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It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.

For an explicit example where $\text{rank}_+^*(V) > \text{rank}_+(V) $, consider the matrix

$$ V= \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$

It is easy to see that $\text{rank}_+^*(V) =4 $ and $\text{rank}_+(V)=3$.

It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.

For an explicit example where $\text{rank}_+^*(V) > \text{rank}_+(V) $, consider the matrix

$$ V= \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$

It is easy to see that $\text{rank}_+^*(V) =4 $ and $\text{rank}_+(V)=3$. This example also shows that your claimed inequality $\text{rank}_+^*(V) \leq n-1 $ does not always hold.

It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.

It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.

For an explicit example where $\text{rank}_+^*(V) > \text{rank}_+(V) $, consider the matrix

$$ V= \begin{bmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \end{bmatrix} $$

It is easy to see that $\text{rank}_+^*(V) =4 $ and $\text{rank}_+(V)=3$.

It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.

For an explicit example that shows that these quantities are not the same, consider the matrix

$$ V= \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix} $$

Every column of $V$ is a conic combination of the two vectors $[1,1,0]^T$ and $[0,1,1]^T$. Thus, $\text{rank}_+(V)=2$. On the other hand, it is easy to see that $\text{rank}_+^*(V)=3$

It is not true that $\text{rank}_+^*(V) \leq \text{rank}_+(V) $. In fact, an equivalent definition of the non-negative rank of $V$ is the minimum number of non-negative vectors (not necessarily columns of $V$) such that every column of $V$ is a conic combination of these vectors. Therefore, the opposite inequality $\text{rank}_+^*(V) \geq \text{rank}_+(V) $ holds.