Mark each statement True or False. Justify each answer. In each part, A represents an n × n matrix.
- a. If A is orthogonally diagonalizable, then A is symmetric.
- b. If A is an orthogonal matrix, then A is symmetric.
- c. If A is an orthogonal matrix, then ||Ax|| = ||x|| for all x in ℝn.
- d. The principal axes of a quadratic form xTAx can be the columns of any matrix P that diagonalizes A.
- e. If P is an n × n matrix with orthogonal columns, then PT = P−l.
- f. If every coefficient in a quadratic form is positive, then the quadratic form is positive definite.
- g. If xTAx > 0 for some x, then the quadratic form xTAx is positive definite.
- h. By a suitable change of variable, any quadratic form can be changed into one with no cross-product term.
- i. The largest value of a quadratic form xTAx, for ||x|| = 1, is the largest entry on the diagonal of A.
- j. The maximum value of a positive definite quadratic form xTAx is the greatest eigenvalue of A.
- k. A positive definite quadratic form can be changed into a negative definite form by a suitable change of variable x = Pu, for some orthogonal matrix P.
- l. An indefinite quadratic form is one whose eigenvalues are not definite.
- m. If P is an n × n orthogonal matrix, then the change of variable x = Pu transforms xTAx into a quadratic form whose matrix is P−1AP.
- n. If U is m × n with orthogonal columns, then UUT x is the orthogonal projection of x onto Col U.
- ○. If B is m × n and x is a unit
vector in ℝn , then ||Bx|| ≤ σ1, where σ1 is the first singular value of B. - p. A singular value decomposition of an m × n matrix B can be written as B = PΣQ, where P is an m × n orthogonal matrix, Q is an n × n orthogonal matrix, and Σ is an m × n “diagonal” matrix.
- q. If A is n × n, then A and ATA have the same singular values.
a)

To mark:
The given statement “If A is orthogonally diagonalizable, then A is symmetric” is true or false.
Answer to Problem 1SE
a.
The given statement is true.
Explanation of Solution
Justification of statement istrue:
Theorem 2:
“An n×n matrix A is orthogonally diagonalizable if and only if A is a symmetric
matrix”
Refer Theorem 2.
The given statement “If A is orthogonally diagonalizable, then A is symmetric” is true.
Proof:
Consider A n×n matrix.
The matrix A is orthogonally diagonalizable if
A=PDPT=PDP−1
Here, P is the orthogonal matrix such that P−1=PT and D is the diagonal matrix.
Calculate the transpose of matrix A as follows:
AT=(PDPT)T=PTTDTPT=PDTPT=PDPT
Substitute A for PDPT.
AT=A
The matrix A and its transpose are equal.
The matrix A is symmetric.
b)

To mark:
The given statement “If A is an orthogonal matrix, then A is symmetric” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Consider an orthogonal matrix A as follows:
A=[0−110]
Consider the transpose of matrix A is denoted by AT.
Show the transpose of the matrix A as follows:
AT=[01−10]
Compare AT and A.
The matrix A and å AT are not equal. Then,
The matrix A is not symmetric.
Thus, the statement “If A is an orthogonal matrix, then A is symmetric” is false.
c)

To mark:
The given statement “If A is an orthogonal matrix, then ‖Ax‖=‖x‖ for all x in ℝn” is true or false.
Answer to Problem 1SE
The given statement is true.
Explanation of Solution
Justification of statement is true:
Apply Theorem 7 of Section 6.2 as shown below.
“Let U be an m×n matrix with orthonormal columns, and let x and y be in Rn. Then
- a. ‖Ux‖=‖x‖
- b. (Ux)⋅(Uy)=x⋅y
- c. (Ux)⋅(Uy)=0 if and only if x⋅y=0.”
Calculate the value of (Ux)⋅(Uy) as follows:
(Ux)⋅(Uy)=(Ux)T(Uy)=xTUTUy
Here, UTU=I.
Substitute I for UTU.
(Ux)⋅(Uy)=xTIy=xTy=x⋅y (1)
Consider y=x.
Substitute x for y in Equation (1).
(Ux)⋅(Ux)=x⋅x‖Ux‖2=‖x‖2‖Ux‖=‖x‖
Thus, The given statement “If A is an orthogonal matrix, then ‖Ax‖=‖x‖ for all x in ℝn” is true.
d.

To mark:
The given statement “The principal axes of a quadratic form xTAx can be the columns of any matrix P that diagonalizes A” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Consider an orthogonal matrix P.
The principal axes of xTAx are the columns of the orthogonal matrix P that diagonalizes A.
Consider A has eigenvalue whose eigenspace has dimension greater than 1, the principal axes are not uniquely determined.
The given statement “The principal axes of a quadratic form xTAx can be the columns of any matrix P that diagonalizes A” is false.
e)

To mark:
The given statement “If P is an n×n matrix with orthogonal columns, then PT=P−1“ is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Apply definition of orthogonal set as shown below.
“A set of vectors {u1, . . . , up} in Rn is said to be an orthogonal set if each pair of distinct vectors from the set is orthogonal, that is, if ui⋅uj=0 whenever i≠j.”
Apply definition of orthonormal set as shown below.
“A set {u1, . . . , up} is an orthonormal set if it is an orthogonal set of unit vectors.”
Consider a matrix P as follows:
P=[1−111]
Consider the columns u=[11] and v=[−11].
Find the vectors are orthogonal as shown below.
u⋅v=[11].[−11]=1×(−1)+1×1=−1+1=0
Hence, the vector set is orthogonal.
Find the vectors are orthonormal as shown below.
‖u‖2=[11].[11]=1×1+1×1=1+1=2
Hence, the vector set is not orthonormal.
The given statement “If P is an n×n matrix with orthogonal columns, then PT=P−1“ is true or false.
f)

To mark:
The given statement “If every coefficient in a quadratic form is positive, thenthequadratic form is positive definite” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Refer Example 6 of Section 7.2.
The each terms in quadratic form Q is positive.
The quadratic form is not positive definite.
g)

To mark:
The given statement “If xTAx>0 for some x, then the quadratic form xTAx is positive definite” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Theorem 5.
“Let Abe an n×n symmetric matrix. Then a quadratic form xTAx is:
a. positive definite if and only if the eigenvalues of A are all positive,
b. negative definite if and only if the eigenvalues of A are all negative, or
c. indefinite if and only if A has both positive and negative eigenvalues.”
Consider a matrix A and x as follows:
A=[200−3]x=[10]
The eigen value of the matrix A is 2 and −3.
Calculate the value of xTAx as follows:
xTAx=[10][200−3][10]=[10][2×1+00]=[10][20]=1×2+0
xTAx=2
Refer Part (c) of the Theorem 5.
The quadratic form xTAx is indefinite.
The value of xTAx>0
The given statement “If xTAx>0 for some x, then the quadratic form xTAx is positive definite” is false.
h)

To mark:
The given statement “By a suitable change of variable, any quadratic form canbe hanged into one with no cross-product term” is true or false.
Answer to Problem 1SE
The given statement is true.
Explanation of Solution
Justification of statement is true:
Theorem 4:
“Let Abe an n×n symmetric matrix. Then there is an orthogonal change of variable, x=Py , that transforms the quadratic form xTAx into a quadratic form yTDy with no cross-product term.”
Refer Theorem 4.
The given statement “By a suitable change of variable, any quadratic form can be hanged into one with no cross-product term” is true.
i)

To mark:
The given statement “The largest value of a quadratic form xTAx, for ‖x‖=1,is the largest entry on the diagonal of A” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Refer Example 3 of Section 7.3.
The given statement “The largest value of a quadratic form xTAx, for ‖x‖=1, is the largest entry on the diagonal of A” is false.
j)

To mark:
The given statement “The maximum value of a positive definite quadratic form xTAx is the greatest eigenvalue of A” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
The maximum value must be computed over the set of unit vectors. Without a restriction on the norm of x , the value of xTAx can be made as large as possible.
Thus, the given statement “The maximum value of a positive definite quadratic form xTAx is the greatest eigenvalue of A” is false.
k)

To mark:
The given statement “A positive definite quadratic form can be changed intoa negative definite form by a suitable change of variable x=Pu, for some orthogonal matrix P” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Consider the orthogonal change of variable x=Py. It changes a positive quadratic form into another quadratic form.
Refer Theorem 5.
The new quadratic form of xTAx is P−1AP.
The value of P−1AP is A.
Thus, it has the same eigen value same as A.
l)

To mark:
The given statement “An indefinite quadratic form is one whose eigenvaluesare not definite” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
The term “definite eigenvalue” is undefined.
Thus, the term “definite eigenvalue” is meaningless.
m)

To mark:
The given statement “If P is an n×n orthogonal matrix, then the change of variable x=Pu transforms xTAx into a quadratic form whose matrix is P−1AP” is true or false.
Answer to Problem 1SE
The given statement is true.
Explanation of Solution
Justification of statement is true:
Consider the value of x=Pu.
Calculate the value of xTAx as follows:
Substitute Pu for x.
xTAx=(Pu)TA(Pu)=uTPTAPu
Substitute P−1 for PT.
xTAx=uTP−1APu
n)

To mark:
The given statement “If U is m×n with orthogonal columns, then UUTx is the orthogonal projection of x onto Col U” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is false:
Consider the value of U=[1−11−1].
For UUTx be the orthogonal projection of x onto Col U, the columns of U must be orthonormal.
o)

To mark:
The given statement “If B is m×n and x is a unit vector in ℝn, then ‖Bx‖≤σ1, where σ1 is the first singular value of B” is true or false.
Answer to Problem 1SE
The given statement is true.
Explanation of Solution
Justification of statement is True:
Consider B be a n×n matrix.
Consider x is aunt matrix in ℝn
Refer Example 2 and Example 1 of Section 7.4.
Get the value of ‖Bx‖≤σ1
The σ1 is the first singular value of B.
p)

To mark:
The given statement “A singular value decomposition of an m×n matrix B can be written as B=PΣQ , where P is an m×m orthogonal matrix, Q is an n×n orthogonal matrix, and ∑ is an m×n “diagonal” matrix” is true or false.
Answer to Problem 1SE
The given statement is true.
Explanation of Solution
Justification of statement is True:
Theorem 10:
“Let Abe an m×n matrix with rank r. Then there exists an m×n matrix ∑ as in (3) for which the diagonal entries in D are the first r singular values of A, σ1≥σ2≥σ3≥⋯≥σr>0, and there exist an m×m orthogonal matrix U and an n×n orthogonal matrix V such that
A=UΣVT”
Refer Theorem10.
The matrices U and V are orthogonal. Then the matrix V is invertible and
V−1=VT
Calculate the value of (VT)−1 as follows:
(VT)−1=(V−1)−1=V
The matrix V is invertible and square.
Then, VT is orthogonal matrix.
q)

To mark:
The given statement “If A is n×n, then A and ATA have the same singular values” is true or false.
Answer to Problem 1SE
The given statement is false.
Explanation of Solution
Justification of statement is False:
Consider the matrix A=[2001].
The singular values of matrix A are 2 and 1.
Consider the value of AT as follows:
AT=[2001]
Calculate the value of ATA as follows:
ATA=[2001][2001]=[4001]
The singular of ATA are 4 and 1.
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