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In this comprehensive blog post, we'll tackle a notoriously difficult Linear Algebra problem, providing a step-by-step solution to help you develop a deeper understanding of the subject. So, let's dive in!
The Challenging Problem: By A University Algebra tutor
Problem Statement:
Let T:R3→R3T: \mathbb{R}^3 \rightarrow \mathbb{R}^3T:R3→R3 be a linear transformation such that:
T3−T2−T+I=0T^3 - T^2 - T + I = 0T3−T2−T+I=0
where III is the identity transformation.
Tasks:
Find the minimal polynomial of TTT.
Determine if TTT is diagonalizable.
Find the Jordan canonical form of TTT.
Step-by-Step Solution
Understanding the Problem
Before we begin solving, it's crucial to understand what the problem is asking:
Minimal Polynomial: The minimal polynomial of a linear transformation TTT is the monic polynomial of least degree such that m(T)=0m(T) = 0m(T)=0.
Diagonalizability: A linear transformation TTT is diagonalizable if there exists a basis of the vector space consisting of eigenvectors of TTT.
Jordan Canonical Form: This is a block-diagonal matrix representing TTT, providing insight into its eigenvalues and the geometric multiplicity of each eigenvalue.
1. Finding the Minimal Polynomial of TTT
We are given the equation:
T3−T2−T+I=0T^3 - T^2 - T + I = 0T3−T2−T+I=0
This can be rewritten as:
T3−T2−T+I=0T^3 - T^2 - T + I = \mathbf{0}T3−T2−T+I=0
where 0\mathbf{0}0 is the zero transformation.
Step 1: Identifying the Characteristic Polynomial
While the characteristic polynomial is not directly given, we can consider that it divides any polynomial satisfied by TTT. However, since R3\mathbb{R}^3R3 is a three-dimensional space, the characteristic polynomial must be of degree 3.
Step 2: Factor the Given Polynomial
Let's factor the polynomial p(x)=x3−x2−x+1p(x) = x^3 - x^2 - x + 1p(x)=x3−x2−x+1.
First, check for rational roots using the Rational Root Theorem. The possible rational roots are ±1\pm1±1.
Testing x=1x = 1x=1:
(1)3−(1)2−(1)+1=1−1−1+1=0(1)^3 - (1)^2 - (1) + 1 = 1 - 1 - 1 + 1 = 0(1)3−(1)2−(1)+1=1−1−1+1=0
So, x=1x = 1x=1 is a root.
Step 3: Perform Polynomial Division
Divide p(x)p(x)p(x) by x−1x - 1x−1:
Using polynomial division:
p(x)=(x−1)(x2−0x−1)p(x) = (x - 1)(x^2 - 0x - 1)p(x)=(x−1)(x2−0x−1)
So,
p(x)=(x−1)(x2−1)p(x) = (x - 1)(x^2 - 1)p(x)=(x−1)(x2−1)
Further factorization:
x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1)x2−1=(x−1)(x+1)
Therefore,
p(x)=(x−1)2(x+1)p(x) = (x - 1)^2(x + 1)p(x)=(x−1)2(x+1)
Step 4: Identify the Minimal Polynomial
The minimal polynomial m(x)m(x)m(x) divides p(x)p(x)p(x), and it is the monic polynomial of least degree such that m(T)=0m(T) = \mathbf{0}m(T)=0.
Since TTT satisfies p(T)=0p(T) = \mathbf{0}p(T)=0, and considering the minimal polynomial must annihilate TTT, the minimal polynomial is:
m(x)=(x−1)(x+1)m(x) = (x - 1)(x + 1)m(x)=(x−1)(x+1)
However, we need to verify if (x−1)(x - 1)(x−1) or (x−1)2(x - 1)^2(x−1)2 is needed.
Step 5: Testing m(x)=(x−1)(x+1)m(x) = (x - 1)(x + 1)m(x)=(x−1)(x+1)
Assuming m(x)=(x−1)(x+1)m(x) = (x - 1)(x + 1)m(x)=(x−1)(x+1), let's test if m(T)=0m(T) = \mathbf{0}m(T)=0:
(T−I)(T+I)=T2−TI+TI−I2=T2−I=?(T - I)(T + I) = T^2 - T I + T I - I^2 = T^2 - I = ?(T−I)(T+I)=T2−TI+TI−I2=T2−I=?
But we need to ensure that T2−IT^2 - IT2−I annihilates TTT.
Given that T3−T2−T+I=0T^3 - T^2 - T + I = \mathbf{0}T3−T2−T+I=0, let's compute T2−IT^2 - IT2−I:
From the given equation:
T3−T2−T+I=0T^3 - T^2 - T + I = \mathbf{0}T3−T2−T+I=0
T3=T2+T−IT^3 = T^2 + T - IT3=T2+T−I
Now, compute T2−IT^2 - IT2−I:
(T2−I)(T+I)=T3+T2−T−I(T^2 - I)(T + I) = T^3 + T^2 - T - I(T2−I)(T+I)=T3+T2−T−I
But this doesn't simplify to 0\mathbf{0}0.
Step 6: Conclude Minimal Polynomial
Given that (x−1)2(x+1)(x - 1)^2(x + 1)(x−1)2(x+1) annihilates TTT, and (x−1)(x+1)(x - 1)(x + 1)(x−1)(x+1) does not, the minimal polynomial is:
m(x)=(x−1)2(x+1)m(x) = (x - 1)^2(x + 1)m(x)=(x−1)2(x+1)
2. Determining if TTT is Diagonalizable
A linear transformation TTT is diagonalizable if the minimal polynomial splits into distinct linear factors, and the geometric multiplicity equals the algebraic multiplicity for each eigenvalue.
Step 1: Identify Eigenvalues
From the minimal polynomial:
m(x)=(x−1)2(x+1)m(x) = (x - 1)^2(x + 1)m(x)=(x−1)2(x+1)
The eigenvalues are:
λ1=1\lambda_1 = 1λ1=1 (algebraic multiplicity 2)
λ2=−1\lambda_2 = -1λ2=−1 (algebraic multiplicity 1)
Step 2: Find Geometric Multiplicities
We need to find the dimension of each eigenspace.
For λ=1\lambda = 1λ=1:
Solve (T−I)v=0(T - I)v = \mathbf{0}(T−I)v=0.
Let E1=ker(T−I)E_1 = \ker(T - I)E1=ker(T−I).
For λ=−1\lambda = -1λ=−1:
Solve (T+I)v=0(T + I)v = \mathbf{0}(T+I)v=0.
Let E−1=ker(T+I)E_{-1} = \ker(T + I)E−1=ker(T+I).
Step 3: Determine Dimensions
Since R3\mathbb{R}^3R3 has dimension 3:
The sum of the dimensions of the eigenspaces cannot exceed 3.
The algebraic multiplicities sum to 3 (2 for λ=1\lambda = 1λ=1 and 1 for λ=−1\lambda = -1λ=−1).
If the geometric multiplicity of λ=1\lambda = 1λ=1 is less than its algebraic multiplicity (i.e., less than 2), then TTT is not diagonalizable.
Step 4: Conclude Diagonalizability
Given that the minimal polynomial has a repeated factor (x−1)2(x - 1)^2(x−1)2, and unless we have enough eigenvectors corresponding to λ=1\lambda = 1λ=1, TTT is not diagonalizable.
3. Finding the Jordan Canonical Form of TTT
Since TTT is not diagonalizable, we'll find its Jordan canonical form.
Step 1: Construct Jordan Blocks
For each eigenvalue, we construct Jordan blocks.
For λ=1\lambda = 1λ=1:
Algebraic multiplicity: 2
Geometric multiplicity: ggg (we need to find ggg)
For λ=−1\lambda = -1λ=−1:
Algebraic multiplicity: 1
Geometric multiplicity: 1
Step 2: Determine the Size of Jordan Blocks
Since the geometric multiplicity of λ=−1\lambda = -1λ=−1 is 1, there's one Jordan block of size 1 for λ=−1\lambda = -1λ=−1.
For λ=1\lambda = 1λ=1, since the minimal polynomial has (x−1)2(x - 1)^2(x−1)2, and TTT is not diagonalizable, there must be at least one Jordan block of size 2.
Step 3: Write the Jordan Canonical Form
The Jordan canonical form JJJ of TTT is:
1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} \] This represents: - A Jordan block of size 2 for \( \lambda = 1 \). - A Jordan block of size 1 for \( \lambda = -1 \). #### Step 4: Verify the Jordan Canonical Form We can verify that this \( J \) satisfies the minimal polynomial \( m(x) \). ## Connecting to Critical Thinking and Visual-Spatial Skills Solving complex Linear Algebra problems like this one requires a blend of critical thinking and visual-spatial reasoning: - Critical Thinking: Breaking down the problem, factoring polynomials, and logically deducing the minimal polynomial demands analytical skills. - Visual-Spatial Skills: Understanding transformations, eigenspaces, and Jordan canonical forms involves visualizing abstract concepts in multiple dimensions. At S.T.E.M. Online, we focus on enhancing these skills, ensuring you not only solve the problem at hand but also develop a deeper comprehension of the underlying concepts. ## Why Choose S.T.E.M. Online as Your University Algebra Tutor As your dedicated University Algebra tutor, S.T.E.M. Online offers: - Personalized tutoring sessions tailored to your learning pace. - Expert tutors with extensive experience in Linear Algebra. - Strategies to strengthen your critical thinking and visual-spatial abilities. Don't let challenging problems hinder your academic progress. With our guidance, you'll tackle even the most difficult questions with confidence. ## Join the Ranks of Our Satisfied Students At S.T.E.M. Online, we're proud to have a track record of success, reflected in our five-star testimonials: - Alex T.: "S.T.E.M. Online transformed my understanding of Linear Algebra. Their approach to teaching is unparalleled!" - Maria L.: "Thanks to my tutor at S.T.E.M. Online, I aced my exams and developed skills that will last a lifetime." - Jason K.: "The best University Algebra tutor I've ever had. Highly recommend S.T.E.M. Online to anyone struggling with math." ## Take the Next Step Towards Mastery Ready to conquer Linear Algebra? Let S.T.E.M. Online be your trusted University Algebra tutor. Contact us today and embark on a journey towards academic excellence. --- S.T.E.M. Online – Empowering students through critical thinking and visual-spatial mastery. Join us and experience the difference that five-star tutoring can make!
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