## Description

General Instructions

The following document contains the solutions to the theory-based questions for

Let T be the transformation corresponding to the reflection of a vector about a given vector u, and A be the corresponding transformation matrix.

A = [T(e1) T(e2) ⋯ T(en)]

where e1,e2 ⋯en represent the orthogonal basis vectors.

Also, we know that the reflection of a vector along a given vector u can be given as:

T ei

yi ei

On substituting these values in A, we get:

A = [y1 y2 ⋯ yn] ⎡y1⎤

y2

AT =

⎣ ⋮ ⎦

yn

⎡

⟨y1,y1⟩ ⟨y1,y2⟩ ⋯ ⟨y1,yn⟩⎤

⟹ AAT =⋮ ⋮ ⋱ ⋮

⎣⟨yn,y1⟩ ⟨yn,y2⟩ ⋯ ⟨yn,yn⟩⎦

For any two integers i and j, 0 ≤ i,j ≤ n

⟨yi,yj⟩

⟨ei,ej⟩

i,ej⟩

When i = j,

Now, when i ≠ j,

Hence,

Thus, we have proved that the transformation matrix is orthoganal. Subsequently,

Given that

1

4

We will compute the inverse by using

= LU L

where

1 ⋯

(Lp) = + 2,⋯

i

1

−4

3 4

L

1

3 4

L

Since A2 is a complete lower traingular matrix, there is no need to compute L3.(L3 = I)

A L

1 4

0

1 ⎡ 1 0 0 0⎤

L L L

0 −5/ ⎣−4 0 0 1⎦

⎤

−1 =

⎦

We can calculate the inverse of the upper triangular matrix using the property that the inverse of an upper triangular matrix is also an upper triangular matrix (the same holds for lower triangular matrices too). This gives us

⎡u 1 0 0 0⎤

U−1U =

0 ⎣ 1⎦

On comparing equations on both sides and solving, we get

⎡ 3 1/5 ⎤

U−1 = 0 −1/4 −2/3 −3/5

⎣

⎤

⟹ A−1 = U =

⎦

⎡

⟹ A−1 = ⎣

The basis of the subspace is {e1 − e2 1 + e2}. The projection of x onto this subspace is given by

Projx = x,e1 − e2⟩ ⟨x,e1 + e2⟩

1 − e2,e1 − e2⟩ ⟨e1 + e2,e1 + 2

= ,e12− e2⟩ (e1 − e2) + ⟨x,e1 + e2⟩ (e1 + e2)

, 1⟩e1 + ⟨x ⋅ 2⟩e2

The matrix A is given by

= ⎡0 1 0⎤

⎣ ⎦

We can verify whether A2 = I by computing A2

2 ⎡1 0 0⎤⎡1 0 0⎤

A =0 1 00 1 0

⎣

0 0 0⎦⎣0 0 0⎦

⎡1 0 0⎤

=0 1 0

⎣0 0 0⎦

≠ I

## Reviews

There are no reviews yet.