Linear Algebra notation: A-Dagger

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I'm a little confused as to some linear algebra notation and would appreciate some clarification.

There exists an $n\times n$ diagonal matrix $Q$ where the entries along the diagonal are the eigenvalues of another $n\times n$ matrix $A$. My confusion is because the question specifically asks for two matrices, $T$ and $T^\dagger$, s.t.

$$ T^{\dagger}A T = Q $$

Up until now $T^\dagger$ has simply been the conjugate transpose of $T$, but wherever I look I find solutions using $T^{-1}$ which I've not seen and seems to be discrete. I've tried using a matrix composed of the eigenvectors of $A$ (as well as a matrix of the normalized eigenvectors of $A$ which was my first stab regardless) and while I've gotten close I haven't quite hit upon a method that works.

So my question is this; Is there a second meaning for $T^\dagger$ or am I correct in using the eigenvector matrices?

(thank you whoever edited this to be prettier)
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2 Answers

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The decomposition $$ T^{\dagger}A T = Q $$ where $T^\dagger $ is the conjugate transpose of $T$ and $Q$ is diagonal exists if and only if $A$ is a normal matrix, i.e. a matrix such that $A^\dagger A=AA^\dagger$.

In this case $T$ is the matrix with columns the eigenvectors of $A$ and $T^{-1}=T^\dagger$.

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Caveat: This answer only works if your matrix $A$ is Hermitian!

From Wikipedia, with respect to the equation $P^{-1}AP=Q$, where $Q$ is a diagonal matrix.

When the matrix $A$ is a Hermitian matrix (resp. symmetric matrix), eigenvectors of $A$ can be chosen to form an orthonormal basis of $\mathbb{C}^n$ (resp. $\mathbb{R}^n$). Under such circumstance $P$ will be a unitary matrix (resp. orthogonal matrix) and $P^{−1}$ equals the conjugate transpose (resp. transpose) of $P$.

And the notation for the conjugate transpose is $P^\dagger$. Hermitian matrices are used quite often in quantum mechanics and quantum computation, so the dagger notation is used with the understanding that it will also be the inverse of the matrix.

So, for this question your doing, I guess choose a set of orthonormal eigenvectors for your matrix $A$.

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