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Recognizing the functions $e_k$ in David's excellent elaboration of Thierry's answer leads to a simple (equivalent) description and a proof for the problem considered here (but probably not the more general one of $n-1$ arbitrary powers.)

Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ x_1 & x_2 & x_3 & \ldots & x_{n-1} & t \\ x_1^2 & x_2^2 & x_3^2 & \ldots & x_{n-1}^2 &t^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_{n-1}^{n-1} & t^{n-1} \end{bmatrix}$$

Claim: The minor obtained from removing the row and column of $t^{n-k}$ is non-invertible exactly when the $x_i$ are the $n-1$ (distinct) roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$

Sketch: Think of $t$ as a variable and the various $x_j$ as parameters. The determinant is a sum of $n!$ terms each with total degree $n^2-n$ in $t$ and the $x_j$. It is also a polynomial $f(t)$ with coefficients polynomals in the $x_i$. This determinant is zero if any two of the columns are equal. Hence $f(t)$ is identically zero if any two $x_i$ are equal so $$f(t)=g(t)\prod_{0<i<j<n}(x_j-x_i)$$ Where $g(t)=e_0t^{n-1}+e_1t^{n-2}+\cdots+e_{n-2}t+e_{n-1}$ and $e_k$ has degree $k$ in the $x_j.$ Observe that

• $e_k$ is $\pm$ the determinant of the minor obtained by deleting the row and column of $t^{n-1-k}$ from $V$.
• $g(t)$ is zero when $t=x_j$ so $g(t)=\prod_{0<j<n}(t-x_j)$

Put this together to establish the claim. We also see that (up to a scalar) $$e_k(x_1,\cdots,x_{n-1}) = \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}.$$

Other notes:

• That expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter. Thus it is possible (with one exception) to choose values for any $n-2$ of the $x_j,$ along with a minor which should have determinant $0,$ and then solve for the value of remaining parameter. (Unless the choices already force that minor to be singular independent of the remaining value.)

It is worth looking at the three cases of the last row, the first row, and the other rows.

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as $k=0$ corresponds to the minor which eliminates the last row. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial form of the claim we see that there can't be enough distinct roots.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first row, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each column and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last row, then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last rows are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated row. Also, this is not a different case than $e_k=0$ since we just have the roots of $t^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k.$

Recognizing the functions $e_k$ in David's excellent elaboration of Thierry's answer leads to a simple (equivalent) description and a proof for the problem considered here (but probably not the more general one of $n-1$ arbitrary powers.)

Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ x_1 & x_2 & x_3 & \ldots & x_{n-1} & t \\ x_1^2 & x_2^2 & x_3^2 & \ldots & x_{n-1}^2 &t^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_{n-1}^{n-1} & t^{n-1} \end{bmatrix}$$

Claim: The minor obtained from removing the row and column of $t^{n-1-k}$ is non-invertible exactly when the $x_i$ are the $n-1$ (distinct) roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$

Sketch: Think of $t$ as a variable and the various $x_j$ as parameters. The determinant is a sum of $n!$ terms each with total degree $\frac{n^2-n}{2}$ in $t$ and the $x_j$. It is also a polynomial $f(t)$ with coefficients polynomals in the $x_i$. This determinant is zero if any two of the columns are equal. Hence $f(t)$ is identically zero if any two $x_i$ are equal so $$f(t)=g(t)\prod_{0<i<j<n}(x_j-x_i)$$ Where $g(t)=e_0t^{n-1}+e_1t^{n-2}+\cdots+e_{n-2}t+e_{n-1}$ and $e_k$ has degree $k$ in the $x_j.$ 

Observe that

• The determinant of the minor obtained by deleting the row and column of $t^{n-1-k}$ from $V\ $ is ${\displaystyle e_k \prod_{0<i<j<n}(x_j-x_i)}$
• $g(t)$ is zero when $t=x_j$ so $g(t)=\prod_{0<j<n}(t-x_j)$

Put this together to establish the claim. We also see that $$e_k(x_1,\cdots,x_{n-1}) = \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}.$$ (I've ignored the possible need for $\pm$ depending on the parity and order of the columns.)

Other notes:

• That expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter. Thus it is generally possible to choose values for any $n-2$ of the $x_j,$ along with a minor which should have determinant $0,$ and then solve uniquely for the value of remaining parameter. (I say generally because there are particular initial choices which force the minor to be singular or non-singular independent of the final value. Then a pertubation, however small, of any one chosen value restores the unique solution for the final value.)

It is worth looking at the three cases of the last row, the first row, and the other rows.

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as $k=0$ corresponds to the minor which eliminates the last row. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial form of the claim we see that there can't be enough distinct roots.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first row, we end up with an all zero column and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each column and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last row, then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last rows are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated row. Also, this is not a different case than $e_k=0$ since we just have the roots of $t^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k.$

This is a few comments on David's excellent elaboration of Thierry's answer.

We may as well assume that the minor eliminates the last row $[1\ x_n\ x_n^2\ \cdots \ x_n^{n-1}]$.

• The expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter so it is easy to set it to zero and solve.

• A cute equivalent description of the condition (for this case) is that the values $x_i$ should be the $n-1$ distinct roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$ This leads to an elementary proof for this case (see below).

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as this corresponds to a minor which eliminates the last column. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial reformulation we see that there can't be enough distinct roots.

I was glad to have this general result mentioned. It is interesting to see what be observed even without it.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first column, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each row and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last column then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last columns are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated column. Also, this is not a different case than $e_k=0$ since we just have the roots of $x^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k$

• Here is a simple proof for the polynomial form of this case (but probably not the more general one of $n-1$ arbitrary powers): Restore the last row and substitute $x_n=t$ thinking of $t$ as a variable and the other $x_i$ as parameters. The determinant is a polynomial of degree $n-1$ in $t$ and we know exactly when it is $0$, when $t=x_j$. Hence the determinant is $\pm\prod_1^{n-1}(t-x_i)$ and the coefficient of $t^{k}$ is $ \pm$ the minor for the last row and $k^\mbox{th}$ column.

Recognizing the functions $e_k$ in David's excellent elaboration of Thierry's answer leads to a simple (equivalent) description and a proof for the problem considered here (but probably not the more general one of $n-1$ arbitrary powers.)

Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ x_1 & x_2 & x_3 & \ldots & x_{n-1} & t \\ x_1^2 & x_2^2 & x_3^2 & \ldots & x_{n-1}^2 &t^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_{n-1}^{n-1} & t^{n-1} \end{bmatrix}$$

Claim: The minor obtained from removing the row and column of $t^{n-k}$ is non-invertible exactly when the $x_i$ are the $n-1$ (distinct) roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$

Sketch: Think of $t$ as a variable and the various $x_j$ as parameters. The determinant is a sum of $n!$ terms each with total degree $n^2-n$ in $t$ and the $x_j$. It is also a polynomial $f(t)$ with coefficients polynomals in the $x_i$. This determinant is zero if any two of the columns are equal. Hence $f(t)$ is identically zero if any two $x_i$ are equal so $$f(t)=g(t)\prod_{0<i<j<n}(x_j-x_i)$$ Where $g(t)=e_0t^{n-1}+e_1t^{n-2}+\cdots+e_{n-2}t+e_{n-1}$ and $e_k$ has degree $k$ in the $x_j.$ Observe that

• $e_k$ is $\pm$ the determinant of the minor obtained by deleting the row and column of $t^{n-1-k}$ from $V$.
• $g(t)$ is zero when $t=x_j$ so $g(t)=\prod_{0<j<n}(t-x_j)$

Put this together to establish the claim. We also see that (up to a scalar) $$e_k(x_1,\cdots,x_{n-1}) = \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}.$$

Other notes:

• That expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter. Thus it is possible (with one exception) to choose values for any $n-2$ of the $x_j,$ along with a minor which should have determinant $0,$ and then solve for the value of remaining parameter. (Unless the choices already force that minor to be singular independent of the remaining value.)

It is worth looking at the three cases of the last row, the first row, and the other rows.

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as $k=0$ corresponds to the minor which eliminates the last row. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial form of the claim we see that there can't be enough distinct roots.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first row, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each column and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last row, then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last rows are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated row. Also, this is not a different case than $e_k=0$ since we just have the roots of $t^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k.$

This is a few comments on David's excellent elaboration of Thierry's answer.

We may as well assume that the minor eliminates the last row $[1\ x_n\ x_n^2\ \cdots \ x_n^{n-1}]$.

• The expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter so it is easy to set it to zero and solve.

• A cute equivalent condition is that the values $x_i$ should be the $n-1$ distinct roots of a polynomial $$a_0x^{n-1}+a_1x^{n-2}+\cdots+a_{n-2}x+a_{n-1}$$ with $a_k=0.$

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as this corresponds to a minor which eliminates the last column. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial reformulation we see that there can't be enough distinct roots.

I was glad to have this general result mentioned. It is interesting to see what be observed even without it.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first column, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each row and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last column then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last columns are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated column. Also, this is not a different case than $e_k=0$ since we just have the roots of $x^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the result so nicely described.

This is a few comments on David's excellent elaboration of Thierry's answer.

We may as well assume that the minor eliminates the last row $[1\ x_n\ x_n^2\ \cdots \ x_n^{n-1}]$.

• The expression $$e_k(x_1,\cdots,x_{n-1}) := \sum_{0<i_1 < i_2 < \cdots < i_k<n} x_{i_1} x_{i_2} \cdots x_{i_k}$$ is linear in each parameter so it is easy to set it to zero and solve.

• A cute equivalent description of the condition (for this case) is that the values $x_i$ should be the $n-1$ distinct roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$ This leads to an elementary proof for this case (see below).

• It seems that $e_0$ would be the zero function. But one should require $k>0$ as this corresponds to a minor which eliminates the last column. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial reformulation we see that there can't be enough distinct roots.

I was glad to have this general result mentioned. It is interesting to see what be observed even without it.

• The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first column, we end up with an all zero row and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each row and what remains is again a Vandermonde matrix and nonsingular.

• If the minor eliminates neither the first nor the last column then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last columns are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated column. Also, this is not a different case than $e_k=0$ since we just have the roots of $x^{n-1}-1.$

That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k$

• Here is a simple proof for the polynomial form of this case (but probably not the more general one of $n-1$ arbitrary powers): Restore the last row and substitute $x_n=t$ thinking of $t$ as a variable and the other $x_i$ as parameters. The determinant is a polynomial of degree $n-1$ in $t$ and we know exactly when it is $0$, when $t=x_j$. Hence the determinant is $\pm\prod_1^{n-1}(t-x_i)$ and the coefficient of $t^{k}$ is $ \pm$ the minor for the last row and $k^\mbox{th}$ column.