Linear Algebraic theft

8/10/2019 Linear Algebraic theft

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1.

Ans.

A =

1+ i 1

2- i 3

A' =

1 - i 2 + i1 3

Adjoint

= 1.0000 + 1.0000i 2.0000 - 1.0000i

1.0000 + 0.0000i 3.0000 + 0.0000i

Conjugate of 1-i =1+i

(1+i)A* =

0.0000 + 2.0000i 3.0000 + 1.0000i

1.0000 + 1.0000i 3.0000 + 3.0000i

(1-i)A=

2.0000 + 0.0000i 1.0000 - 1.0000i

1.0000 - 3.0000i 3.0000 - 3.0000i

((1-i)A)*=

0.0000 + 2.0000i 3.0000 + 1.0000i

1.0000 + 1.0000i 3.0000 + 3.0000i

We have just verifed that

(cA)*=conj(c) A*

Next wed show Adjoint of Adjoint of A is A

Adjoint of AAdjoint

= 1.0000 + 1.0000i 2.0000 - 1.0000i

1.0000 3.0000

(adjoint of A)=

1.0000 - 1.0000i 1.0000 + 0.0000i

2.0000 + 1.0000i 3.0000 + 0.0000i

adjoint of adjoint of A = conjugate of A =

1+ i 1

2- i 3


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Let B =

1.0000 + 3.0000i 2.0000 + 3.0000i

4.0000 - 2.0000i 6.0000 + 3.0000iB=

1.0000 - 3.0000i 4.0000 + 2.0000i

2.0000 - 3.0000i 6.0000 - 3.0000i

B*=

1.0000 + 3.0000i 4.0000 - 2.0000i

2.0000 + 3.0000i 6.0000 + 3.0000i

A*+B*=

2.0000 + 4.0000i 6.0000 - 3.0000i

3.0000 + 3.0000i 9.0000 + 3.0000i

(A+B) =

2.0000 + 4.0000i 3.0000 + 3.0000i6.0000 - 3.0000i 9.0000 + 3.0000i

(A+B)=

2.0000 - 4.0000i 6.0000 + 3.0000i

3.0000 - 3.0000i 9.0000 - 3.0000i

(A+B)*=

2.0000 + 4.0000i 6.0000 - 3.0000i

3.0000 + 3.0000i 9.0000 + 3.0000i

=A*+B* (verified)

AB=

2.0000 + 2.0000i 5.0000 + 8.0000i

17.0000 - 1.0000i 25.0000 +13.0000i

(AB)=

2.0000 - 2.0000i 17.0000 + 1.0000i

5.0000 - 8.0000i 25.0000 -13.0000i

(AB)*=

2.0000 + 2.0000i 17.0000 - 1.0000i

5.0000 + 8.0000i 25.0000 +13.0000i

B*A* .. (from previous computed B* and A*)

= 2.0000 + 2.0000i 17.0000 - 1.0000i

5.0000 + 8.0000i 25.0000 +13.0000i

=(AB)* (thus verified)

3.


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Ans. = [ + , + , (+ ) + ] = [ , , ( ) + ]5.

A=1 0 0

0 2 00 0 3

1 =1.0000 0 0

0 0.5000 0

0 0 0.3333

L =

3 0 5

1 1 0

0 -1 1

(L*)A=ALL*= ALA-1

L*=(A-1)LA

=

3.0000 2.0000 0

0 1.0000 -1.5000

1.6667 0 1.0000

7.


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Set G(t) =g(t)*e^(-t^2)Claim D is adjoint of D

Replace G(t) for g(t) in the proof below:

7.2

1 1 + 1 2 = 2 3 + 2 2 = 0 , = 0 = 3 1 1 +

1 2 =33(1-i)x =y (Check the other is redundant as (1+i)y=(1+i)(1-i)x=2x)

1 11 4 1 1 + 1 2 =00

Again 2 112 Orthonormailation yeilds:

1 = 11 + 11 =1 + 2 =3, similarly c2 =1/3


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We get,3^(1/2)/3

3^(1/2)*(1/3 - i/3)

as an orthonormal basis

2 0 0 1 0 0 2 = (1 )((2 )2 1) = 0, = 1 , 1, 3

2 0

0 1 0 0 2 = 2 0

0 1 0 0 2

Corresponding Eigen Vectors are:

0 0.7071i 0.7071i

i 0 00 -0.7071 0.7071

Now we need to find an orthogonal basis

Applying Gram-Schmidth, we get: (Setting u1=w1,u1=-1*w2, and

u3=w3)

0 - 0.7071i +0.7071i

- i 0 0

0 0.7071 0.7071

as the orthogonal basis.


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We are done.

(A*A)*=A*A** (adjoint of products)

=A*A (Involutiveness of adjoint)

We can set apart eigenvalues of A as:

Those which are positive. These would be positive in AA(since

eig(A^2)=

^2 for each in A), ditto in A*A

Those which are negative, shall also be positive in AA, ditto in A*A

Those which are complex. Now since we take onjugates their product

is also positive, in AA* (not in AA) .

7.3

The symmetric Positive definite matrix is A

6 22 3

Constraint, cos^2 theta +sin^2 theta =1:

5,7.2


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XX=1 = 2 =, = 2 =(6

)(3

)

4 = 0 ,

= 2,7

Eigen value are 2 and 7Min is 2 max is 7

However, we see that the vector is of cos theta and sin theta,

We know that trigonometric functions are oscillating functions there arecountably infinite values of maxima and minima over Rn

Normalised eigen vectors are:

0.4472 -0.8944

-0.8944 -0.4472

such that theta =cos-1(0.4472) sin-1(-0.8944), accounting for osscilatingnatures, there are infinite of such points.

Similarly, theta =cos-1 (-0.8944) sin-1(-(0.4472), also representcountably infinite of such points.

5.

In case all diagonal elements are positive:

f is a unit circle or sphere in form a^2+b^2=1

In case they are not all positive they can represent a parabola in C, which

means a parabola, provided sum of squares of all positive entries

outweigh the sum square of all non-negative entries by more than1.

5


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We have an entire family of polynomials of power three can be covered

in 3*3 matrix. F=1 fixes a curve, given coefficients.

This can also be understood in terms of eigenvalues.

f(x) has one optima, and at most 3 eigen values. Then at most 2

saddle points. Else 1 saddle point.7.

The relevant symmetric positive definite matrix

A =1 0 0

0 2 00 0 3

= 2 =By defn of unit circle, = 2

=det(A)=(1-)(2-)(3-)=0=1,2 OR 31 is ,minima, 2 is saddle point 3 is maxima.

5,7.3


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Finding eigen vectors:0.4472 is for -0.8944 for max.

-0.8944 min -0.4472

max x1^2 +x2^2+x3^2

A=1 0 00 1 00 0 1

G =

1 1 0

0 1 10 0 1

G-1A=

1 -1 1

0 1 -1

0 0 1

eigenvalues of G-1A

=1

Corresponding eigen vectors0.4472 -0.8944-0.8944 -0.4472

Give us the maxima.

Proof. Consider a finite dimensional orthonormal basis B =

{e1, . . . , en} in V . Ris a linear operator. Then [R]B= ([R(e1)]B, . . . , [T

(en)]B).

Recall that for an orthonormal basis, the scalar product of vectors in V isequal to the standard scalar product in Rn of their respective coordinate

column (or row) vectors. If Ris orthogonal, the set of vectors R(e1), .


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. . , [R (en)is orthonormal (since e1, . . . , en is orthonormal). Hence thecolumns [R (e1)]B, . . . , R (en)]Bare orthonormal, so A = [R]Bis an

orthogonal matrix. Conversely, assume that the matrix A = [R ]Bis

orthogonal. That implies that the vectors R (e1), . . . ,[R(en) is an

orthonormal set. Thus, Rpreserves all pairwise scalar products of theelements of the basis B. It follows then that Rpre- serves all scalar

products of vectors of V , i.e., T is an orthogonal operator.

If:

Only IF:

. To see this, set x = eiin (5.6.1) to observe ui*ui= 1 for each i, and then

set x = ej+ ejfor j kto obtain 0 = +=2()By setting x = ej+ iek in the eqn above, it also follows

that 0 = 2Im()This guarantees that U is unitary

Let D = andBbe

By definition, the change of basis matrix Mhas the property that


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Had vi been othonormal would be , .

Let A be the matrix. Det(A-)= (1/5)(1-)(-1)(i+)(-4i)=(4/5)i(1-)(i+)=0= 0.8944 + 0.4472i-0.4472 - 0.8944i

Eigen Vector for 1

15 2525

5

=


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Solving, and normalising, yields:

0.8944

-0.4472

1

52

525 5 = 0

0

Solving yields:

0.4472i

-0.8944i

P=

0.8944 + 0.4472i

-0.4472 - 0.8944i

PAP-1=

-0.2683 - 0.7155i 0

0 0.7155 + 0.2683i

Ans.

The solution to this matrix As axis cn be given by AX=X

Implies(A-I)x=

*x =0

The solution is (by method of simple substituition)0.4472

-0.8944

Which is the required axis.

Eigenvectors of A corresponding to 1=

0.4472

-1 1 0

0.7071 -1 0.70710.7071 0 -1.707


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-0.8944Now acos(.4722)and asin(-.8944)=-1.1071

Which is the required rotation.

Eig vector corresponding to axis of rotation-0.8944

-0.4472

eigenvalues =

-0.5000 + 0.8660i 0.0000 + 0.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i -0.5000 - 0.8660i 0.0000 + 0.0000i

0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i

Sum of eigenvalues= 0 =1 +cos(2theta)theta=arcos(0)= 1.5708

In case the underlying space is Real, U*=U shall have the same

eigenvalues as U. U and U* commute, so eigenvalues of sum =sum of

eigenvalues.In real space this means {2*eigenvalues of U }

In complex space it means {2*Real(eigenvalues of U)}

7.51

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7.5.3

1


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Solving the eigenvalue equations we get3 = 0( 1) = 0, = 1,0,1-1 0 1

-ix2=-x1 -ix2=0 -ix2=x1

ix1=-x2 (x2 free) ix1=0 ix1=x2 (x2 free)

x3 free x3 free x3 free

-0.5000 + 0.8660i 0.0000 + 0.0000i 0.0000 + 0.0000i

0.0000 + 0.0000i -0.5000 - 0.8660i 0.0000 + 0.0000i

0.0000 + 0.0000i 0.0000 + 0.0000i 1.0000 + 0.0000i

could be a basis

Enough to evaluate 4=14 = 4=1 , =1, || = 1But then it is Unitary.

3


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7.

Linear Algebraic theft

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