A related question is Why are the residuals in $R^{n−p}$?.

A geometric interpretation of OLS is that the model space is the linear span of vectors and contains all the potential fitted values.

We need that each point in that space has a unique coordinate in terms of the model parameters (otherwise a prediction can be explained by multiple parameter values).

This means that the columns in the design matrix, that represents the predictors in the OLS model, are linearly independent. This requires that there are at least as many observations as parameters.

It is necessary condition but not a sufficient condition. A counterexample is when we have an intercept $x_0 =(1,1,1,1)$ and two predictors $x_1 = (1,1,0,0)$ and $x_2 = (0,0,1,1)$ with a model $y = a + b x_1 + c x_2$ then we can have multiple model values with the se solution. For example $a=1,b=1,c=1$ gives $y = (2,2,2,2)$ but also $a=2,b=0,c=0$ will give that $y = (2,2,2,2)$ answer.

The problem in that counterexample is that there is a linear dependence and we can write $x_0-x_1-x_2=0$.

A related question is Why are the residuals in $R^{n−p}$?.

A geometric interpretation of OLS is that the model space is the linear span of vectors and contains all the potential fitted values.

We need that each point in that space has a unique coordinate in terms of the model parameters (otherwise a prediction can be explained by multiple different sets of parameter values).

This means that the columns in the design matrix, that represents the predictors in the OLS model, are linearly independent. This requires that there are at least as many observations as parameters.

It is necessary condition but not a sufficient condition. A counterexample is when we have an intercept $x_0 =(1,1,1,1)$ and two predictors $x_1 = (1,1,0,0)$ and $x_2 = (0,0,1,1)$ with a model $y = a + b x_1 + c x_2$ then we can have multiple model values with the same solution. For example $a=1,b=1,c=1$ gives $y = (2,2,2,2)$ but also $a=2,b=0,c=0$ will give that $y = (2,2,2,2)$ answer.

The problem in that counterexample is that there is a linear dependence and we can write $x_0-x_1-x_2=0$.

Another complicating issue is that the observations need to be in a space that has sufficiently high dimensions. With OLS we often imagine independent normal distributed variables, but this may not need to be the case. The observations can be for instance discrete values and it might be possible that all the observed outcome values are equal to zero. Then we can have more than $k+1$ observations and a model space without dependent columns, and the fitting still doesn't work.