Sure! Let's tackle part (a) in two steps, as you requested. --- ## (a) Find the characteristic polynomial and eigenvalues of matrix \( A \). ### **Step 1: Find the characteristic polynomial of \( A \).** The characteristic polynomial of a matrix \( A \) is defined as \( \det(A - \lambda I) \). Given: \[ A = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 4 & 1 \\ 1 & -1 & 2 \end{bmatrix} \] Subtract \( \lambda \) from the diagonal entries: \[ A - \lambda I = \begin{bmatrix} 2-\lambda & 1 & 1 \\ -1 & 4-\lambda & 1 \\ 1 & -1 & 2-\lambda \end{bmatrix} \] Now, compute the determinant: \[ \begin{align*} \det(A - \lambda I) &= \begin{vmatrix} 2-\lambda & 1 & 1 \\ -1 & 4-\lambda & 1 \\ 1 & -1 & 2-\lambda \end{vmatrix} \\ &= (2-\lambda)\begin{vmatrix}4-\lambda & 1 \\ -1 & 2-\lambda \end{vmatrix} - 1\begin{vmatrix}-1 & 1 \\ 1 & 2-\lambda\end{vmatrix} + 1\begin{vmatrix}-1 & 4-\lambda \\ 1 & -1\end{vmatrix} \end{align*} \] Calculate each minor: - \( (4-\lambda)(2-\lambda) - (1)(-1) = (4-\lambda)(2-\lambda) + 1 \) - \( (-1)(2-\lambda) - (1)(1) = -(2-\lambda) - 1 = -2 + \lambda - 1 = \lambda - 3 \) - \( (-1)(-1) - (4-\lambda)(1) = 1 - (4-\lambda) = \lambda - 3 \) Plug in the values: \[ \begin{align*} \det(A - \lambda I) &= (2-\lambda)[(4-\lambda)(2-\lambda) + 1] - 1[\lambda-3] + 1[\lambda-3] \\ &= (2-\lambda)[(4-\lambda)(2-\lambda) + 1] \\ \end{align*} \] Now, expand \( (4-\lambda)(2-\lambda) = 8 - 4\lambda - 2\lambda + \lambda^2 = 8 - 6\lambda + \lambda^2 \) So, \[ (4-\lambda)(2-\lambda) + 1 = (8 - 6\lambda + \lambda^2) + 1 = 9 - 6\lambda + \lambda^2 \] Now, multiply: \[ \det(A - \lambda I) = (2-\lambda)(9 - 6\lambda + \lambda^2) \] Expand: \[ (2-\lambda)(\lambda^2 - 6\lambda + 9) = 2(\lambda^2 - 6\lambda + 9) - \lambda(\lambda^2 - 6\lambda + 9) \] \[ = 2\lambda^2 - 12\lambda + 18 - (\lambda^3 - 6\lambda^2 + 9\lambda) \] \[ = 2\lambda^2 - 12\lambda + 18 - \lambda^3 + 6\lambda^2 - 9\lambda \] \[ = -\lambda^3 + 8\lambda^2 - 21\lambda + 18 \] **Explanation:** We subtracted \(\lambda\) from the diagonal entries of \(A\), calculated the determinant, and expanded to obtain the characteristic polynomial. **Characteristic Polynomial:** \[ \boxed{-\lambda^3 + 8\lambda^2 - 21\lambda + 18} \] --- ### **Step 2: Find the eigenvalues of \( A \).** The eigenvalues are the roots of the characteristic polynomial: \[ -\lambda^3 + 8\lambda^2 - 21\lambda + 18 = 0 \] Or, \[ \lambda^3 - 8\lambda^2 + 21\lambda - 18 = 0 \] Let’s try rational roots (factors of 18): \( \pm1, \pm2, \pm3, \pm6, \pm9, \pm18 \). Try \(\lambda = 1\): \( 1 - 8 + 21 - 18 = -4 \) (not zero) Try \(\lambda = 2\): \( 8 - 32 + 42 - 18 = (8 - 32) + (42 - 18) = -24 + 24 = 0 \) So, \(\lambda = 2\) is a root. Now, perform synthetic or polynomial division to factor out \((\lambda - 2)\): Divide \( \lambda^3 - 8\lambda^2 + 21\lambda - 18 \) by \( (\lambda - 2) \): \[ \lambda^3 - 8\lambda^2 + 21\lambda - 18 = (\lambda - 2)(\lambda^2 - 6\lambda + 9) \] Factor \( \lambda^2 - 6\lambda + 9 = (\lambda - 3)^2 \) So, characteristic equation factors as: \[ (\lambda - 2)(\lambda - 3)^2 = 0 \] So, eigenvalues are: \[ \boxed{2, 3, 3} \] **Explanation:** We solved the cubic polynomial and factored it to find the eigenvalues. --- ## **Final Answers for (a):** - **Characteristic polynomial:** \( \boxed{-\lambda^3 + 8\lambda^2 - 21\lambda + 18} \) - **Eigenvalues:** \( \boxed{2, 3, 3} \) --- Would you like to proceed to (b)?

Certainly! Let's proceed to part (b) with two detailed steps, including explanations. --- ## (b) What are the algebraic and geometric multiplicities of the eigenvalues of \(A\)? ### **Step 1: Determine the algebraic multiplicities of each eigenvalue.** From part (a), the eigenvalues are: - \(\lambda = 2\) - \(\lambda = 3\) (with multiplicity 2) **Algebraic multiplicity** is the multiplicity of each eigenvalue as a root of the characteristic polynomial. - \(\lambda = 2\) appears once, so **algebraic multiplicity** of 2 is **1**. - \(\lambda = 3\) appears twice, so **algebraic multiplicity** of 3 is **2**. **Explanation:** Eigenvalues' algebraic multiplicities are obtained directly from the roots of the characteristic polynomial. --- ### **Step 2: Find the geometric multiplicities by computing the dimension of the eigenspaces.** The **geometric multiplicity** of an eigenvalue is the dimension of the null space (kernel) of \(A - \lambda I\). --- **For \(\lambda = 2\):** Calculate \(A - 2I\): \[ A - 2I = \begin{bmatrix} 2-2 & 1 & 1 \\ -1 & 4-2 & 1 \\ 1 & -1 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 1 \\ -1 & 2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \] Solve \((A - 2I)\mathbf{x} = 0\): Row reduce: \[ \begin{bmatrix} 0 & 1 & 1 \\ -1 & 2 & 1 \\ 1 & -1 & 0 \end{bmatrix} \] Swap row 1 with row 3 to get a leading 1: \[ \begin{bmatrix} 1 & -1 & 0 \\ -1 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} \] Add row 1 to row 2: \[ \text{Row 2} = \text{Row 2} + \text{Row 1} \Rightarrow [-1+1, 2+(-1), 1+0] = [0, 1, 1] \] Now the matrix: \[ \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix} \] Subtract row 2 from row 3: \[ \text{Row 3} - \text{Row 2} = [0, 1-1, 1-1] = [0, 0, 0] \] The augmented matrix indicates: - \(x_1 - x_2 = 0 \Rightarrow x_1 = x_2\) - \(x_2 + x_3 = 0 \Rightarrow x_3 = -x_2\) Let \(x_2 = t\), then: \[ x_1 = t,\quad x_3 = -t \] Eigenvector corresponding to \(\lambda=2\): \[ \mathbf{x} = t \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \] **Dimension of eigenspace:** 1 (since only one free parameter) **Therefore,** the **geometric multiplicity** of \(\lambda=2\) is **1**. --- **For \(\lambda = 3\):** Calculate \(A - 3I\): \[ A - 3I = \begin{bmatrix} -1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{bmatrix} \] Solve \((A - 3I)\mathbf{x} = 0\): Observe rows: - Rows 1 and 2 are identical. - Row 3 is \(-\) row 1. Reduce: \[ \begin{bmatrix} -1 & 1 & 1 \\ -1 & 1 & 1 \\ 1 & -1 & -1 \end{bmatrix} \] Replace row 3 with row 3 + row 1: \[ [1 + (-1), -1 + 1, -1 + 1] = [0, 0, 0] \] So, the system reduces to two equations: \[ -1 x_1 + 1 x_2 + 1 x_3 = 0 \] and the second identical. Express \(x_1\): \[ x_1 = x_2 + x_3 \] Choose free parameters \(x_2 = s\), \(x_3 = t\), then: \[ x_1 = s + t \] Eigenvectors: \[ \mathbf{x} = s \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \] The eigenspace has dimension **2** (two free parameters), so geometric multiplicity of \(\lambda=3\) is **2**. --- ## **Final answers for (b):** | Eigenvalue | Algebraic multiplicity | Geometric multiplicity | |--------------|-------------------------|-------------------------| | 2 | 1 | 1 | | 3 | 2 | 2 | **Explanation:** - The algebraic multiplicities come from the roots of the characteristic polynomial. - The geometric multiplicities are found by computing the nullity of \((A - \lambda I)\) for each eigenvalue. --- Would you like to proceed with part (c)?

Certainly! Let's proceed to part (c) with detailed steps and explanations. --- ## (c) Show that the matrix \(A\) is diagonalizable and find matrices \(P\) and \(D\) such that \(P^{-1}AP = D\). ### **Step 1: Verify diagonalizability** A matrix is diagonalizable if it has enough linearly independent eigenvectors to form a basis of \(\mathbb{R}^3\) (or \(\mathbb{C}^3\)). From part (b), the eigenvalues and their algebraic and geometric multiplicities are: - \(\lambda = 2\): algebraic multiplicity 1, geometric multiplicity 1 - \(\lambda = 3\): algebraic multiplicity 2, geometric multiplicity 2 Since the sum of geometric multiplicities equals 3 (the size of the matrix), and for each eigenvalue, the geometric multiplicity equals or exceeds the algebraic multiplicity (here, it equals), the matrix \(A\) is diagonalizable. **Explanation:** Having enough linearly independent eigenvectors (total 3 for a \(3 \times 3\) matrix) ensures diagonalizability. --- ### **Step 2: Find eigenvectors and construct \(P\) and \(D\)** **Eigenvector for \(\lambda=2\):** From part (b), the eigenvector (up to scalar multiples) is: \[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \] **Eigenvectors for \(\lambda=3\):** The eigenspace has basis vectors (from previous calculations): - For \(x_2 = 1, x_3=0\): \[ \mathbf{v}_{3,1} = \begin{bmatrix} 1 + 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \] - For \(x_2=0, x_3=1\): \[ \mathbf{v}_{3,2} = \begin{bmatrix} 0 + 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \] **Construct matrix \(P\):** Arrange eigenvectors as columns: \[ P = \begin{bmatrix} | & | & | \\ \mathbf{v}_{2} & \mathbf{v}_{3,1} & \mathbf{v}_{3,2} \\ | & | & | \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix} \] **Construct \(D\):** Since the eigenvalues are on the diagonal: \[ D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \] **Verification:** - \(A P = P D\) - \(P\) is invertible (determinant non-zero). **Explanation:** By constructing \(P\) with eigenvectors and \(D\) with eigenvalues, we show \(A\) is similar to \(D\), confirming diagonalizability. --- ## **Final answer:** \[ \boxed{ P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} } \] and \[ \text{such that } P^{-1}AP = D \] --- Would you like to proceed with part (d)?

Certainly! Let's proceed to part (d) with detailed steps and explanations. --- ## (d) Write \(b\) as a linear combination of eigenvectors of \(A\). Use the properties of eigenvalues and eigenvectors to find \(A \mathbf{b}\). --- ### **Step 1: Express \(b\) as a linear combination of eigenvectors** Given: \[ b = \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] Eigenvectors (from previous parts): - For \(\lambda=2\): \[ \mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \] - For \(\lambda=3\): \[ \mathbf{v}_{3,1} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \] \[ \mathbf{v}_{3,2} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \] Since the eigenvectors form a basis (they are linearly independent), \(b\) can be written as: \[ b = c_1 \mathbf{v}_2 + c_2 \mathbf{v}_{3,1} + c_3 \mathbf{v}_{3,2} \] which expands to: \[ \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} = c_1 \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + c_3 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \] Write the system: \[ \begin{cases} 4 = c_1 (1) + c_2 (1) + c_3 (1) \\ 4 = c_1 (1) + c_2 (1) + c_3 (0) \\ 0 = c_1 (-1) + c_2 (0) + c_3 (1) \end{cases} \] Simplify: \[ \begin{cases} 4 = c_1 + c_2 + c_3 \quad (1) \\ 4 = c_1 + c_2 \quad (2) \\ 0 = -c_1 + c_3 \quad (3) \] From equation (2): \[ c_1 + c_2 = 4 \Rightarrow c_2 = 4 - c_1 \] From equation (3): \[ 0 = -c_1 + c_3 \Rightarrow c_3 = c_1 \] Substitute into equation (1): \[ 4 = c_1 + (4 - c_1) + c_1 = c_1 + 4 - c_1 + c_1 = 4 + c_1 \] Solve: \[ 4 = 4 + c_1 \Rightarrow c_1 = 0 \] Then: \[ c_2 = 4 - 0 = 4 \] \[ c_3 = c_1 = 0 \] **Therefore:** \[ b = 0 \cdot \mathbf{v}_2 + 4 \cdot \mathbf{v}_{3,1} + 0 \cdot \mathbf{v}_{3,2} = 4 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \] --- ### **Step 2: Find \(A \mathbf{b}\) using eigenvalues and eigenvectors** Using the property: For an eigenvector \(\mathbf{v}\) with eigenvalue \(\lambda\): \[ A \mathbf{v} = \lambda \mathbf{v} \] Since \(b\) is expressed as a linear combination of eigenvectors: \[ \mathbf{b} = 4 \mathbf{v}_{3,1} \] and \[ A \mathbf{v}_{3,1} = 3 \mathbf{v}_{3,1} \] then: \[ A \mathbf{b} = A (4 \mathbf{v}_{3,1}) = 4 A \mathbf{v}_{3,1} = 4 \times 3 \mathbf{v}_{3,1} = 12 \mathbf{v}_{3,1} \] Recall \(\mathbf{b} = 4 \mathbf{v}_{3,1}\), so: \[ A \mathbf{b} = 3 \times \mathbf{b} \] **Alternatively**, directly: \[ A \mathbf{b} = 12 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix} \] --- ## **Final answers:** - \(b\) can be written as: \[ b = 4 \mathbf{v}_{3,1} \] - \(A \mathbf{b} = 12 \mathbf{v}_{3,1} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix}\) - Equivalently, \(A \mathbf{b} = 3 \mathbf{b}\) **Explanation:** Expressing \(b\) as a linear combination of eigenvectors allows us to easily compute \(A \mathbf{b}\) using eigenvalues, leveraging the property \(A \mathbf{v} = \lambda \mathbf{v}\). --- Would you like to proceed with part (e)?

Certainly! Let's proceed to part (e) with detailed steps and explanations. --- ## (e) Find a formula for \(A^k \mathbf{b}\) for all non-negative integers \(k\) (see example 4.29). ### **Step 1: Recall the eigen-decomposition approach** Since \(A\) is diagonalizable, we can write: \[ A = P D P^{-1} \] where \(D\) is the diagonal matrix of eigenvalues, and \(P\) is the matrix of eigenvectors. Then, \[ A^k = P D^k P^{-1} \] and for any vector \(b\), \[ A^k b = P D^k P^{-1} b \] From part (d), we expressed \(b\) as a linear combination of eigenvectors, specifically: \[ b = 4 \mathbf{v}_{3,1} \] and we found that: \[ A \mathbf{b} = 3 \mathbf{b} \] which indicates that \(\mathbf{b}\) is an eigenvector associated with eigenvalue 3. **However**, note that in part (d), the eigenvector \(\mathbf{v}_{3,1}\) corresponds to eigenvalue 3. Thus, the key idea is: \[ A^k \mathbf{b} = 3^k \mathbf{b} \] since \(\mathbf{b}\) is in the eigenspace associated with eigenvalue 3. --- ### **Step 2: Write the formula for \(A^k \mathbf{b}\)** Because \(A\) acts on \(\mathbf{b}\) as scalar multiplication by 3, for any non-negative integer \(k\): \[ \boxed{ A^k \mathbf{b} = 3^k \mathbf{b} } \] In explicit form: \[ A^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] **Explanation:** Since \(\mathbf{b}\) is an eigenvector with eigenvalue 3, applying \(A^k\) multiplies \(\mathbf{b}\) by \(3^k\). --- ## **Final answer:** \[ \boxed{ A^k \mathbf{b} = 3^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} } \] --- Would you like to proceed with part (f)? Q: F A: Certainly! Let's proceed to part (f) with detailed steps and explanations. --- ## (f) Use the formula from (e) to compute \(A^k \mathbf{b}\) for specific \(k\) and compare with the result from part (d). ### **Step 1: Compute \(A^k \mathbf{b}\) using the formula** From part (e), we have: \[ A^k \mathbf{b} = 3^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] Let's pick specific values of \(k\): - For \(k=0\): \[ A^0 \mathbf{b} = \mathbf{b} = \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] - For \(k=1\): \[ A^1 \mathbf{b} = 3 \times \mathbf{b} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix} \] - For \(k=2\): \[ A^2 \mathbf{b} = 3^2 \times \mathbf{b} = 9 \times \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} = \begin{bmatrix} 36 \\ 36 \\ 0 \end{bmatrix} \] - For \(k=3\): \[ A^3 \mathbf{b} = 3^3 \times \mathbf{b} = 27 \times \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} = \begin{bmatrix} 108 \\ 108 \\ 0 \end{bmatrix} \] --- ### **Step 2: Compare with the direct computation in part (d)** In part (d), we found: \[ A \mathbf{b} = 3 \mathbf{b} \Rightarrow A^k \mathbf{b} = 3^k \mathbf{b} \] which matches exactly with the formula derived here. **Therefore**, the formula aligns perfectly with the previous calculation and confirms the eigen-decomposition approach. --- ## **Final conclusion:** \[ \boxed{ A^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} } \] for all non-negative integers \(k\). --- Would you like to proceed with part (g)? Q: G

Certainly! Let's proceed to part (g) with detailed reasoning. --- ## (g) Is \(A\) similar to \(B\)? If yes, find an invertible matrix \(P\) such that \(P^{-1} A P = B\). (Hint: First check if \(B\) is diagonalizable.) ### **Step 1: Examine the matrix \(B\)** Given: \[ B = \begin{bmatrix} 0 & 0 & 2 \\ 0 & 2 & -1 \\ 4 & 0 & 0 \end{bmatrix} \] **Check if \(B\) is diagonalizable:** - Find eigenvalues of \(B\). --- ### **Step 2: Find eigenvalues of \(B\)** Compute \(\det(B - \lambda I)\): \[ B - \lambda I = \begin{bmatrix} -\lambda & 0 & 2 \\ 0 & 2 - \lambda & -1 \\ 4 & 0 & -\lambda \end{bmatrix} \] Calculate determinant: \[ \det(B - \lambda I) = \begin{vmatrix} -\lambda & 0 & 2 \\ 0 & 2 - \lambda & -1 \\ 4 & 0 & -\lambda \end{vmatrix} \] Use cofactor expansion along the first row: \[ = (-\lambda) \begin{vmatrix} 2 - \lambda & -1 \\ 0 & -\lambda \end{vmatrix} - 0 + 2 \begin{vmatrix} 0 & 2 - \lambda \\ 4 & 0 \end{vmatrix} \] Compute minors: - First minor: \[ (2 - \lambda)(-\lambda) - (0)(-1) = (2 - \lambda)(-\lambda) = -\lambda(2 - \lambda) = -2\lambda + \lambda^2 \] - Second minor: \[ 0 \times 0 - 4 \times (2 - \lambda) = -4(2 - \lambda) = -8 + 4 \lambda \] Putting it all together: \[ \det(B - \lambda I) = (-\lambda)(-2\lambda + \lambda^2) + 2(-8 + 4 \lambda) = \lambda(2 \lambda - \lambda^2) + (-16 + 8 \lambda) \] Simplify: \[ \lambda (2 \lambda - \lambda^2) = 2 \lambda^2 - \lambda^3 \] So, \[ \det(B - \lambda I) = 2 \lambda^2 - \lambda^3 - 16 + 8 \lambda \] Rewrite: \[ = -\lambda^3 + 2 \lambda^2 + 8 \lambda - 16 \] Factor out \(-1\): \[ = -(\lambda^3 - 2 \lambda^2 - 8 \lambda + 16) \] Now, find roots of: \[ \lambda^3 - 2 \lambda^2 - 8 \lambda + 16 = 0 \] Try rational roots: factors of 16: \(\pm1, \pm2, \pm4, \pm8, \pm16\). Test \(\lambda=1\): \[ 1 - 2 - 8 + 16 = 7 \neq 0 \] Test \(\lambda=2\): \[ 8 - 8 - 16 + 16 = 0 \] Yes, \(\lambda=2\) is an eigenvalue. Divide polynomial by \(\lambda - 2\): Perform polynomial division: \[ \lambda^3 - 2 \lambda^2 - 8 \lambda + 16 = (\lambda - 2)(\lambda^2 + 0 \lambda - 8) \] Check: \[ (\lambda - 2)(\lambda^2 - 8) = \lambda^3 - 8 \lambda - 2 \lambda^2 + 16 \] which matches the original polynomial. Thus, \[ \lambda^3 - 2 \lambda^2 - 8 \lambda + 16 = (\lambda - 2)(\lambda^2 - 8) \] Set equal to zero: \[ (\lambda - 2)(\lambda^2 - 8) = 0 \] Eigenvalues: \[ \boxed{ \lambda = 2,\quad \lambda^2=8 \Rightarrow \lambda = \pm 2 \sqrt{2} } \] So, eigenvalues are: \[ \boxed{ \lambda = 2,\quad \lambda= 2 \sqrt{2},\quad \lambda= -2 \sqrt{2} } \] --- ### **Step 3: Diagonalizability of \(B\)** Since all eigenvalues are distinct, \(B\) is diagonalizable (distinct eigenvalues imply diagonalizability). --- ### **Step 4: Compare with \(A\)** Recall from part (a), the eigenvalues of \(A\) are: \[ 2, 3, 3 \] and from earlier parts, the eigenvalues of \(A\) are \(2, 3, 3\), with multiplicities. In particular: - \(A\) has eigenvalues \(2, 3, 3\), with repeated eigenvalues. - \(B\) has eigenvalues \(2, 2 \sqrt{2}, -2 \sqrt{2}\), all distinct. **Since the eigenvalues are different**, \(A\) and \(B\) **cannot be similar** (similar matrices have the same eigenvalues with the same algebraic multiplicities). --- ## **Final conclusion:** \[ \boxed{ \text{Matrix } A \text{ is *not* similar to } B } \] because they have different eigenvalues (and multiplicities). Specifically, \(A\) has eigenvalues \(2, 3, 3\), whereas \(B\) has eigenvalues \(2, 2 \sqrt{2}, -2 \sqrt{2}\). --- # **Summary:** - \(A\) and \(B\) are **not similar** because their eigenvalues do not match. - \(A\) is diagonalizable over \(\mathbb{R}\), but \(B\) is diagonalizable over \(\mathbb{R}\) with distinct eigenvalues. - Since eigenvalues differ, no invertible \(P\) exists such that \(P^{-1}AP = B\). --- **Would you like any further explanation or assistance with another part?**

Certainly! Here is a summarized version of the answers for all parts (a) through (g): --- ### **(a) Characteristic Polynomial and Eigenvalues of \(A\):** - **Characteristic Polynomial:** \(-\lambda^3 + 8 \lambda^2 - 21 \lambda + 18\) - **Eigenvalues:** \(2,\; 3,\; 3\) --- ### **(b) Algebraic and Geometric Multiplicities:** | Eigenvalue | Algebraic Multiplicity | Geometric Multiplicity | |--------------|-------------------------|-------------------------| | 2 | 1 | 1 | | 3 | 2 | 2 | --- ### **(c) Diagonalizability and Matrices \(P\) and \(D\):** - \(A\) is diagonalizable. - An invertible matrix \(P\) and diagonal matrix \(D\): \[ P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \] such that \(P^{-1}AP = D\). --- ### **(d) Express \(b\) as a Linear Combination of Eigenvectors and Find \(A \mathbf{b}\):** - \(b = 4 \times \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\), which is an eigenvector for \(\lambda=3\). - Therefore, \(A \mathbf{b} = 3 \mathbf{b} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix}\). --- ### **(e) Formula for \(A^k \mathbf{b}\):** \[ A^k \mathbf{b} = 3^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] for all non-negative integers \(k\). --- ### **(f) Computation of \(A^k \mathbf{b}\) for specific \(k\):** - \(k=0:\; \mathbf{b} = \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix}\) - \(k=1:\; \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix}\) - \(k=2:\; \begin{bmatrix} 36 \\ 36 \\ 0 \end{bmatrix}\) - \(k=3:\; \begin{bmatrix} 108 \\ 108 \\ 0 \end{bmatrix}\) Confirmed the formula \(A^k \mathbf{b} = 3^k \mathbf{b}\). --- ### **(g) Is \(A\) similar to \(B\)?** **No**, \(A\) and \(B\) are **not similar** because their eigenvalues differ: - \(A\): eigenvalues \(2, 3, 3\). - \(B\): eigenvalues \(2, 2 \sqrt{2}, -2 \sqrt{2}\). --- If you'd like further details or explanations, feel free to ask!

Certainly! Here's the complete in-text version of the answers for parts (a) through (g): --- **(a)** The characteristic polynomial of matrix \(A\) is \(-\lambda^3 + 8 \lambda^2 - 21 \lambda + 18\). The eigenvalues are the roots of this polynomial, which are \(\boxed{2,\, 3,\, 3}\). **(b)** The algebraic and geometric multiplicities are as follows: | Eigenvalue | Algebraic Multiplicity | Geometric Multiplicity | |--------------|-------------------------|-------------------------| | 2 | 1 | 1 | | 3 | 2 | 2 | **(c)** Since the sum of geometric multiplicities equals 3, the size of the matrix, \(A\) is diagonalizable. An explicit matrix \(P\) of eigenvectors and diagonal matrix \(D\) are: \[ P = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \] such that \(P^{-1} A P = D\). **(d)** The vector \(b = \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix}\) can be expressed as a scalar multiple of the eigenvector \(\mathbf{v}_{3,1} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}\): \[ b = 4 \times \mathbf{v}_{3,1} \] Since \(\mathbf{v}_{3,1}\) corresponds to eigenvalue \(\lambda=3\), we have: \[ A \mathbf{b} = 3 \mathbf{b} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix} \] **(e)** The formula for \(A^k \mathbf{b}\) for all non-negative integers \(k\) is: \[ A^k \mathbf{b} = 3^k \mathbf{b} = 3^k \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix} \] **(f)** For specific values: - When \(k=0\), \(A^0 \mathbf{b} = \mathbf{b} = \begin{bmatrix} 4 \\ 4 \\ 0 \end{bmatrix}\). - When \(k=1\), \(A \mathbf{b} = 3 \mathbf{b} = \begin{bmatrix} 12 \\ 12 \\ 0 \end{bmatrix}\). - When \(k=2\), \(A^2 \mathbf{b} = 9 \mathbf{b} = \begin{bmatrix} 36 \\ 36 \\ 0 \end{bmatrix}\). - When \(k=3\), \(A^3 \mathbf{b} = 27 \mathbf{b} = \begin{bmatrix} 108 \\ 108 \\ 0 \end{bmatrix}\). This confirms that \(A^k \mathbf{b} = 3^k \mathbf{b}\). **(g)** The matrices \(A\) and \(B\) are **not similar** because their eigenvalues differ. \(A\) has eigenvalues \(2, 3, 3\), while \(B\) has eigenvalues \(2, 2 \sqrt{2}, -2 \sqrt{2}\). Since similar matrices must have identical eigenvalues with the same algebraic multiplicities, \(A\) is **not** similar to \(B\). --- Let me know if you'd like any further clarification!