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The square matrix A has an eigenvalue λ with corresponding eigenvector e - CIE - A-Level Further Maths - Question 9 - 2013 - Paper 1
Question 9
The square matrix A has an eigenvalue λ with corresponding eigenvector e. The non-singular matrix M is of the same order as A. Show that Me is an eigenvector of the ... show full transcript
Worked Solution & Example Answer:The square matrix A has an eigenvalue λ with corresponding eigenvector e - CIE - A-Level Further Maths - Question 9 - 2013 - Paper 1
Step 1
Write down the eigenvalues of A and obtain corresponding eigenvectors.
Answer
To find the eigenvalues of matrix A, we compute the characteristic polynomial given by ( \text{det}(A - \lambda I) = 0 ).
Calculating ( A - \lambda I ):
[ A - \lambda I = \begin{pmatrix} -1 - \lambda & 2 \ 0 & 4 - \lambda \end{pmatrix} ]
The determinant is calculated as follows:
[ \text{det}(A - \lambda I) = (-1 - \lambda)(4 - \lambda) = 0 ]
This gives us the eigenvalues:
- ( \lambda_1 = -1 )
- ( \lambda_2 = 2 )
- ( \lambda_3 = 4 )
Now to find the corresponding eigenvectors:
- For ( \lambda = -1 ): Solve ( (A - (-1)I)e = 0 ).
- The equation reduces to finding [ \begin{pmatrix} 0 & 2 \ 0 & 5 \end{pmatrix} ], yielding eigenvector ( e_1 = \begin{pmatrix} 1 \ 0 \end{pmatrix} ).
- For ( \lambda = 2 ): Solve ( (A - 2I)e = 0 ).
- Eigenvector found is ( e_2 = \begin{pmatrix} 1 \ -1 \end{pmatrix} ).
- For ( \lambda = 4 ): Solve ( (A - 4I)e = 0 ).
- Eigenvector found is ( e_3 = \begin{pmatrix} 0 \ 1 \end{pmatrix} ).
Step 2
find the eigenvalues and corresponding eigenvectors of B.
Answer
Given that ( B = MAM^{-1} ), we first need to find the eigenvalues of B.
The relationship for eigenvalues shows that if ( A \) has eigenvalue ( \lambda ), then ( B ) will have the same eigenvalue ( \lambda ) given the non-singularity of M. Thus, the eigenvalues of B are:
- ( \lambda_1 = -1 )
- ( \lambda_2 = 2 )
- ( \lambda_3 = 4 )
Next, we determine the corresponding eigenvectors of B. Each eigenvector is transformed by the matrix M:
- For ( \lambda_1 = -1 ): The eigenvector is obtained by applying ( Me_1 ):
- ( e^B_1 = M \begin{pmatrix} 1 \ 0 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ 0 \end{pmatrix} ).
- For ( \lambda_2 = 2 ): Applying M yields:
- ( e^B_2 = M \begin{pmatrix} 1 \ -1 \end{pmatrix} = \begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} ).
- For ( \lambda_3 = 4 ): Applying M gives:
- ( e^B_3 = M \begin{pmatrix} 0 \ 1 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} ).