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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?

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I'm not sure that your induction step is valid. You replace $\det(B_{ij})$ with $\det(A_{ij})$ simply because these are $(n-1)\times(n-1)$ matrices. But the indexing on the cofactors does not preserve the pattern of $i-j$ as the exponent on $t$. – alex.jordan May 15 '12 at 18:46
That never occurred to me but you're right. – hauptbenutzer May 15 '12 at 21:31
up vote 3 down vote accepted

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

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I think this is the most elegant solution. Thank you. – hauptbenutzer May 15 '12 at 21:29
My pleasure. $\phantom{XXXX}$ – Martin Argerami May 16 '12 at 16:55

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

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

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