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I got a couple questions regarding a proof of a well known property of the determinant. I'm not sure if the proof is correct (found it online):

Proposition: If $B$ is a matrix gotten from $A$ by swapping two columns, then $\det(B) = - \det(A)$.

Proof: Let $\tau$ be the transposition which gives the column swap taking $A$ to $B$. Note, that since $\tau$ is a transposition, we have $sgn(\tau) = -1$. Then we can write \begin{align*} \det(B) &= \sum_{\sigma \in \mathcal{S}_n} sgn(\sigma) a_{1 \tau \circ \sigma(1)} \ldots a_{n \tau \circ \sigma(n)} \\ &= - \sum_{\sigma \in \mathcal{S}_n} sgn(\tau) \cdot sgn(\sigma) a_{1 \tau \circ \sigma(1)} \ldots a_{n \tau \circ \sigma(n)} \\ &= - \sum_{\sigma \in \mathcal{S}_n} sgn(\tau \circ \sigma) a_{1 \tau \circ \sigma(1)} \ldots a_{n \tau \circ \sigma(n)} \end{align*}But $\tau$ is fixed, and as $\sigma$ runs through the elements of $\mathcal{S}_n$, $\sigma \circ \tau$ also runs through the elements of $\mathcal{S}_n$. Replacing $\sigma \circ \tau$ with $\sigma$ gives us $-\det(A)$. Q.E.D.

I wonder what he means when he says: "Let $\tau$ be the transposition which gives the column swap." Can this be done with only one transposition? Also, I'm not sure if the last part of the proof is correct. The author speaks of replacing $\sigma \circ \tau$ with $\sigma$. But composition is not commutative, and in his proof he spoke of $\tau \circ \sigma$.

Any clarifications and/or improvements would be appreciated.

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  • $\begingroup$ Swapping two columns amounts to replacing $\sigma $ with $\beta =\tau \sigma $ and $sgn\beta =-sgn\sigma$. $\beta $ is what the author is referring to by saying "we may replace..."etc... $\endgroup$ – Matematleta Jun 11 '15 at 22:08
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A transposition is by definition a permutation that exchanges two elements, so yes, two columns can be exchanged with exactly one transposition.

For the remainder of your question, there appears to be an error, but not a serious one. The author surely meant to use $\tau\circ \sigma$ instead of $\sigma\circ \tau$. After correcting that error the proof is correct.

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  • $\begingroup$ Ok, but why can we replace $\tau \circ \sigma$ with just $\sigma$ at the end of the proof? $\endgroup$ – Kamil Jun 12 '15 at 7:34
  • $\begingroup$ @Kamilnl b because both indices ranges over all permutations. Summing over the same set. $\endgroup$ – Matt Samuel Jun 12 '15 at 10:32

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