Relationship between null-space and determinant

Question

Assume a real-valued square $n\times n$ matrix $A$ has a non-trivial nullspace.

Why is the following true?

Since A has a non-trivial nullspace, it is not invertible and thus $\det A = 0$

I understand that a matrix is not invertible if $\det A = 0$, but I don't understand how having a non-trivial nullspace implies that a matrix has no inverse.

Answer 1

This is the key insight to this proof: If we have $Ax=b$, then by multiplying both sides by $A^{-1}$ on the left, we get $x=A^{-1}b$. Thus, if $A$ is invertible, then $Ax=b$ has only one solution.

Now, let's consider $Ax=\mathbf{0}$. If $A$ has a non-trivial null space, that, by definition, means that there are multiple possible solutions for this equation. However, if $A$ is invertible, then $Ax=\mathbf{0}$ only has one solution, as shown in the first paragraph. Thus, $A$ can not be invertible and have a non-trivial null space at the same time, so if it has a non-trivial null space, it must be not invertible.

Answer 2

If $A$ has nontrivial null space then there exists a vector $v\neq 0$ such that $Av=0$. As matrix multiplication is a linear operation this implies that $A(\lambda v)=0$ for every scalar $\lambda$. Particularly $A$ is not injective, and hence not invertible.