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