1. **State the problem:** We are given the matrix \(A = \begin{pmatrix} 3 & 1 & 3 & 4 \\ -1 & 1 & 0 & -1 \\ 0 & 0 & 5 & 3 \\ 0 & 0 & -3 & -1 \end{pmatrix}\) and need to find its Jordan normal form. 2. **Recall the process:** To find the Jordan normal form, we: - Find the eigenvalues by solving \(\det(A - \lambda I) = 0\). - For each eigenvalue, find the algebraic and geometric multiplicities. - Determine the sizes of Jordan blocks from these multiplicities. - Construct the Jordan normal form matrix with Jordan blocks on the diagonal. 3. **Find eigenvalues:** Calculate \(\det(A - \lambda I)\): \[ A - \lambda I = \begin{pmatrix} 3-\lambda & 1 & 3 & 4 \\ -1 & 1-\lambda & 0 & -1 \\ 0 & 0 & 5-\lambda & 3 \\ 0 & 0 & -3 & -1-\lambda \end{pmatrix} \] Since the matrix is block upper-triangular, the determinant is the product of the determinants of the diagonal blocks: \[ \det(A - \lambda I) = \det\begin{pmatrix} 3-\lambda & 1 \\ -1 & 1-\lambda \end{pmatrix} \times \det\begin{pmatrix} 5-\lambda & 3 \\ -3 & -1-\lambda \end{pmatrix} \] Calculate each determinant: \[ (3-\lambda)(1-\lambda) - (-1)(1) = (3-\lambda)(1-\lambda) + 1 \] \[ = 3 - 3\lambda - \lambda + \lambda^2 + 1 = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 \] and \[ (5-\lambda)(-1-\lambda) - (3)(-3) = -(5-\lambda)(1+\lambda) + 9 \] \[ = -[5 + 5\lambda - \lambda - \lambda^2] + 9 = -[5 + 4\lambda - \lambda^2] + 9 = \lambda^2 - 4\lambda + 4 = (\lambda - 2)^2 \] 4. **Eigenvalues:** Both blocks give \((\lambda - 2)^2\), so the characteristic polynomial is \((\lambda - 2)^4\). The only eigenvalue is \(\lambda = 2\) with algebraic multiplicity 4. 5. **Find geometric multiplicity:** Solve \((A - 2I)\mathbf{x} = 0\). \[ A - 2I = \begin{pmatrix} 1 & 1 & 3 & 4 \\ -1 & -1 & 0 & -1 \\ 0 & 0 & 3 & 3 \\ 0 & 0 & -3 & -3 \end{pmatrix} \] Row reduce: - Add row 1 and row 2: row 2 becomes zero vector. - Rows 3 and 4 are dependent (row 4 = -row 3). Rank is 2, so nullity (dimension of eigenspace) is \(4 - 2 = 2\). 6. **Jordan blocks:** Algebraic multiplicity = 4, geometric multiplicity = 2. This means the Jordan normal form has two Jordan blocks for eigenvalue 2, sizes summing to 4, with two blocks indicating at least one block of size greater than 1. Possible block sizes: 2 and 2. 7. **Jordan normal form:** \[ J = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix} \] This matrix has two Jordan blocks of size 2 for eigenvalue 2. **Final answer:** The Jordan normal form of \(A\) is $$ \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix} $$