Hermitian Operators and Projectors in Linear Algebra

[LagrangeEuler]

Matrix
$$\left[<br /> \begin{array}{rr}<br /> 1 & 1 \\<br /> 0& 0 \\<br /> \end{array} \right]$$
is not symmetric. When we find eigenvalues of that matrix we get $\lambda_1=1$, $\lambda_2=0$, or we get matrix
$$\left[<br /> \begin{array}{rr}<br /> 1 & 0 \\<br /> 0& 0 \\<br /> \end{array} \right]$$.
First matrix is not hermitian, whereas second one it is. How it is possible that some operator is hermitian in one basis, and is not in the other one. Second matrix is also a projector, and the first one it is not. How that is possible.

 

[fresh_42]

Have you considered how a basis transformation has to look like, in order to conserve symmetry? Does yours look like one? And what do you mean by a projector? The first matrix is also a projection, i.e. surjective, just not at the same angle.

 

[George Jones]

And what do you mean by a projector?.

In English linear algebra references, a projector (on an inner product space) typically is a linear transformation that is a projection, and that has orthogonal null space and range, i.e., idempotent and Hermitian. I am not sure if other languages make this distinction.