# Let lambda be an eigenvalue of A, such that no eigenvector of A associated with lambda has a zero entry. Then prove that every list of n−1 columns of A−lambda I is linearly independent.

Let $\lambda$ be an eigenvalue of A, such that no eigenvector of A associated with $\lambda$ has a zero entry. Then prove that every list of n−1 columns of $A-\lambda I$ is linearly independent.
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The null space of ${A}_{n×n}-\lambda I$ is the eigenspace of $\lambda$. Since no eigenvector corresponding to $\lambda$ contains a 0 entry, the basis of the eigenspace must contain a single eigenvector of A. This means the nullity of $A-\lambda I$ is 1, and by the rank-nullity theorem, its rank n−1.
Now recall that $\left(A-\lambda I\right)v$ is a linear combination of the columns of $A-\lambda I$. This means their exists a tuple $\left({v}_{1},{v}_{2},...,{v}_{n}\right)$ with ${v}_{i}\ne 0$ that satisfies ${v}_{1}{C}_{1}+{v}_{2}{C}_{2}+\cdot \cdot \cdot +{v}_{n}{C}_{n}=0$. Since the rank of these vectors is n−1, there is one linearly dependent column ${C}_{j}$, and since ${v}_{j}\ne 0$ for all j, that ${C}_{j}$ could be any column