Question

Let $A$ be a $3 \times 3$ matrix. If $\lambda_1, \lambda_2, \lambda_3$ are the eigenvalues of $A$, then the eigenvalues of the matrix $(I + aA)^{-1}(I + bA)$, where $a, b$ are scalars such that $a \lambda_i \neq -1$ and $b \lambda_i \neq -1$ for $i = 1, 2, 3$, are:

• A. \(\frac{b \lambda_1}{1 + a \lambda_1}, \frac{b \lambda_2}{1 + a \lambda_2}, \frac{b \lambda_3}{1 + a \lambda_3}\)
• B. \(\frac{1 + b \lambda_1}{1 + a \lambda_1}, \frac{1 + b \lambda_2}{a \lambda_2}, \frac{1 + b \lambda_3}{a \lambda_3}\)
• C. \(\frac{1 + a \lambda_1}{1 + b \lambda_1}, \frac{1 + a \lambda_2}{1 + b \lambda_2}, \frac{1 + a \lambda_3}{1 + b \lambda_3}\)
• D. \(\frac{1 + b \lambda_1}{1 + a \lambda_1}, \frac{1 + b \lambda_2}{1 + a \lambda_2}, \frac{1 + b \lambda_3}{1 + a \lambda_3}\)

Hint:

For matrices, if \(\lambda\) is an eigenvalue of \(A\), then for any polynomial \(p(A)\), \(p(\lambda)\) is the corresponding eigenvalue of \(p(A)\). Also, for invertible matrices, eigenvalues of inverses are reciprocals of the eigenvalues.

Answer 1

The Correct Option is D

Since \(\lambda_i\) are eigenvalues of \(A\), for each \(i\), \(A \mathbf{v}_i = \lambda_i \mathbf{v}_i\), where \(\mathbf{v}_i\) are the eigenvectors. Consider the matrix \(M = (I + aA)^{-1}(I + bA)\). We want eigenvalues \(\mu_i\) of \(M\). For the eigenvector \(\mathbf{v}_i\), \[ (I + aA) \mathbf{v}_i = (I + a \lambda_i) \mathbf{v}_i, \] and hence \[ (I + aA)^{-1} \mathbf{v}_i = \frac{1}{1 + a \lambda_i} \mathbf{v}_i. \] Similarly, \[ (I + bA) \mathbf{v}_i = (I + b \lambda_i) \mathbf{v}_i. \] Therefore, \[ M \mathbf{v}_i = (I + aA)^{-1}(I + bA) \mathbf{v}_i = (I + aA)^{-1} (1 + b \lambda_i) \mathbf{v}_i = \frac{1 + b \lambda_i}{1 + a \lambda_i} \mathbf{v}_i. \] Thus the eigenvalues of \(M\) are: \[ \mu_i = \frac{1 + b \lambda_i}{1 + a \lambda_i}, \quad i = 1, 2, 3. \]