Is x in the nullspace of A an eigenvector of A?

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    Eigenvalue Proof
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SUMMARY

If x is a non-zero vector in the nullspace of matrix A, then x is definitively an eigenvector of A. This conclusion arises from the fact that A is singular, which implies that the eigenvalue λ associated with x is 0. The relationship Ax = 0 confirms that the action of A on x results in the zero vector, aligning with the definition of an eigenvector where A operates on x to yield a scalar multiple of x.

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  • Understanding of linear algebra concepts, particularly eigenvalues and eigenvectors.
  • Familiarity with matrix properties, specifically singular and invertible matrices.
  • Knowledge of nullspaces and their significance in linear transformations.
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sana2476
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Let x not equal to zero be a vector in the nullspace of A. Then x is an eigenvector of A.


I'm not sure how to start this proof
 
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If x is a non-zero vector in the null space of A, then you know that A is singular, and you also know that [tex]\lambda = 0[/tex] is an eigenvalue of A since A is invertible if and only if zero is not an eigenvalue of A. That should start you off.
 
Saying that x is in the null space of A means that Ax= 0= 0x.
 

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