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.
This problem is originally from the book Matrix Analysis (2ed) by Roger A. Horn. It's 1.4.P12 on page 82. I know how to prove the reverse direction, but I have no idea for above direction. Any suggestions?
The null space of $A_{n\times n}-\lambda I$ 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 $(A-\lambda I)v$ is a linear combination of the columns of $A-\lambda I$. This means their exists a tuple $(v_1,v_2,...,v_n)$ with $v_i\ne0$ that satisfies $v_1C_1+v_2C_2+\cdot\cdot\cdot+v_nC_n=0$. Since the rank of these vectors is $n-1$, there is one linearly dependent column $C_j$, and since $v_j\ne0$ for all $j$, that $C_j$ could be any column