4
$\begingroup$

Given $A, B \in \mathbb{R}^{n\times n}, t \in \mathbb{R}\setminus \{0\}$ with $b_{ij} = t^{i-j}\cdot a_{ij}$. Prove $\det(A) = \det(B)$.

I first thought of induction. I can easily prove this for $n \le 2$.

My induction hypothesis: $\det(A) = \det(B)$ with $A, B \in \mathbb{R}^{n\times n}$

Induction step: $\det(B) = \sum_{i=1}^{n+1} b_{ij} \cdot (-1)^{i+j} \cdot \det(B_{ij})\overset{IH}{=} \sum_{i=1}^{n+1} t^{i-j} a_{ij} \cdot (-1)^{i+j} \cdot \det(A_{ij})$

So far, so good, but I can't seem to get rid of the exponentiation of $t$. Any thoughts?

  • 0
    That never occurred to me but you're right.2012-05-15

2 Answers 2

3

Let $ T=\begin{bmatrix}t \\ & t^2\\ & & \ddots\\ & & & t^n\end{bmatrix} $ (and zeroes off-diagonal). Then $ B=TAT^{-1}, $ so $\det B=\det T\; \det A\;(\det T)^{-1}=\det A$.

  • 0
    My pleasure. $\phantom{XXXX}$2012-05-16
2

In my answer here I give the permutations-based definition of the determinant, which is equivalent to all other standard definitions.

This result is almost immediate using the permutations-based definition.

$ \begin{align} \det(B) & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^nB_{i,\sigma(i)}\\ & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^n(t^{i-\sigma(i)}A_{i,\sigma(i)})\\ & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^nt^{i-\sigma(i)}\prod_{i=1}^nA_{i,\sigma(i)}\\ \end{align} $

And $\prod_{i=1}^nt^{i-\sigma(i)}=1$, since the sum over all $i$ gives $t^{\binom{n+1}{2}-\binom{n+1}{2}}=t^0=1$. So

$ \begin{align} \det(B) & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^nA_{i,\sigma(i)}\\ &= \det(A) \end{align} $