Math Problem Statement

Give me the answer

Give me the answer

Solution

Let’s break down the questions and address them step by step.


1. Transformation TT: R2R3\mathbb{R}^2 \to \mathbb{R}^3

The transformation TT takes (1,2)(1, 2) to (3,1,5)(3, -1, 5) and (3,1)(3, 1) to (3,1,1)(3, 1, -1).
We are tasked to find T(1,1)T(-1, 1).

Solution:

The transformation is linear, so T(a,b)=aT(1,2)+bT(3,1)T(a, b) = a \cdot T(1, 2) + b \cdot T(3, 1).
Let’s compute T(1,1)T(-1, 1):
T(1,1)=(1)(3,1,5)+(1)(3,1,1)T(-1, 1) = (-1) \cdot (3, -1, 5) + (1) \cdot (3, 1, -1)
=(3,1,5)+(3,1,1)=(0,2,6)= (-3, 1, -5) + (3, 1, -1) = (0, 2, -6)
Thus, T(1,1)=(0,2,6)T(-1, 1) = (0, 2, -6).


2. Transformation PP: R3R2\mathbb{R}^3 \to \mathbb{R}^2

Defined as P(x,y,z)=(x+z,y+z)P(x, y, z) = (x + z, y + z).

We are asked to find the kernel, image, nullity, and rank of PP.

Kernel of PP:

The kernel is the set of all vectors (x,y,z)R3(x, y, z) \in \mathbb{R}^3 such that P(x,y,z)=(0,0)P(x, y, z) = (0, 0).
From P(x,y,z)=(x+z,y+z)P(x, y, z) = (x + z, y + z):
x+z=0andy+z=0x + z = 0 \quad \text{and} \quad y + z = 0
This implies x=zx = -z and y=zy = -z, so (x,y,z)=(z,z,z)=z(1,1,1)(x, y, z) = (-z, -z, z) = z(-1, -1, 1).
The kernel is therefore:
ker(P)=span{(1,1,1)}.\ker(P) = \text{span}\{(-1, -1, 1)\}.

Image of PP:

The image is the set of all vectors (a,b)R2(a, b) \in \mathbb{R}^2 that can be written as P(x,y,z)=(x+z,y+z)P(x, y, z) = (x + z, y + z).
Setting a=x+za = x + z and b=y+zb = y + z, we see that any (a,b)(a, b) is possible. Thus:
Im(P)=R2.\text{Im}(P) = \mathbb{R}^2.

Nullity of PP:

The nullity is the dimension of the kernel:
nullity(P)=1.\text{nullity}(P) = 1.

Rank of PP:

The rank is the dimension of the image:
rank(P)=2.\text{rank}(P) = 2.

From the Rank-Nullity Theorem, we confirm:
rank(P)+nullity(P)=3.\text{rank}(P) + \text{nullity}(P) = 3.


3. Verification of Sylvester’s Law of Inertia:

Sylvester’s law states that the number of positive, negative, and zero eigenvalues of a symmetric matrix remains invariant under congruence transformations.

In this case, verifying Sylvester's law depends on the quadratic form represented by the transformation P(x,y,z)P(x, y, z). Let me know if you want detailed eigenvalue verification for this.


If you'd like me to explore any part of this in greater depth, feel free to ask!


Questions to Expand Your Understanding:

  1. What are the geometric implications of the kernel and image of a linear transformation?
  2. How does the rank-nullity theorem help classify linear transformations?
  3. Can you provide another example of a transformation to apply Sylvester’s law?
  4. How do we compute eigenvalues for a transformation matrix to verify its properties?
  5. Can you explore the relationship between the dimensions of the domain and codomain of PP?

Tip:

Always use the Rank-Nullity Theorem to double-check your kernel and image computations!

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Math Problem Analysis

Mathematical Concepts

Linear Transformations
Kernel
Image
Rank-Nullity Theorem
Sylvester's Law of Inertia

Formulas

T(a, b) = a * T(1, 2) + b * T(3, 1)
Kernel: Ax + Bz = 0 and By + Cz = 0
Rank-Nullity Theorem: rank(T) + nullity(T) = dim(domain)

Theorems

Rank-Nullity Theorem
Sylvester's Law of Inertia

Suitable Grade Level

Undergraduate Mathematics