The Eigenvalue problem

Question

In page 32 of R. Shankar's Principles of Quantum Mechanics is given the eigenvalue problem:

We begin by rewriting Eq. (1.8.2)as $$(\Omega - \omega I)|V\rangle =|0\rangle \tag{1.8.3}$$ Opening both sides with $(\Omega - \omega I)^{-1}$, assuming it exists, we get $$|V\rangle =(\Omega - \omega I)^{-1}|0\rangle \tag{1.8.4}$$ Now any finite operator(an operator with finite matrix elements) acting on the null vector can only give us a null vector. It therefore seems that in asking for a nonzero eigenvector $|V\rangle$, we are trying to get something for nothing out of Eq. (1.8.4). This is impossible. It follows that our assumption that the operator $(\Omega - \omega I)^{-1}$ exists(as a finite operator) is false. So we ask when this situation will obtain. Basic matrix theory tells us (see Appendix $\mathrm A. 1$) that the inverse of any matrix $M$ is given by $$M^{-1}= \frac{\mathrm{cofactor} M}{\mathrm{det}(M)} \tag{1.8.5}$$ Now the cofactor of $M$ is finite if $M$ is. Thus what we need is the vanishing of the determinant. The condition for nonzero eigenvectors is therefore $$\mathrm{det}(\Omega - \omega I) =0 \tag{1.8.6}$$

Why is a vanishing determinant necessary to find nonzero eigenvectors?

Answer 1

The operator $(\Omega - \omega I)$ can be seen as a matrix. But to be inversible, a matrix needs its determinant to be non-zero. Why does it need to be non-zero ?

Let's take a matrix A, which we'll assume is inversible and let's demonstrate that its determinant can't be zero : $$ det(A) \cdot det (A^{-1}) = det(A \cdot A^{-1}) = det (I) = 1 $$ (with the property of multiplication of determinants) thus $det(A)$ can't be zero.

Answer 2

For, for notational clarity, you want to understand $\vert 0\rangle$ as the $0$ vector.

Start with $$ (\Omega-\omega I)\vert V\rangle=\vert 0\rangle \, . \tag{1} $$ If $(\Omega-\omega I)$ as a matrix has an inverse - call it $(\Omega-\omega I)^{-1}$ as you do, then multiply both sides of (1) by this inverse to get $$ (\Omega-\omega I)^{-1}(\Omega-\omega I)\vert V\rangle = \vert V\rangle =\vert 0\rangle \tag{2} $$ so that your eigenvector $\vert V\rangle $ is the $\vert 0\rangle$ vector. If you do not want $\vert V\rangle$ to be $\vert 0\rangle$, the assumption that $(\Omega-\omega I)$ has an inverse must be false. For this to be false, it must be that $$ \hbox{Det}\left(\Omega-\omega I\right)=0\, . $$ Alternatively, if the determinant is NOT $0$, then the inverse will exists and you find from (2) that $\vert V\rangle$ is the $0$ vector.