I know that the eigenvalues are $\lambda$ which satisfy $A\vec{v} = \lambda \vec{v}$. I know that to find the eigenvalues of a matrix, we find the characteristic polynomial, set it to zero, and solve for $\lambda$. But I don't understand why this is the 'recipe' for finding eigenvalues -- what is the intuition? Thanks.


If $v$ is an eigenvector for $\lambda$ then $Av = \lambda v$ or $(A - \lambda I)v = 0$. In other words, $v$ belongs to the kernel or null space of $A - \lambda I$.

The null space of $A - \lambda I$ is non-trivial iff $\det(A - \lambda I) = 0$ iff the characteristic polynomial $p(\lambda) = \det(A - \lambda I) = 0$.

Therefore finding $\lambda$ such $p(\lambda) = 0$ is both a necessary and sufficient condition to there being eigenvectors for such $\lambda$.

That's the theory for the 'recipe'.


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