Let's break down the solution step-by-step as shown in the image: --- ## (a) **Characteristic Polynomial** Given: \[ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & -4 & 1 \\ 0 & 0 & 5 \end{bmatrix} \] The characteristic polynomial is: \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{bmatrix} 1-\lambda & 1 & 0 \\ 0 & -4-\lambda & 1 \\ 0 & 0 & 5-\lambda \end{bmatrix} \] Taking the determinant (upper triangular matrix): \[ \det(A - \lambda I) = (1-\lambda)[(-4-\lambda)(5-\lambda) - (1)(0)] - 1[0 - (0)(5-\lambda)] + 0 \] But since it's upper triangular, determinant is the product of diagonal elements: \[ (1-\lambda)(-4-\lambda)(5-\lambda) \] Given in the image, the characteristic polynomial is: \[ -(\lambda - 5)(\lambda - 1)(\lambda + 4) \] --- ## (b) **Eigenvalues and Eigenspaces** ### **Eigenvalues (from the characteristic polynomial):** - \(\lambda_1 = -4\) - \(\lambda_2 = 1\) - \(\lambda_3 = 5\) ### **Eigenspaces:** For each eigenvalue \(\lambda\), solve \((A - \lambda I)\vec{x} = 0\): #### **For \(\lambda_1 = -4\):** \[ A - (-4)I = \begin{bmatrix} 5 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 9 \end{bmatrix} \] Row reduce: - Second row: \(0x + 0y + 1z = 0 \implies z=0\) - First row: \(5x + y = 0 \implies y = -5x\) So, the eigenvector: \(\begin{pmatrix} x \\ -5x \\ 0 \end{pmatrix}\) Basis: \(\text{span} \left\{ \begin{pmatrix} 1 \\ -5 \\ 0 \end{pmatrix} \right\}\) #### **For \(\lambda_2 = 1\):** \[ A - 1I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & 4 \end{bmatrix} \] Row reduce: - Third row: \(4z = 0 \implies z = 0\) - Second row: \(-5y = 0 \implies y = 0\) - First row: \(0x + 1y = 0 \implies y = 0\) So, the eigenvector: \(\begin{pmatrix} x \\ 0 \\ 0 \end{pmatrix}\) Basis: \(\text{span} \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right\}\) #### **For \(\lambda_3 = 5\):** \[ A - 5I = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -9 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] Row reduce: - Third row: \(0z = 0\) - Second row: \(-9y + z = 0 \implies z = 9y\) - First row: \(-4x + y = 0 \implies y = 4x\) So, the eigenvector: \(\begin{pmatrix} x \\ 4x \\ 36x \end{pmatrix}\) Basis: \(\text{span} \left\{ \begin{pmatrix} 1 \\ 4 \\ 36 \end{pmatrix} \right\}\) --- ## (c) **Algebraic and Geometric Multiplicities** From the characteristic polynomial, all eigenvalues have multiplicity 1: - Algebraic multiplicity for each: **1** - Geometric multiplicity for each: **1** --- ### **Final Answers:** **(a)** \[ \boxed{-(\lambda - 5)(\lambda - 1)(\lambda + 4)} \] **(b)** - \(\lambda_1 = -4\), eigenspace: \(\text{span} \left\{ \begin{pmatrix} 1 \\ -5 \\ 0 \end{pmatrix} \right\}\) - \(\lambda_2 = 1\), eigenspace: \(\text{span} \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right\}\) - \(\lambda_3 = 5\), eigenspace: \(\text{span} \left\{ \begin{pmatrix} 1 \\ 4 \\ 36 \end{pmatrix} \right\}\) **(c)** All eigenvalues have algebraic multiplicity 1 and geometric multiplicity 1.

Let's proceed step-by-step, providing detailed explanations and calculations for each part. --- ## **Given matrix:** \[ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & -4 & 1 \\ 0 & 0 & 5 \end{bmatrix} \] --- # **Part (a): Compute the characteristic polynomial of \(A\)** ### **Step 1: Form \(A - \lambda I\)** \[ A - \lambda I = \begin{bmatrix} 1 - \lambda & 1 & 0 \\ 0 & -4 - \lambda & 1 \\ 0 & 0 & 5 - \lambda \end{bmatrix} \] ### **Step 2: Find the determinant \(\det(A - \lambda I)\)** Since \(A - \lambda I\) is an **upper triangular matrix**, its determinant is the product of its diagonal entries: \[ \det(A - \lambda I) = (1 - \lambda) \times (-4 - \lambda) \times (5 - \lambda) \] Thus, the **characteristic polynomial** is: \[ p(\lambda) = (1 - \lambda)(-4 - \lambda)(5 - \lambda) \] or, rewritten: \[ p(\lambda) = - (\lambda - 1)(\lambda + 4)(\lambda - 5) \] **Final answer for (a):** \[ \boxed{ \textbf{Characteristic Polynomial}:\quad - (\lambda - 1)(\lambda + 4)(\lambda - 5) } \] --- # **Part (b): Find eigenvalues and eigenvectors (basis of eigenspaces)** ### **Step 1: Eigenvalues** From the characteristic polynomial, the eigenvalues are the roots: \[ \boxed{ \lambda_1 = -4, \quad \lambda_2 = 1, \quad \lambda_3 = 5 } \] --- ### **Step 2: Eigenvectors and eigenspaces** We solve \((A - \lambda I)\mathbf{x} = 0\) for each eigenvalue. --- ## Eigenvalue 1: \(\lambda = -4\) ### **Step 2.1: Compute \(A + 4I\)** \[ A + 4I = \begin{bmatrix} 1 + 4 & 1 & 0 \\ 0 & -4 + 4 & 1 \\ 0 & 0 & 5 + 4 \end{bmatrix} = \begin{bmatrix} 5 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 9 \end{bmatrix} \] ### **Step 2.2: Solve \((A + 4I)\mathbf{x} = 0\)** Let \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\). From the system: \[ \begin{cases} 5x_1 + x_2 = 0 \quad (1) \\ 0x_1 + 0x_2 + x_3 = 0 \quad (2) \\ 9x_3 = 0 \quad (3) \end{cases} \] From (3): \[ 9x_3 = 0 \implies x_3 = 0 \] From (2): \[ x_3 = 0 \] From (1): \[ 5x_1 + x_2 = 0 \implies x_2 = -5x_1 \] Let \(x_1 = t\). Then: \[ x_2 = -5t, \quad x_3 = 0 \] Eigenvector: \[ \boxed{ \mathbf{v}_{-4} = t \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} } \] **Basis for eigenspace \(E_{-4}\):** \[ \boxed{ \text{Span}\left\{ \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} \right\} } \] --- ## Eigenvalue 2: \(\lambda = 1\) ### **Step 2.1: Compute \(A - I\)** \[ A - I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & 4 \end{bmatrix} \] ### **Step 2.2: Solve \((A - I)\mathbf{x} = 0\)** Let \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\). System: \[ \begin{cases} 0x_1 + 1x_2 + 0x_3 = 0 \Rightarrow x_2=0 \\ 0x_1 - 5x_2 + x_3= 0 \Rightarrow 0 - 0 + x_3= 0 \Rightarrow x_3=0 \\ 0x_1 + 0x_2 + 4x_3= 0 \Rightarrow 4x_3=0 \Rightarrow x_3=0 \end{cases} \] No restriction on \(x_1\). Set \(x_1 = s\). Eigenvector: \[ \boxed{ \mathbf{v}_1 = s \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} } \] **Basis for eigenspace \(E_{1}\):** \[ \boxed{ \text{Span}\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\} } \] --- ## Eigenvalue 3: \(\lambda=5\) ### **Step 2.1: Compute \(A - 5I\)** \[ A - 5I = \begin{bmatrix} -4 & 1 & 0 \\ 0 & -9 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] ### **Step 2.2: Solve \((A - 5I)\mathbf{x} = 0\)** Let \(\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\). System: \[ \begin{cases} -4x_1 + x_2= 0 \Rightarrow x_2= 4x_1 \\ 0x_1 - 9x_2 + x_3= 0 \Rightarrow -9(4x_1) + x_3= 0 \Rightarrow -36x_1 + x_3= 0 \Rightarrow x_3=36x_1 \\ 0=0 \] Let \(x_1 = r\). Then: \[ x_2=4r, \quad x_3=36r \] Eigenvector: \[ \boxed{ \mathbf{v}_5 = r \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} } \] **Basis for eigenspace \(E_5\):** \[ \boxed{ \text{Span}\left\{ \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} \right\} } \] --- # **Summary of eigenspaces:** | Eigenvalue | Eigenvector basis | Eigen-space span | |--------------|---------------------|------------------| | \(-4\) | \(\begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} \right\}\) | | \(1\) | \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\}\) | | \(5\) | \(\begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} \right\}\) | --- # **Part (c): Algebraic and geometric multiplicities** Since each eigenvalue appears as a simple root in the characteristic polynomial: \[ p(\lambda) = - (\lambda - 1)(\lambda + 4)(\lambda - 5) \] All roots are **distinct**, so: - **Algebraic multiplicity (AM):** 1 for each eigenvalue. - **Geometric multiplicity (GM):** 1 for each eigenvalue (since each eigenspace is one-dimensional). --- ## **Final Summary:** | Eigenvalue | Algebraic multiplicity | Geometric multiplicity | Eigenvector basis | |--------------|--------------------------|------------------------|-------------------| | \(-4\) | 1 | 1 | \(\left\{ \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} \right\}\) | | \(1\) | 1 | 1 | \(\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\}\) | | \(5\) | 1 | 1 | \(\left\{ \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} \right\}\) | --- **This completes the detailed step-by-step solution.**

Certainly! In part (c), the three questions are: 1. **Compute the algebraic multiplicity of each eigenvalue.** 2. **Compute the geometric multiplicity of each eigenvalue.** 3. **State the algebraic and geometric multiplicities for each eigenvalue.** Let's address each systematically. --- ## Part (c): Algebraic and Geometric Multiplicities of Each Eigenvalue ### 1. **Algebraic Multiplicity (AM):** - The algebraic multiplicity of an eigenvalue is its multiplicity as a root of the characteristic polynomial. **From part (a), the characteristic polynomial:** \[ p(\lambda) = - (\lambda - 1)(\lambda + 4)(\lambda - 5) \] - Each factor appears **once**, so: | Eigenvalue | Algebraic Multiplicity (AM) | |--------------|------------------------------| | \(-4\) | **1** | | \(1\) | **1** | | \(5\) | **1** | ### 2. **Geometric Multiplicity (GM):** - The geometric multiplicity is the dimension of the eigenspace corresponding to each eigenvalue, i.e., the number of linearly independent eigenvectors. From part (b), for each eigenvalue, the eigenspace is **one-dimensional**, spanned by a single eigenvector: | Eigenvalue | Eigenvectors (basis) | Geometric Multiplicity (GM) | |--------------|----------------------------------------------------------------|------------------------------| | \(-4\) | \(\begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix}\) | **1** | | \(1\) | \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) | **1** | | \(5\) | \(\begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix}\) | **1** | --- ## **Final answer:** | Eigenvalue | Algebraic Multiplicity | Geometric Multiplicity | |--------------|--------------------------|------------------------| | \(-4\) | **1** | **1** | | \(1\) | **1** | **1** | | \(5\) | **1** | **1** | **Note:** Since algebraic and geometric multiplicities are equal for each eigenvalue, the matrix is diagonalizable. --- **This completes the detailed answers to all three questions in part (c).**

Certainly! Let's carefully explain how we determine the algebraic and geometric multiplicities for each eigenvalue, and why they are all equal to 1 in this case. --- ## Understanding Algebraic and Geometric Multiplicities ### **1. Algebraic Multiplicity (AM):** - It is the **multiplicity of the eigenvalue as a root** of the characteristic polynomial. - To find it, we factor the characteristic polynomial and see how many times each eigenvalue appears. ### **2. Geometric Multiplicity (GM):** - It equals the **dimension of the eigenspace** associated with that eigenvalue. - The eigenspace is found by solving \((A - \lambda I) \mathbf{x} = 0\). - The number of linearly independent eigenvectors (the basis vectors) gives the geometric multiplicity. --- ## Step-by-Step explanation for each eigenvalue: ### **Eigenvalues and the characteristic polynomial** From part (a), the characteristic polynomial is: \[ p(\lambda) = - (\lambda - 1)(\lambda + 4)(\lambda - 5) \] This polynomial has **three distinct roots**, each appearing **once** (since each factor is to the power 1). **Therefore:** \[ \boxed{ \text{Algebraic multiplicity of each eigenvalue} = 1 } \] --- ### **Why?** - Roots are \(\lambda = -4, 1, 5\). - Each root appears only once (no repeated factors). --- ### **Eigenvector calculation (from part (b))** For each eigenvalue, we solve: \[ (A - \lambda I) \mathbf{x} = 0 \] and find **one linearly independent eigenvector**. Because each eigenspace is spanned by **a single vector**, the **dimension of the eigenspace** (geometric multiplicity) is **1** for each eigenvalue. --- ## **Detailed calculations for each eigenvalue:** ### **Eigenvalue \(\lambda = -4\):** - \(A + 4I\): \[ \begin{bmatrix} 5 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 9 \end{bmatrix} \] - Solving \((A + 4I) \mathbf{x} = 0\): \[ \begin{cases} 5x_1 + x_2 = 0 \Rightarrow x_2 = -5 x_1 \\ x_3 = 0 \quad (\text{from the second row: } x_3=0) \\ 9x_3=0 \Rightarrow x_3=0 \end{cases} \] - Free parameter: \(x_1 = t\). - Eigenvector: \(\begin{bmatrix} t \\ -5t \\ 0 \end{bmatrix}\). - Basis: \(\left\{ \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} \right\}\). **Dimension of eigenspace (GM):** 1. --- ### **Eigenvalue \(\lambda=1\):** - \(A - I\): \[ \begin{bmatrix} 0 & 1 & 0 \\ 0 & -5 & 1 \\ 0 & 0 & 4 \end{bmatrix} \] - Solve: \[ \begin{cases} x_2=0 \quad (\text{from } 0x_1 + 1x_2 = 0) \\ -5x_2 + x_3= 0 \Rightarrow x_3=0 \\ 4x_3=0 \Rightarrow x_3=0 \end{cases} \] - \(x_1\) is free: let \(x_1 = s\). - Eigenvector: \(\begin{bmatrix} s \\ 0 \\ 0 \end{bmatrix}\). - Basis: \(\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\}\). **Dimension of eigenspace (GM):** 1. --- ### **Eigenvalue \(\lambda=5\):** - \(A - 5I\): \[ \begin{bmatrix} -4 & 1 & 0 \\ 0 & -9 & 1 \\ 0 & 0 & 0 \end{bmatrix} \] - Solve: \[ \begin{cases} -4x_1 + x_2= 0 \Rightarrow x_2=4x_1 \\ -9x_2 + x_3= 0 \Rightarrow -9(4x_1) + x_3= 0 \Rightarrow x_3=36x_1 \end{cases} \] - \(x_1\) is free: \(x_1 = r\). - Eigenvector: \(\begin{bmatrix} r \\ 4r \\ 36r \end{bmatrix}\). - Basis: \(\left\{ \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} \right\}\). **Dimension of eigenspace (GM):** 1. --- ## **Summary:** - **Algebraic multiplicity:** Since the polynomial factors are all simple, each eigenvalue appears **once** → **AM = 1**. - **Geometric multiplicity:** For each eigenvalue, the eigenspace is **one-dimensional**, spanned by a single eigenvector → **GM = 1**. ### **Conclusion:** The eigenvalues are **distinct**, with algebraic multiplicities and geometric multiplicities both equal to 1, which indicates the matrix is **diagonalizable**. --- **In summary:** | Eigenvalue | Algebraic multiplicity | Geometric multiplicity | |--------------|--------------------------|------------------------| | \(-4\) | 1 | 1 | | \(1\) | 1 | 1 | | \(5\) | 1 | 1 | --- Hope this detailed explanation clarifies how we determine the multiplicities!

Certainly! Here are all the final answers summarized: --- ### **(a) Characteristic Polynomial:** \[ \boxed{ -(\lambda - 5)(\lambda - 1)(\lambda + 4) } \] --- ### **(b) Eigenvalues and Eigenvectors:** | Eigenvalue | Eigenvector basis | Eigenspace span | |--------------|---------------------|-----------------| | \(-4\) | \(\begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix} \right\}\) | | \(1\) | \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\}\) | | \(5\) | \(\begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix}\) | \(\text{Span}\left\{ \begin{bmatrix} 1 \\ 4 \\ 36 \end{bmatrix} \right\}\) | --- ### **(c) Algebraic and Geometric Multiplicities:** | Eigenvalue | Algebraic Multiplicity | Geometric Multiplicity | |--------------|--------------------------|------------------------| | \(-4\) | 1 | 1 | | \(1\) | 1 | 1 | | \(5\) | 1 | 1 | --- These are the concise final answers.