From the characteristic equation (2-)³ = 0, we find that = 2 is the only eigenvalue. Its algebraic multiplicity is 3, as it is a root of multiplicity 3.

The geometric multiplicity of an eigenvalue is the dimension of its corresponding eigenspace, which is given by n - rank(A - I). For = 2, we calculate A - 2I. The rank of this matrix is 2 (since it has two linearly independent rows). Therefore, the geometric multiplicity is 3 - 2 = 1.

The geometric multiplicity of an eigenvalue is the dimension of its corresponding eigenspace, which is given by n - rank(A - I). For = 2, we calculate A - 2I. The rank of this matrix is 2 (since it has two linearly independent rows). Therefore, the geometric multiplicity is 3 - 2 = 1.

Since the algebraic multiplicity (3) is greater than the geometric multiplicity (1), the matrix is not diagonalizable. We need to find the Jordan canonical form. With one eigenvalue = 2 and a geometric multiplicity of 1, there must be only one Jordan block corresponding to = 2. Since the size of the matrix is 3 × 3, this single Jordan block must be of size 3 × 3.

Since the algebraic multiplicity (3) is greater than the geometric multiplicity (1), the matrix is not diagonalizable. We need to find the Jordan canonical form. With one eigenvalue = 2 and a geometric multiplicity of 1, there must be only one Jordan block corresponding to = 2. Since the size of the matrix is 3 × 3, this single Jordan block must be of size 3 × 3.

Find the Jordan canonical form of the matrix $A = [[2, 1, 0], [0, 2, 1], [0, 0, 2]]$ .

Answer: The Jordan canonical form of the matrix $A$ is $J = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$ .

Step-by-step solution

Step 1: Find the eigenvalues

To find the eigenvalues of the matrix $A$ , we need to solve the characteristic equation, which is given by the determinant of $(A - \lambda I)$ set to zero. Here, $I$ is the identity matrix and $\lambda$ represents the eigenvalues.

Step 2: Calculate the determinant

The given matrix $A$ is an upper triangular matrix. For any triangular matrix (upper or lower), the eigenvalues are simply the entries on its main diagonal. Thus, the characteristic polynomial is $(2-\lambda)^3 = 0$ .

Step 3: Determine the eigenvalues and algebraic multiplicity

From the characteristic equation $(2-\lambda)^3 = 0$ , we find that $\lambda = 2$ is the only eigenvalue. Its algebraic multiplicity is 3, as it is a root of multiplicity 3.

Step 4: Find the geometric multiplicity

The geometric multiplicity of an eigenvalue is the dimension of its corresponding eigenspace, which is given by $n - \text{rank}(A - \lambda I)$ . For $\lambda = 2$ , we calculate $A - 2I$ . The rank of this matrix is 2 (since it has two linearly independent rows). Therefore, the geometric multiplicity is $3 - 2 = 1$ .

Step 5: Construct the Jordan blocks

Since the algebraic multiplicity (3) is greater than the geometric multiplicity (1), the matrix is not diagonalizable. We need to find the Jordan canonical form. With one eigenvalue $\lambda = 2$ and a geometric multiplicity of 1, there must be only one Jordan block corresponding to $\lambda = 2$ . Since the size of the matrix is $3 \times 3$ , this single Jordan block must be of size $3 \times 3$ .

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Find the Jordan canonical form of the matrix A=[[2,1,0],[0,2,1],[0,0,2]]A = [[2, 1, 0], [0, 2, 1], [0, 0, 2]]A=[[2,1,0],[0,2,1],[0,0,2]].