Question

If \( \lambda_1, \lambda_2 \) and \( \lambda_3 \) are the eigen values of a square matrix \( A \), then the eigen values of \( A^{-1} + 2I + A \) are

• A. \( \frac{1}{\lambda_1} + 2 + \lambda_1 \), \( \frac{1}{\lambda_2} + 2 + \lambda_2 \) and \( \frac{1}{\lambda_3} + 2 - \lambda_3 \)
• B. \( \frac{1}{\lambda_1} + 2 + \lambda_1 \), \( \frac{1}{\lambda_2} - 2 - \lambda_2 \) and \( \frac{1}{\lambda_3} - 2 - \lambda_3 \)
• C. \( \frac{1}{\lambda_1} - 2 - \lambda_1 \), \( \frac{1}{\lambda_2} - 2 - \lambda_2 \) and \( \frac{1}{\lambda_3} - 2 - \lambda_3 \)
• D. \( \frac{1}{\lambda_1} + 2 + \lambda_1 \), \( \frac{1}{\lambda_2} + 2 + \lambda_2 \) and \( \frac{1}{\lambda_3} + 2 + \lambda_3 \)

Hint:

If \( \lambda \) is an eigenvalue of a matrix \( A \), then any matrix polynomial or functional expression \( f(A) \) will have corresponding eigenvalues given directly by substituting \( \lambda \) into the function, resulting in \( f(\lambda) \).

Answer 1

The Correct Option is D

Concept: Let \( A \) be an \( n \times n \) square matrix. If \( \lambda \) is an eigenvalue of \( A \) corresponding to a non-zero eigenvector \( X \), then by definition: \[ AX = \lambda X \] Using matrix algebra and spectral properties, we can determine the eigenvalues of functions of matrices:

• Eigenvalues of an inverse matrix (\( A^{-1} \)): If \( A \) is non-singular, multiplying both sides of \( AX = \lambda X \) by \( A^{-1} \) gives \( X = \lambda A^{-1} X \), which transforms directly to \( A^{-1}X = \frac{1}{\lambda} X \). Thus, the eigenvalues are \( \frac{1}{\lambda} \).

• Eigenvalues of a scalar multiple matrix (\( kI \)): Since \( IX = X \), we have \( (kI)X = kX \). The eigenvalue of the identity matrix scaled by \( k \) is identically \( k \).

• Eigenvalues of matrix polynomials/combinations: If \( P(A) = c_m A^m + \cdots + c_1 A + c_0 I \), then the eigenvalues of \( P(A) \) are given exactly by \( P(\lambda) = c_m \lambda^m + \cdots + c_1 \lambda + c_0 \).

Step 1: Application of spectral mapping theorem to each component.
We are asked to find the eigenvalues of the linear matrix combination expression: \[ B = A^{-1} + 2I + A \] Let \( X_i \) be the non-zero eigenvector associated with the specific eigenvalue \( \lambda_i \) of the square matrix \( A \). This means: \[ AX_i = \lambda_i X_i \quad \text{for } i = 1, 2, 3. \]

Step 2: Operating the matrix equation onto the eigenvector \( X_i \).
Let us apply the matrix combination expression \( B \) directly onto the eigenvector \( X_i \): \[ BX_i = (A^{-1} + 2I + A)X_i \] Distributing the eigenvector vector multiplication over the matrix addition linearly: \[ BX_i = A^{-1}X_i + 2IX_i + AX_i \]

Step 3: Substituting known eigenvalue relations.
We substitute the fundamental eigenvalue properties for each standalone term inside the equation:

• For the matrix vector product \( A^{-1}X_i \), since \( A \) has eigenvalue \( \lambda_i \), its inverse operation satisfies \( A^{-1}X_i = \frac{1}{\lambda_i}X_i \).

• For the identity component product, \( 2IX_i = 2X_i \).

• For the primary matrix product, \( AX_i = \lambda_i X_i \).
Substituting these direct scalar equations back into our operational matrix-vector expansion: \[ BX_i = \left(\frac{1}{\lambda_i}\right)X_i + 2X_i + \lambda_i X_i \]

Step 4: Factoring out the common vector component to finalize the eigenvalue format.
We now safely factor out the common non-zero eigenvector \( X_i \) from the right side of the expression: \[ BX_i = \left( \frac{1}{\lambda_i} + 2 + \lambda_i \right) X_i \] This precisely fits the standard eigenvalue-eigenvector functional definition \( BX_i = \mu_i X_i \), showing that the scalar multiplier value is the new eigenvalue. Repeating this exact linear mapping process independently for all three eigenvalues \( \lambda_1, \lambda_2, \lambda_3 \), the complete set of new eigenvalues for the matrix \( A^{-1} + 2I + A \) is uniquely evaluated as: \[ \frac{1}{\lambda_1} + 2 + \lambda_1, \quad \frac{1}{\lambda_2} + 2 + \lambda_2, \quad \text{and} \quad \frac{1}{\lambda_3} + 2 + \lambda_3 \] This maps explicitly and without any sign alterations to Option (D).