# Thread: inner product space, complex case

1. ## inner product space, thank you!

i'd be extremely grateful if anyone can help me out with this
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Suppose V is a ﬁnite dimensional inner product space and T : V → V a linear operator.

1. Prove im(T∗ ) = (kerT )⊥.
2. Prove rank(T ) = rank(T∗ ).

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T∗ should be displayed as T^*, as is (kerT )⊥ should be (kerT)^⊥

2. Originally Posted by ericmik
Suppose V is a ﬁnite dimensional inner product space and T : V → V a linear operator.

1. Prove im(T∗ ) = (kerT )⊥.
2. Prove rank(T ) = rank(T∗ ).
From the relation $\displaystyle \langle Tx,y\rangle = \langle x, T^*y\rangle$ it follows that if Tx=0 then $\displaystyle x\perp T^*y$. In other words, $\displaystyle \text{ker}(T)\subseteq(\text{Im}(T^*))^\perp$.

The same argument with T and T* interchanged shows that $\displaystyle \text{ker}(T^*)\subseteq(\text{Im}(T))^\perp$. Therefore $\displaystyle \text{null}(T^*)\leqslant n-\text{rank}(T) = \text{null}(T)$ by the rank plus nullity theorem, where n = dim(V). Interchanging T and T* again, you see that $\displaystyle \text{null}(T)\leqslant \text{null}(T^*)$, and therefore $\displaystyle \text{null}(T) = \text{null}(T^*)$. That shows (by the rank plus nullity theorem again) that $\displaystyle \text{rank}(T) = \text{rank}(T^*)$.

Finally, it follows that $\displaystyle \text{null}(T) = \text{rank}((T^*)^\perp)$. But we know that $\displaystyle \text{ker}(T)\subseteq(\text{Im}(T^*))^\perp$, and if these spaces have the same dimension then they must be equal.

3. thank u so much