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I am just getting started with Linear Algebra and I have a conceptual doubt regarding eigenvalues and eigenvectors.

Suppose we have $$Ax = \lambda x$$ where $\lambda$ is a scalar , $x\not= 0$ is called the eigenvector of matrix $A$ corresponding to eigenvalue $\lambda$. If we multiply $x$ by some scalar $\alpha$ times then we have that

$$A(\alpha x)=\alpha Ax = \alpha \lambda x = \lambda (\alpha x)$$

Now we have that $(\alpha x,\lambda)$ is an eigenvector-eigenvalue pair of $A$. Then isn't it the case that we have found all the eigenvectors just by taking scalar multiples of eigenvectors? In other words, why do we specifically solve the equation $ det(A-\lambda I)=0$ if we can get them as above?

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    $\begingroup$ Some eigenspaces are 2-dimensional or higher. Scaling alone won't find all the eigenvectors. $\endgroup$
    – Randall
    Oct 5, 2021 at 12:58

2 Answers 2

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We solve the equation $\det(A-\lambda\operatorname{Id})=0$ in order to know which scalars turn out to be eigenvalues. Then, after having them, for each each $\lambda$ which turns out to be an eigenvalue, we search for the corresponding eigenvectors by solving the equation $Av=\lambda v$.

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What you have found is that if $x$ is an eigenvector of the eigenvalue $\lambda$ than also $\alpha x$ is an eigenvector of the same eigenvalue. In other words: to a given eigenvalue corresponds an entire linear space of eigenvectors, spanned by $x$. This is called the eigenspace of $\lambda$. But, for a different eigenvalue the eigenspace is, in general, different, and can be determined only if you know the eigenvalue.

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