Question

Which of the following statement(s) is/are true with respect to eigenvalues and eigenvectors of a matrix?

• A. The sum of the eigenvalues of a matrix equals the sum of the elements of the principal diagonal.
• B. If \(\lambda\) is an eigenvalue of a matrix \(A\), then \(\dfrac{1}{\lambda}\) is always an eigenvalue of its transpose \((A^{\mathsf T})\).
• C. If \(\lambda\) is an eigenvalue of an \emph{orthogonal} matrix \(A\), then \(\dfrac{1}{\lambda}\) is also an eigenvalue of \(A\).
• D. If a matrix has \(n\) distinct eigenvalues, it also has \(n\) independent eigenvectors.

Hint:

Remember three handy facts: \(\mathrm{tr}(A)=\sum \lambda_i\); \(A\) and \(A^{\mathsf T}\) have the same eigenvalues; distinct eigenvalues \(\Rightarrow\) independent eigenvectors. For orthogonal \(A\), eigenvalues lie on the unit circle, so reciprocals are also eigenvalues.

Answer 1

The Correct Option is A, C, D

Step 1: Trace equals sum of eigenvalues.
For any \(n\times n\) matrix \(A\), its characteristic polynomial has coefficients related to power sums of eigenvalues; in particular, \[ \mathrm{tr}(A) \;=\; \sum_{i=1}^{n} \lambda_i \] (counting algebraic multiplicities). Since \(\mathrm{tr}(A)\) is the sum of diagonal entries, (A) is true.

Step 2: About the transpose and reciprocals.
\(A^{\mathsf T}\) is similar to \(A\) over \(\mathbb{C}\) only for special classes; in general, \(A\) and \(A^{\mathsf T}\) always have the same eigenvalues (their characteristic polynomials are identical). Thus if \(\lambda\) is an eigenvalue of \(A\), then \(\lambda\) (not \(1/\lambda\)) is an eigenvalue of \(A^{\mathsf T}\). Counterexample: \[ A=\begin{bmatrix}2&0 \\ 0&3\end{bmatrix}. \] Eigenvalues of both \(A\) and \(A^{\mathsf T}\) are \(\{2,3\}\), whereas \(1/\lambda\in\{1/2,1/3\}\) are not eigenvalues. Hence (B) is false.

Step 3: Orthogonal matrices.
If \(A\) is orthogonal, \(A^{-1}=A^{\mathsf T}\). For any eigenpair \((\lambda,\mathbf v)\) of \(A\), \(A\mathbf v=\lambda \mathbf v\). Then \[ A^{-1}\mathbf v=\frac{1}{\lambda}\mathbf v. \] So \(1/\lambda\) is an eigenvalue of \(A^{-1}\). But \(A^{-1}\) and \(A\) have eigenvalues related by reciprocals, and since \(A^{-1}=A^{\mathsf T}\) and \(A^{\mathsf T}\) has the same spectrum as \(A\), it follows that \(1/\lambda\) is also an eigenvalue of \(A\). (Indeed, for real orthogonal \(A\), \(|\lambda|=1\).) Thus (C) is true.

Step 4: Distinct eigenvalues \(\Rightarrow\) independent eigenvectors.
For any linear operator on an \(n\)-dimensional space, eigenvectors associated with distinct eigenvalues are linearly independent. Therefore an \(n\times n\) matrix with \(n\) distinct eigenvalues has \(n\) linearly independent eigenvectors and is diagonalizable. Hence (D) is true.

Final Answer:
\[ \boxed{(A),\ (C),\ (D)} \]