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.



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


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @LebronJames, true, we hit roadblock. See the other answer. $\endgroup$ – Jesse P Francis Nov 13 '15 at 5:35

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