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Let $A=\begin{pmatrix} 1&0\\ 0&1\\ \end{pmatrix}$. Find the bases for the eigenspaces of the matrix $A$. I know the bases for the eigenspace corresponding to each eigenvector is a vector (or system) that can scale to give any other vector contained in that said eigenspace. Thus, we see that the identity matrix has only one distinct eigenvalue $\lambda=1$. Thus the eigenvector satisfies the equation $(A-\lambda I)\vec{x}=\vec{0}$. Then we must solve $\begin{pmatrix} 0&0&0\\ 0&0&0\\ \end{pmatrix}$. I say that any vector satisfies this equation however the key says that the basis is $ \lbrace (1,0),(0,1) \rbrace$. How do they come up with this solution. Thanks in advance!

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It is true that any vector satisfies the equation. So, our eigenspace is the whole space $\mathbb{R}^2$. Since $\{(1,0),(0,1)\}$ is a basis for $\mathbb{R}^2$, it is also a basis for this eigenspace.

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    $\begingroup$ Yes. This is precisely the explanation I needed. Thank you! $\endgroup$ – coreyman317 Apr 5 '18 at 8:50

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