Question:

Let \(A\) be a square matrix. Which of the following statement about eigenvectors is correct?

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If \[ \boxed{ A\mathbf{x}=\lambda\mathbf{x}, } \] then every non-zero scalar multiple \[ \boxed{ k\mathbf{x},\;k\neq0, } \] is also an eigenvector corresponding to the same eigenvalue.
Updated On: Jul 14, 2026
  • Eigen vectors are always unique
  • Eigen vectors are unique only if the eigen value has multiplicity \(1\)
  • Eigen vectors are unique up to a scalar multiple
  • Eigen vectors cannot be scaled
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The Correct Option is C

Solution and Explanation

Step 1: Recall the definition of an eigenvector. If \[ A\mathbf{x}=\lambda\mathbf{x}, \] then \[ \mathbf{x} \] is an eigenvector corresponding to the eigenvalue \[ \lambda. \]

Step 2:
Apply the scaling property. If \[ \mathbf{x} \] is an eigenvector, then for any non-zero scalar \[ k, \] \[ A(k\mathbf{x}) = kA\mathbf{x} = k\lambda\mathbf{x} = \lambda(k\mathbf{x}). \] Hence, \[ k\mathbf{x} \] is also an eigenvector corresponding to the same eigenvalue. Therefore, \[ \boxed{ \text{Eigenvectors are unique only up to a non-zero scalar multiple.} } \] Thus, \[ \boxed{(C)} \] is the correct answer.
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