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Show that

$$det \begin{bmatrix} 1 & 1 & \cdots &1 \\ \lambda_1 & \lambda_2 & \cdots &\lambda_n \\ \lambda^2_1 & \lambda^2_2 & \cdots &\lambda^2_n \\ \vdots & \vdots & \vdots & \vdots \\ \lambda^{n-1}_1 & \lambda^{n-1}_2 & \cdots &\lambda^{n-1}_n \\ \end{bmatrix} =\prod_{n\ge i>j\ge1}(λ_i−λ_j)$$

My work so far,

I am trying to use induction on $n$. For the base case n=2, the claim is true, since we have $(\lambda_2 - \lambda_1)$.

I'm not sure how to carry out the inductive step.

Any hints or suggestions are welcome.

Thanks,

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The matrix (or its transpose) is called the Vandermonde matrix. One proof of its determinant is given in the article here.

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Hint: Adding a scalar multiple of one column to another column leaves the determinant unchanged.

Use transformation $C_i=C_i-C_1, 2\le i\le n$ (leaving $(1,0,0,...0)$ in first row) and then try induction?

Edit: As mentioned in the other answer, the matrix is called Vandermonde Determinant and proof in lines of what I have mentioned above is there in this proofwiki article (Proof 1)

There a second operation, namely $a_{ij}=a_{ij}-x_1\times a_{i-1,j-1}$ after above transformation is used to prove $V_n=\prod_{k=2}^n (x_k-x_1)V_{n-1}$, where we know $V_{n-1}$ by induction hypothesis.

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  • $\begingroup$ Hi @JessePFrancis, I have my first row now as [1 0 0 ... 0], with determinant unchanged, as you said. Now, if I use the Laplace expansion and expand along the first row, the only non-zero contribution is 1* det[(nx1) x (nx1)] submatrix, from which I can apply the induction hypothesis. Am I done? Seems way too simple, I must have missed something... $\endgroup$ – User001 Nov 13 '15 at 4:38
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    $\begingroup$ @LebronJames, true, we hit roadblock. See the other answer. $\endgroup$ – Jesse P Francis Nov 13 '15 at 5:35

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