# Proving or disproving this matrix V is invertible.

• MHB
• kaienfr
In summary: V$will also be an$m\times m$matrix. Since all diagonal elements of$V$are equal to$1$, the diagonal elements of the inverse of$V$will also be equal to$1$. Additionally, since all off-diagonal elements of$V$are strictly greater than$1$, the off-diagonal elements of the inverse of$V$will also be strictly greater than$1$. Therefore, the inverse of$V$will have all diagonal elements equal to$1$and all off-diagonal elements strictly greater than$1$, which means it will have a non-zero determinant. This further supports the conclusion that$V$is invertible.In summary, based on the observations about kaienfr Let us take$n\in \mathbb{N}^*$bins and$d\in \mathbb{N}^*$balls. Denote the set$B = \{\alpha^1, \ldots, \alpha^m\}$to be all possible choices for putting$d$balls into$n$bins, such as $$\alpha^1 = (d,0,\ldots, 0), ~ \alpha^2 = (0,d,\ldots, 0), \ldots$$ Let us define the matrix$V$as: $$V = \begin{bmatrix} (\alpha^1)^{\alpha^1} & \cdots & (\alpha^1)^{\alpha^m}\\ (\alpha^2)^{\alpha^1} & \cdots & (\alpha^2)^{\alpha^m}\\ \vdots & \vdots & \vdots\\ (\alpha^m)^{\alpha^1} & \cdots & (\alpha^m)^{\alpha^m} \end{bmatrix}$$ where the notation$(\alpha^i)^{\alpha^j} = \displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^j_k}$, under assumption that$0^0=1$. Now, I'd like to know: "is the matrix$V$invertible?" I have tested several examples, and it seems that$V$is always invertible, but I have no idea how to prove it or find a counterexample. Here are two facts I understood: 1. all diagonal elements are strictly positives, so the trace of$V$is strictly positive. 2. the matrix$V$is not symmetric. Does anyone can help to prove or disprove the invertibility of$V$? Thanks a lot in advance for sharing any idea. I would like to offer some insights on this problem. Firstly, let's consider the dimensions of the matrix$V$. Since there are$m$possible choices for putting$d$balls into$n$bins, the matrix$V$will be an$m\times m$matrix. Next, let's take a closer look at the diagonal elements of$V$. The diagonal elements are of the form$(\alpha^i)^{\alpha^i}$, which is equal to$\displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^i_k}$. Since$0^0=1$, we can rewrite this as$\displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^i_k} = \displaystyle\prod_{k=1}^{n}1^{\alpha^i_k} = 1$. Therefore, all diagonal elements of$V$are equal to$1$. Now, let's consider the off-diagonal elements of$V$. These elements are of the form$(\alpha^i)^{\alpha^j}$, where$i\neq j$. Recall that$(\alpha^i)^{\alpha^j} = \displaystyle\prod_{k=1}^{n}(\alpha^i_k)^{\alpha^j_k}$. Since$i\neq j$, there must exist at least one$k$such that$\alpha^i_k \neq \alpha^j_k$. This means that at least one term in the product will be greater than$1$, since$\alpha^i_k > \alpha^j_k$or$\alpha^i_k < \alpha^j_k$. Therefore, all off-diagonal elements of$V$are strictly greater than$1$. Based on these observations, we can conclude that all diagonal elements of$V$are equal to$1$, and all off-diagonal elements are strictly greater than$1$. This means that the determinant of$V$will be strictly greater than$1$. In general, a matrix with all diagonal elements equal to$1$and all off-diagonal elements strictly greater than$1$will have a non-zero determinant. Therefore, we can conclude that$V$is invertible. To further support this conclusion, let's consider the inverse of$V\$. The

## 1. Can you explain what it means for a matrix to be invertible?

A matrix is invertible if it has an inverse matrix, which when multiplied together, result in the identity matrix. In simpler terms, an invertible matrix can be "undone" or reversed.

## 2. How do you prove that a matrix is invertible?

To prove that a matrix is invertible, you can use a variety of methods such as finding the determinant, calculating the rank, or performing row operations to reduce the matrix to the identity matrix. If any of these methods result in a non-zero value, then the matrix is invertible.

## 3. What happens if a matrix is not invertible?

If a matrix is not invertible, it is called a singular matrix. This means that it does not have an inverse and cannot be "undone". In practical terms, this means that the matrix cannot be used to solve certain equations or perform certain operations.

## 4. Can a non-square matrix be invertible?

No, a non-square matrix cannot be invertible. In order for a matrix to have an inverse, it must have the same number of rows and columns, meaning it must be a square matrix.

## 5. Is the inverse of a matrix always unique?

Yes, the inverse of a matrix is always unique. This means that there can only be one inverse matrix for a given matrix. If a matrix has more than one inverse, it is not considered to be invertible.

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