( 3 )
```

```
1 1
```

```
1 3
```

#### 1 1

```
5 3
```

#### + 5

```
1 1
```

#### 5 1

```
= ( 3 )
```

```
h
```

```
(1)( 3 )(1)(1)
```

```
i
```

```
```

```
h
```

```
(1)( 3 )(1)(5)
```

```
i
```

```
+ 5
```

```
h
```

```
(1)(1)(1)(5)
```

```
i
```

```
= ( 3 )(
```

```
2
+ 24)( 8 ) + 5(4 + 5)
```

```
=
```

```
3
5
```

```
2
+ 24
```

```
=(
```

```
2
+ 524)
```

```
=(+ 8)(3).
```

```
The eigenvalues are 1 = 3, 2 = 0, and 3 =8.
```

```
Next, find an orthonormal basis for each eigenspace. We start with the eigenvalue 1 = 3. We
```

```
want to solve the equationAx= 3x, or (A 3 I)x= 0. Row reduce the augmented matrix:
```

h

```
A 3 0
```

```
i
```

```
=
```

#### 6 1 5 0

#### 1 2 1 0

#### 5 1 6 0

```
R 1 R 2
```

#### 1 2 1 0

#### 6 1 5 0

#### 5 1 6 0

```
R 2 =R 2 +6R 1
```

#### 1 2 1 0

#### 0 11 11 0

#### 5 1 6 0

```
R 3 =R 3 5 R 1
```

#### 1 2 1 0

#### 0 11 11 0

#### 0 11 11 0

```
R 3 =R 3 +R 2
```

#### 1 2 1 0

#### 0 11 11 0

#### 0 0 0 0

```
R 2 =
```

```
1
11
```

```
R 2
```

#### 1 2 1 0

#### 0 1 1 0

#### 0 0 0 0

```
R 1 =R 1 +2R 2
```

#### 1 0 1 0

#### 0 1 1 0

#### 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
```

```
x 1 x 3 = 0
```

```
x 2 x 3 = 0
```

```
x 3 = free
```

```
x 1 =x 3
```

```
x 2 =x 3
```

```
x 3 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

#### =

```
x 3
```

```
x 3
```

```
x 3
```

```
=x 3
```

#### 1

#### 1

#### 1

```
, x 3 free.
```

```
The 3-eigenspace is one-dimensional, and spanned by
```

#### 1

#### 1

#### 1

. Normalize to makeu 1 =

```
1
3
```

#### 1

#### 1

#### 1

#### .

```
Repeat for the 0-eigenspace:
```

```
h
```

```
A 0 0
```

```
i
```

```
=
```

#### 3 1 5 0

#### 1 1 1 0

#### 5 1 3 0

```
R 1 R 2
```

#### 1 1 1 0

#### 3 1 5 0

#### 5 1 3 0

```
R 2 =R 2 +3R 1
```

#### 1 1 1 0

#### 0 4 8 0

#### 5 1 3 0

R 3 =R 3 5 R 1

#### 1 1 1 0

#### 0 4 8 0

#### 0 4 8 0

```
R 3 =R 3 +R 2
```

#### 1 1 1 0

#### 0 4 8 0

#### 0 0 0 0

```
R 2 =
```

```
1
4
```

```
R 2
```

#### 1 1 1 0

#### 0 1 2 0

#### 0 0 0 0

```
R 1 =R 1 R 2
```

#### 1 0 1 0

#### 0 1 2 0

#### 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
```

```
x 1 x 3 = 0
```

```
x 2 + 2x 3 = 0
```

```
x 3 = free
```

```
x 1 =x 3
```

```
x 2 = 2 x 3
```

```
x 3 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

#### =

```
x 3
```

```
2 x 3
```

```
x 3
```

```
=x 3
```

#### 1

#### 2

#### 1

```
, x 3 free.
```

```
The 0-eigenspace is one-dimensional, and spanned by
```

#### 1

#### 2

#### 1

. Normalize to makeu 2 =

```
1
6
```

#### 1

#### 2

#### 1

#### .

```
Repeat for the8-eigenspace:
```

```
h
```

```
A+ 8 0
```

```
i
```

```
=
```

#### 5 1 5 0

#### 1 9 1 0

#### 5 1 5 0

```
R 3 =R 3 R 1
```

#### 5 1 5 0

#### 1 9 1 0

#### 0 0 0 0

```
R 1 R 2
```

#### 1 9 1 0

#### 5 1 5 0

#### 0 0 0 0

```
R 2 =R 2 5 R 1
```

#### 1 9 1 0

#### 0 44 0 0

#### 0 0 0 0

```
R 2 =
```

```
1
44
```

```
R 2
```

#### 1 9 1 0

#### 0 1 0 0

#### 0 0 0 0

```
R 1 =R 1 9 R 2
```

#### 1 0 1 0

#### 0 1 0 0

#### 0 0 0 0

#### .

```
Reinterpret as equations, and solve for basic variables in terms of free variables:
```

```
```

```
x 1 +x 3 = 0
```

```
x 2 = 0
```

```
x 3 = free
```

```
x 1 =x 3
```

```
x 2 = 0
```

```
x 3 = free
```

```
Writexas a vector in parametric form, using the free variable as a parameter:
```

```
x=
```

```
x 1
```

```
x 2
```

```
x 3
```

#### =

```
x 3
```

#### 0

```
x 3
```

```
=x 3
```

#### 1

#### 0

#### 1

```
, x 3 free.
```

```
The8-eigenspace is one-dimensional, and spanned by
```

#### 1

#### 0

#### 1

. Normalize to makeu 3 =

```
1
2
```

#### 1

#### 0

#### 1

#### .

```
Finally, put all the orthonormal bases for the different eigenspaces together to make an orthogonal
```

```
matrix:
```

#### P=

```
h
```

```
u 1 u 2 u 3
```

```
i
```

```
=
```

```
1
3
```

```
1
6
```

```
1
2
1
3
```

```
2
6
```

#### 0

```
^1
3
```

```
^1
6
```

```
^1
2
```

#### .

```
The diagonal matrix has the corresponding eigenvalues in the corresponding places:
```

#### D=

```
1 0 0
```

```
0 2 0
```

```
0 0 3
```

#### =

#### 3 0 0

#### 0 0 0

#### 0 0 8

#### .

- Topic 9:Consider the quadratic formQ(x) = 4x

```
2
1 ^4 x^1 x^2 + 7x
```

```
2
2.
```

```
1) Find a unit vectorxthat maximizes the value ofQ(x) subject to the constraintx= 1.
```

```
2) ClassifyQas positive definite, positive semidefinite, negative definite, negative semidefinite, or
```

```
indefinite.
```

#### SOLUTION:

```
Find the matrix of the quadratic form. We haveQ(x) =x
Axfor
```

#### A=

#### "

#### 4 2

#### 2 7

#### .

```
Find the eigenvalues ofA:
```

0 = det(AI) =

```
4 2
```

```
2 7
```

```
= (4)(7)(2)(2) =
```

```
2
11 + 24 = (8)(3).
```

```
The eigenvalues (in order from greatest to least) are 1 = 8 and 2 = 3. The top eigenvalue gives the
```

```
maximum value ofQ(x) subject to the constraintx= 1. Furthermore, the maximum is achieved
```

```
at a corresponding eigenvector. Find a unit vectorxthat satisfiesAx= 8x, i.e., (A 8 I)x= 0:
```

```
h
```

```
A 8 0
```

```
i
```

```
=
```

#### "

#### 4 2 0

#### 2 1 0

```
R 1 R 2
```

#### "

#### 2 1 0

#### 4 2 0

```
R 2 =R 2 2 R 1
```

#### "

#### 2 1 0

#### 0 0 0

```
R 1 =
```

```
1
2
R 1
```

#### "

#### 1

```
1
2
```

#### 0

#### 0 0 0

#### .

```
This saysx 1 +
```

```
1
2
x 2 = 0, orx 1 =
```

```
1
2
x 2. That is,
```

```
x=
```

#### "

```
x 1
```

```
x 2
```

#### =

#### "

```
1
2
```

```
x 2
```

```
x 2
```

#### =

```
1
2
```

```
x 2
```

#### "

#### 1

#### 2

#### .

```
One eigenvector is
```

#### "

#### 1

#### 2

. Normalize to make it

```
x=
^1
5
```

#### "

#### 1

#### 2

#### .

```
This is our answer. (You can check thatQ(x) = 8.)
```

```
2) SinceAhas only positive eigenvalues,Qis positive definite. (It is also positive semidefinite.)
```

Version: April 30, 2022