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Reduce the matrix \( \left(\begin{array}{lll}1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right) \) to Jordan Canonical form

Ask by Donnelly Lang. in Nepal
Jan 13,2025

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Answer

The Jordan Canonical Form of the matrix is: \[ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

Solution

To reduce the matrix \[ A = \begin{pmatrix} 1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1 \end{pmatrix} \] to its Jordan Canonical Form (JCF), follow these steps: ### 1. **Find the Eigenvalues** The eigenvalues of an upper triangular matrix are the entries on its diagonal. Thus, the eigenvalues of \( A \) are: - \( \lambda_1 = 1 \) (with algebraic multiplicity 2) - \( \lambda_2 = 4 \) (with algebraic multiplicity 1) ### 2. **Determine the Geometric Multiplicity** For each eigenvalue, determine the dimension of its eigenspace (geometric multiplicity). #### For \( \lambda = 1 \): Solve \( (A - I)\mathbf{x} = 0 \): \[ A - I = \begin{pmatrix} 0 & 5 & 7 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \end{pmatrix} \] This system reduces to: - \( 5x_2 + 7x_3 = 0 \) - \( 3x_2 + 3x_3 = 0 \) From the second equation: \( x_2 = -x_3 \). Substituting into the first equation: \( 5(-x_3) + 7x_3 = 2x_3 = 0 \) leads to \( x_3 = 0 \) and consequently \( x_2 = 0 \). Thus, the only free variable is \( x_1 \), meaning the eigenspace for \( \lambda = 1 \) is one-dimensional. #### For \( \lambda = 4 \): Solve \( (A - 4I)\mathbf{x} = 0 \): \[ A - 4I = \begin{pmatrix} -3 & 5 & 7 \\ 0 & 0 & 3 \\ 0 & 0 & -3 \end{pmatrix} \] This system reduces to: - \( -3x_1 + 5x_2 + 7x_3 = 0 \) - \( 3x_3 = 0 \) From the second equation: \( x_3 = 0 \). Substituting into the first equation: \( -3x_1 + 5x_2 = 0 \) leads to \( x_1 = \frac{5}{3}x_2 \). Thus, the eigenspace for \( \lambda = 4 \) is one-dimensional. ### 3. **Construct the Jordan Canonical Form** - **For \( \lambda = 1 \)**: Since the algebraic multiplicity is 2 and the geometric multiplicity is 1, there is **one Jordan block of size 2**. - **For \( \lambda = 4 \)**: There is **one Jordan block of size 1**. Therefore, the Jordan Canonical Form \( J \) of matrix \( A \) is: \[ J = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix} \] This form consists of a \( 2 \times 2 \) Jordan block for the eigenvalue \( 1 \) and a \( 1 \times 1 \) block for the eigenvalue \( 4 \). **Final Answer:** After reduction, a Jordan matrix with a size‑2 block for 1 and a 1×1 block for 4. Thus, J =  1 1 0  0 1 0  0 0 4

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The Deep Dive

To reduce the matrix \( A = \left(\begin{array}{ccc}1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right) \) to Jordan Canonical Form (JCF), we first need to find the eigenvalues and their respective geometric and algebraic multiplicities. 1. **Eigenvalue Calculation:** We can find the eigenvalues by determining the characteristic polynomial, which is given by: \[ \det(A - \lambda I) = 0 \] where \( I \) is the identity matrix. This leads to: \[ \det \left(\begin{array}{ccc} 1 - \lambda & 5 & 7 \\ 0 & 4 - \lambda & 3 \\ 0 & 0 & 1 - \lambda \end{array}\right) = (1 - \lambda)(4 - \lambda)(1 - \lambda) \] The characteristic polynomial is: \[ (1 - \lambda)^2 (4 - \lambda) = 0 \] Thus, the eigenvalues are \( \lambda_1 = 1 \) (with algebraic multiplicity 2) and \( \lambda_2 = 4 \) (with algebraic multiplicity 1). 2. **Geometric Multiplicity:** To determine the geometric multiplicity for each eigenvalue: For \( \lambda_1 = 1 \): \[ A - I = \left(\begin{array}{ccc} 0 & 5 & 7 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \end{array}\right) \] The rank of this matrix is 2, thus the nullity (the geometric multiplicity) is \( 3 - 2 = 1 \). For \( \lambda_2 = 4 \): \[ A - 4I = \left(\begin{array}{ccc} -3 & 5 & 7 \\ 0 & 0 & 3 \\ 0 & 0 & -3 \end{array}\right) \] The rank is again 2, so the nullity (the geometric multiplicity) is \( 3 - 2 = 1 \). 3. **Constructing the Jordan Form:** Since the algebraic multiplicity of \( \lambda_1 = 1 \) is 2 and the geometric multiplicity is 1, there will be one Jordan block of size 2 for eigenvalue 1. For \( \lambda_2 = 4 \), since both its algebraic and geometric multiplicities are 1, it will correspond to a 1x1 Jordan block. Therefore, the Jordan Canonical form \( J \) is: \[ J = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix} \] In summary, the Jordan Canonical Form of the matrix \( \left(\begin{array}{lll}1 & 5 & 7 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right) \) is given by: \[ J = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{pmatrix} \]

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