# Show $X^TX$ is not invertible [duplicate]

Suppose that the first column of $$X \in R^{N \times (p+1)}$$ is full of $$1$$s. Show that in the following cases, there exists no inverse matrices for $$X^TX$$. (a). $$N < p+1$$ (b). $$N \ge p+1$$ and two columns of $$X$$ are the same.

So far, I know 1. for a matrix to be invertible, it has to be a square matrix. 2.$$A^{-1}A = I = AA^{-1}$$ for an invertible matrix. I do not understand the statement "the first column is full of $$1$$" How should this information be used to solve the problem? Could someone explain the idea?

• – Trevor Gunn Dec 8 '19 at 2:57
• Not sure what the first column has to do with anything. If $N < p + 1$ or $X$ has two identical columns then it has a non-trivial null space. Then if $Xv = 0$ we have $X^TXv = 0$ so $X^TX$ cannot be invertible. – Trevor Gunn Dec 8 '19 at 2:58