Do you need other results to prove this? Could you use proof by induction?
I don't know where to begin regarding how to prove this.
Asked
Active
Viewed 550 times
0
-
2A good place to start is the definition. With that being said... what is your definition of determinant? Because if you were to be using the Leibniz definition of determinants then this should be rather trivial. If you were instead taught using the Laplace definition then it takes a bit more effort. – JMoravitz Jun 23 '20 at 18:31
-
It depends on your definition of determinant. It is straightforward, if tedious, to do from the algebraic definition. – copper.hat Jun 23 '20 at 18:31
-
Using the explicit formula for the determinant of an $n \times n$ matric $A = (a_{i,j})$, it seems fairly easy to derive it: $$\det(A)=\sum_{\sigma \in S_n} \left( \text{sgn}(\sigma) \prod_{i=1}^n a_{i, \sigma(i)} \right)$$ – PrincessEev Jun 23 '20 at 18:32
-
I have the Laplace definition of the determinant. – Henry Brown Jun 23 '20 at 18:39
2 Answers
0
The determinant can be defined as (Using Einstein's summation notation):
$\det(\mathbf{A}) = \varepsilon_{i_1\dots i_n} a_{1i_1} \dots a_{ki_k} \dots a_{mi_m} \dots a_{ni_n}$
Therefore exchanging the files k and m
$\det(\mathbf{A'}) = \varepsilon_{i_1\dots i_k\dots i_m\dots i_n} a_{1i_1} \dots a_{mi_k} \dots a_{ki_m} \dots a_{ni_n}$
Using the antisymmetry of the levi-civita tensor and renaming the dummy index you get:
$\det(\mathbf{A'}) = -\varepsilon_{i_1\dots i_m\dots i_k\dots i_n} a_{1i_1} \dots a_{mi_k} \dots a_{ki_m} \dots a_{ni_n}=-\varepsilon_{i_1\dots i_k\dots i_m\dots i_n} a_{1i_1} \dots a_{mi_m} \dots a_{ki_k} \dots a_{ni_n}=-\det(\mathbf{A})$
carlorop
- 308
0
Let $A'$ be $A$ with exchanged $i$-th and $j$-th column.
Let $I'$ be the $n * n$ identity matrix with $i$-th and $j$-th rows exchanged. Using both Leibniz or Laplace determinant definition it's easy to prove that $\det(I') = -1$.
Because exchanging $i$-th and $j$-th column of matrix $A$ is basically multiplying matrix $A$ by $I'$ so $\det(A') = \det(A \cdot I')$, we get $$\det(A') = \det(A \cdot I') = \det(A) \cdot \det(I') = -det(A)$$
We proved that exchanging two columns multiplies determinant by $-1$. Because $\det(A) = \det(A^T)$, so for exchanging rows it's the same.
