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Here, $T$ and $U$ are self-adjoint operators on an $n$-dimensional inner product space.

I have wasted 2 hours on this stupid question but didn't get any headstart.

The main goal is to show that $T=U$. So I thought all I need to show is $T-U=0$. But I really don't know how to use the hypothesis to reach the conclusion. One answer to this problem that I came across claims that we need to find an orthonormal basis for $V$ consisting of eigenvectors of $T^2=U^2$ and since $T$ and $U$ are positive semidefinite, so are their squares. I can prove this but then it claims that the same basis would contain the eigenvectors of $T$ and $U$ as well with eigenvalues that are square root of eigenvalues of $T^2=U^2$, and since it's the same basis, $T$ and $U$ would be equal on that basis and hence they are equal.

I find that latter claim to be absurd and non-sensical, because if we assume an orthonormal basis $\gamma$ consisting of eigenvectors of $T$ and hence $U$, then if $\lambda$ is an eigenvalue of the corresponding eigenvector $v \in \gamma$ of $T^2$, then I can only show that $ \|\sqrt{\lambda} v\| = \| T(v) \|$. It's impossible for me to show that the two vectors are also equal, besides it's norm being equal.

I really need a help in this problem, because I banged my head on the wall, yet I couldn't find the correct method to solve this. If anyone has a better answer to this problem, please suggest.

Minto P
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    The given condition implies that $T$ and $U$ have the same range. By restricting them to their common range, we may assume that they are positive definite. Now see https://math.stackexchange.com/q/4774695 – user1551 Sep 08 '24 at 07:41
  • If $PTP^{-1} = \Lambda$ ($P$ an orthonormal basis of eigenvectors and $\Lambda = \mathrm{diag}(\text{eigenvalues})$), then $$PT^2P^{-1} = PTP^{-1}PTP^{-1} = \Lambda^2 \text{.}$$ – Eric Towers Sep 08 '24 at 10:04
  • @CalvinLin I am not satisfied with the answer. The person just created a new vector out of thin air to solve this problem. – Minto P Sep 08 '24 at 11:13

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