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What would be the correct way to explain that $\lambda = 0$ implies $(\lambda I-A)\vec{x}=\vec{0}$ has a non-trivial solution?
Could I just say, if $\lambda = 0$, then $\det(A) = 0$ and $A$ is singular and $(\lambda I-A)\vec{x}=\vec{0}$ will not have just the trivial solution.
The question is asking if $A$ has an eigenvalue of $\lambda$ then $(\lambda I-A)\vec{x}=\vec{0}$ will only have the trivial solution.
Okay, the first thing I recall is, like you said, definition of eigenvalues as the determinant of a matrix, as well as the invertible matrix theorem (IMT). IMT has a condition that says: if the determinant of a matrix is zero, then it is not invertible. Therefore, its null-space (what you have mentioned) is not trivial.
Explanation: $$\det (A-\lambda I)=(\lambda-\lambda_1)\cdot (\lambda-\lambda_2)\cdot ...(\lambda-\lambda_n)$$ Where $\lambda_i$ is an eigenvalue of the matrix $A$. $\lambda$ is the free variable. If we let $\lambda=0$, then we get the following: $$\det (A-0\cdot I)=\det(A)=(0-\lambda_1)\cdot (0-\lambda_2)\cdot ...(0-\lambda_n)=(-1)^n*\prod_{i=1}^{n}\lambda_i$$ Therefore, we have shown that the determinant of a matrix is the product of its eigenvalues. If at least one of the $\lambda_i=0$ (We don't care which), then we know that $\det A=0$. If that is true, then your hypothesis follows from the statement of the IMT given above. The null-space of $A$ has a non-trivial solution, since there will be at least one free variable in the reduced-row-echelon form of $A$, because the matrix is rank-deficient.
