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I'm trying to prove that above $\Bbb C$ it is true that $\mathrm{Ker}(T)=\mathrm{Ker}(T^* T)$ where $T^*$ is the conjugate transpose of $T$... Is it actually true? How can I prove it? (One side of containing is obvious, yet I couldn't find a way to prove that the dimensions are equal)

Thanks!

Cameron Buie
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ORBOT Inc.
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1 Answers1

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Let $x \in \ker (T^*T)$, then $$ \|Tx\|^2 = (Tx, Tx) = (x, T^*Tx) = 0 $$ so $Tx = 0$. As you wrote, the other inclusion is obvious.

martini
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