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Let $A$ be an $n \times n$ matrix and $v$ be an eigen vector of $A$ corresponding to eigen value $\lambda$. I would like know the following geometrically (intuition). If we substract $\lambda$ from all the diagonal entries then the resulting matrix $B$ is non singular. Can some one explain this process geometrically. I understand it mathematically why rows/columns of $B$ are linearly dependent.

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  • $\begingroup$ $B=A-\lambda I$ is "singular" because $Bv=0$ and $v$ is not equal to $0$. $\endgroup$ – Srinivas K Feb 6 '15 at 11:30
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you can interpret $Ax = \lambda x$ geometrically. that says that if a vector and its transform are collinear, then it is an eigenvector. this is only true for a real eigenvalue. that the condition about the determinant of $A - \lambda I$ is zero is a consequence of this. of course, you can take the geometric interpretation of the determinant as the volume and connect to $(A - \lambda I) x = 0$

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    $\begingroup$ In fact $\;A-\lambda I\;$ almost never equals zero. What is true is that $\;\lambda\;$ is an eigenvalue of $\;A\;$ iff $\;A-\lambda I\;$ is singular, or not regular, or has non-zero-kernel, or its determinant is zero, etc. $\endgroup$ – Timbuc Feb 6 '15 at 11:01

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