Finite Dimensional Vector Space proof: No linear maps such that $PQ-QP=I$

Question

Question: In quantum mechanics, physicists make use of linear maps P and Q on an infinite dimensional vector space so that $PQ-QP=I$. Prove that if V is a finite dimensional vector space, and $P:V \rightarrow V$ and $Q: V \rightarrow V$ are linear maps, then $PQ-QP \ne I $ (there is no quantum mechanical theory in finite dimensions).

Attempt: I was a bit put off by this question but I think I might have worked out a solution and I'm not sure if it's acceptable. I decided to use the trace property of matrices. $$Tr(PQ) = \sum_{i}\sum_{j}p_{ij}q_{ji} $$ $$Tr(QP)=\sum_{j}\sum_{i}q_{ji}p_{ij}$$ $$Tr(PQ-QP) = \Big( \sum_{i}\sum_{j}p_{ij}q_{ji} \Big)-\Big(\sum_{j}\sum_{i}q_{ji}p_{ij}\Big) = 0 < Tr(I)$$

Is there something I'm missing?

Yup, it is correct...because I am assuming you work over $;\Bbb R;$ or $;\Bbb C;$, right? In fact, over any field with char$,F\neq\dim V;$ it will be fine. — DonAntonio, Nov 30 '16 at 11:17

score 1 · Answer 1 · answered Nov 30 '16 at 11:17

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Assuming that $V$ is a vector space over some field with characteristic 0 (e.g. $\mathbb{R}$ or $\mathbb{C}$), your solution is correct.

answered Nov 30 '16 at 11:17

Dominik

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Finite Dimensional Vector Space proof: No linear maps such that $PQ-QP=I$

1 Answers1