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I am trying to understand eigen values for my control systems class , I will try to explain it below if someone could be kind enough to point out my mistakes ?

Lets $A$ be a transformation matrix , If we apply this transformation certain vector , which doesn't change its direction but rather just scale in length with that transformation and that scaling can be represented in $\lambda$ such $$Ax = \lambda x$$

where $\lambda$ is a scalar , and x is the eigenvector of transformation $A$ , if we say that $A$ has eigenvalues then $A-\lambda I$ should be a singlular matrix

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  • $\begingroup$ That’s nicely said. $\endgroup$ Jul 6, 2018 at 11:48

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That's correct. To be very precise, if $\lambda = \pm 1$, then the length of a corresponding eigenvector $x$ doesn't change.

But note also that a same matrix can have different eigenvalues with different eigenvectors. For instance, look at $$A = \begin{bmatrix} 2 & 0\\ 0 & 3 \end{bmatrix}$$ For this matrix, $(1,0)$ is an eigenvector for the eigenvalue $2$ and $(0,1)$ is an eigenvector for the eigenvalue $3$.

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Consider a matrix A For the definition of the eigenvector v of A we can also say that Av=(lambda)v it's the action of the matrix A on v which is equals to the some scaler(lambda) times of v i.e in the direction of the vector v.

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