Are Both Eigenvectors Correct?

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SUMMARY

Both eigenvectors derived from an eigenvalue are valid, as scalar multiplication of an eigenvector results in another eigenvector. For instance, the eigenvector \(\begin{bmatrix} 1 \\ \frac{1}{3} \end{bmatrix}\) is equivalent to \(\begin{bmatrix} 3 \\ 1 \end{bmatrix}\) when multiplied by the scalar 3. This principle is grounded in the definition of eigenvectors, confirming that multiple representations of the same eigenvector are acceptable.

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  • Understanding of eigenvalues and eigenvectors
  • Familiarity with linear algebra concepts
  • Knowledge of scalar multiplication in vector spaces
  • Basic proficiency in matrix representation
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  • Study the properties of eigenvectors and eigenvalues in linear algebra
  • Explore scalar multiplication and its implications in vector spaces
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Students and professionals in mathematics, particularly those studying linear algebra, as well as data scientists and engineers applying eigenvector concepts in computational algorithms.

Cpt Qwark
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say for example when I calculate an eigenvector for a particular eigenvalue and get something like
\begin{bmatrix}<br /> 1\\<br /> \frac{1}{3}<br /> \end{bmatrix}

but the answers on the book are

\begin{bmatrix}<br /> 3\\<br /> 1<br /> \end{bmatrix}

Would my answers still be considered correct?
 
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Yes, multiplying an eigenvector with a scalar yields another eigenvector (makes kind of sense right? Think about the definition). Usually people will multiply the result with whatever scalar makes all the entries integers for representation purposes, but both results are correct.
 
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