eigenvalues get squared but eigenvectors are retained
eigenvalues are retained but eigenvectors change
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The Correct Option isC
Solution and Explanation
If a matrix $A$ has an eigenvalue $\lambda$ with corresponding eigenvector $\mathbf{v}$, then by definition:
\[
A\mathbf{v} = \lambda \mathbf{v}
\]
Now consider the squared matrix $A^2$:
\[
A^2 \mathbf{v} = A(A\mathbf{v}) = A(\lambda \mathbf{v}) = \lambda (A\mathbf{v}) = \lambda (\lambda \mathbf{v}) = \lambda^2 \mathbf{v}
\]
This shows that $\mathbf{v}$ remains an eigenvector of $A^2$, but the corresponding eigenvalue becomes $\lambda^2$.
Therefore:
– Eigenvectors are retained.
– Eigenvalues get squared.
Hence, option (C) is correct.