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As a foreword, this is part of my homework so I have left all the calculations out: I just need an explanation.

I was trying to show that D = U†AU is diagonal, where D is the diagonalized matrix, A is a Hermitian matrix, U is found using the eigenvectors and U† is of course the conjugate transpose of U. I got D = [1,-3 ; -3,1] but my question is how do I know if this is diagonal or not? Is it diagonal because it reduces to [1,0 ; 0,1] and so the non diagonal entries are zero?

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  • $\begingroup$ A matrix $M$ is called diagonal if all its non-diagonal entries are zero. If $M$ is similar to a diagonal matrix, i.e. if $M=T^{-1}DT$ for some diagonal matrix $D$, then $M$ is called diagonalizable. So your matrix $D$ is not diagonal, but it is diagonalizable. $\endgroup$ – Inactive - Objecting Extremism May 26 '14 at 6:46
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A diagonal matrix is an a square matrix $D = [d_{ij}]$ with $d_{ij} = 0$ for $i \neq j$; that is, a square matrix which can only have non-zero entries on the diagonal. The matrix $$\left[\begin{matrix}1 & -3\\ -3 & 1\end{matrix}\right]$$ is not diagonal. Your proposed definition of diagonal is equivalent to invertibility. You must have made a mistake in your evaluation of the product $U^{\dagger}AU$, or possibly in determining $U$ itself.

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