Chapter 3: Problem 30
Let \(A=\left[\begin{array}{cc}B & 0 \\ 0 & C\end{array}\right]\) where \(B\) and \(C\) are square matrices. a. Show that \(c_{A}(x)=c_{B}(x) c_{C}(x)\). b. If \(\mathbf{x}\) and \(\mathbf{y}\) are eigenvectors of \(B\) and \(C,\) respectively, show that \(\left[\begin{array}{l}\mathbf{x} \\\ 0\end{array}\right]\) and \(\left[\begin{array}{l}0 \\\ \mathbf{y}\end{array}\right]\) are eigenvec- tors of \(A,\) and show how every eigenvector of \(A\) arises from such eigenvectors.
Short Answer
Step by step solution
Understanding the Block Matrix Structure
Deriving the Characteristic Polynomial of A
Establishing Eigenvectors of A from B and C
Applying the Same for C
Concluding Eigenvector Representation of A
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Polynomial
In the context of block matrices, such as \( A = \begin{bmatrix} B & 0 \ 0 & C \end{bmatrix} \), the characteristic polynomial takes a particularly intriguing form. Here, \( B \) and \( C \) are square matrices themselves. The structure of the block matrix allows for a simplification in the calculation: the characteristic polynomial of \( A \) can be expressed as the product of the characteristic polynomials of \( B \) and \( C \).
More formally:
- Calculate \( c_B(x) \) as \( \det(xI_B - B) \).
- Calculate \( c_C(x) \) as \( \det(xI_C - C) \).
- The characteristic polynomial of \( A \) is \( c_A(x) = c_B(x) \cdot c_C(x) \).
Eigenvectors
When dealing with block matrices, the eigenvectors of the overall matrix \( A = \begin{bmatrix} B & 0 \ 0 & C \end{bmatrix} \) can be constructed from the eigenvectors of the submatrices \( B \) and \( C \). If \( \mathbf{x} \) is an eigenvector of \( B \) with eigenvalue \( \lambda \):-
- Then the vector \( \begin{bmatrix} \mathbf{x} \ 0 \end{bmatrix} \) becomes an eigenvector of \( A \).
- Similarly, if \( \mathbf{y} \) is an eigenvector of \( C \) with eigenvalue \( \mu \), the vector \( \begin{bmatrix} 0 \ \mathbf{y} \end{bmatrix} \) is an eigenvector of \( A \).
Determinant Calculation
For block matrices where off-diagonal blocks are zero, the determinant of the whole matrix \( A \) is simply the product of the determinants of the blocks on the diagonal:
- Calculate \( \det(B) \), the determinant of one block.
- Calculate \( \det(C) \), the determinant of the other block.
- The determinant of \( A \) is \( \det(A) = \det(B) \times \det(C) \).