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If I have a Hermitian operator $H:V \to V$ on a finite-dimensional vector space $V$, and I write down its matrix representation in some basis $B$ with matrix representation being $[H]_B$, then in what cases is $[H]_B$ invertible? ( I assume being invertible is a property of the linear operator independent of the basis, if I am not wrong )

What I tried was that, I know the operator $H$ will have a spectral decomposition $\sum_i \lambda_i |i\rangle \langle i|$. Now an operator is defined by how it acts on a basis ( any ). The assumption I took was if an operator on a space maps one basis to another ( not necessarily orthonormal ) then the operator is invertible. Thus if a Hermitian operator has to be invertible $\lambda_i \neq 0 \; \forall \;i$. But I am still not satisfied, is there a simple proof?

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    $\begingroup$ This looks like a pure math question. $\endgroup$ – ACuriousMind Apr 7 '15 at 13:11
  • $\begingroup$ @ACuriousMind I was reading about something else from Neilson chuang when this doubt occurred so I posted here.Later I thought about removing but one answer was posted. $\endgroup$ – sashas Apr 7 '15 at 13:59
  • $\begingroup$ Related meta post: meta.physics.stackexchange.com/q/6698/2451 $\endgroup$ – Qmechanic Apr 16 '15 at 19:58
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It is invertible iff its determinate doesn't vanish $$ \det([H]_B) \ne 0 $$ Note that this property of the determinate is invariant under a change of basis since: \begin{align} \det(S^{-1} \cdot [H]_B \cdot S) & = \det(S^{-1}) \cdot \det([H]_B) \cdot \det(S) = \frac{1}{\det(S)} \cdot \det([H]_B) \cdot \det(S) \\ & = \det([H]_B) \end{align} with $S$ the matrix representing the change of basis. Also a hermitian matrix is always diagonalizable, such that $$ \det(H) = \prod_i \lambda_i $$.

So a hermitian matrix/operator H on a vector space $V$ is invertiable if and only if none of its eigenvalues $\lambda_i$ are zero.

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Well, this is a problem of linear algebra substantially. The basic idea is that a square matrix $A$ is invertible if and only if $detA\neq 0$. Then, if the matrix is diagonalizable, since the determinant is invariant under coordinate transformations $A'=C^{-1}AC^1$, when you compute the determinant you get that, you get $detA=\lambda_1\dots\lambda_n$. (Note that not all the Eigenvalues on the diagonal should be different and that's why the other answer is wrong (indeed by definition Eigenvalues can't be zero)). Then you want that the matrix is diagonalizable. Now suppose $A$ was the representation in some basis of your operator $H$. Then, what you want is that $H$ is diagonalizable. This happens when the concatenation of the basis vectors of the eigenspace of each Eigenvalue is a basis for $V$. This happens if $H$ is a complete operator (is a single complete measurement). For example, the operator $x$ corresponding to the measurement of the x component of a particle's position, is not a complete operator since it makes part of a bigger measurement. On the other hand, the operators $x,y,z$ do form a complete set of operators.

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