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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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