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Multiplying the eigenvalues by the eigenvectors. Does this give any meaningful result?

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closed as off-topic by Leucippus, Henrik, pjs36, Juniven, Shailesh Mar 24 '17 at 23:58

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Well, the definition of an eigenvector is:

Suppose that $T$ is a linear transformation on some vector space and $\mathbf{v}$ is a vector in that space. Then $\mathbf{v}$ is an eigenvector of $T$ if and only if $\mathbf{v}$ is not the zero vector, and there is some scalar $\lambda$ such that

$$T(\mathbf{v}) = \lambda \mathbf{v}.$$

If $\mathbf{v}$ is an eigenvector, then $\lambda$ is its eigenvalue.

So if you have an eigenvector $\mathbf{v}$ with an eigenvalue $\lambda$, the significance of the product $\lambda \mathbf{v}$ is that it's the same as $T(\mathbf{v})$.

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