Why is the nullity of an invertible matrix 0?

Question

I am aware that nullity is the dimension of the null space. So I assume that the null space is empty? But why?

Answer 1

The null space isn't empty, but it is the zero space (the subspace consisting of only the origin).

As to why a matrix is invertible if is has zero nullity, this comes back to what it means for a matrix (or more specifically a linear map) to be invertible. It means that you can reverse its effects. If a matrix has nullity above $0$, that means there is more than one vector that is sent to $\vec 0$. And if there is more than one vector which is sent to $\vec 0$, then you can't reverse the effects of the map, since given $\vec 0$, you can't know whether it's the result of applying the map to $\vec 0$, or some other vector in the null space.

The fact that if you can uniquely reverse $\vec 0$, then you can uniquely reverse any other result as well, and therefore the map / matrix is invertible, is an important theorem, although not a very difficult one (it's part of the invertible matrix theorem).

Answer 2

The unique vector space of dimension zero is the trivial space $\{0\}$ which includes the zero vector. This space is nonempty, and in fact the empty set is not a vector space because any vector space must have the zero vector.

By the invertible matrix theorem, one of the equivalent conditions to a matrix being invertible is that its kernel is trivial, i.e. its nullity is zero. I will prove one direction of this equivalence and leave the other direction for you to prove.

$(\Rightarrow)$ Suppose $A$ is an invertible $n\times n$ matrix. Let $v\in\ker A$ so that $Av=0$. Then $A^{-1}Av=A^{-1}0\iff Iv=0\iff v=0$. Thus, $\ker A=\{0\}$ so $A$ has nullity zero.

Answer 3

Let A be an $n\times n$ invertible matrix. The null space of the matrix is given by (the solution set of) $$ A\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}$$

Then$$A^{-1}\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}$$

This is the null space of A, containing only the point zero.

Answer 4

$(\Longleftarrow)$ Suppose that the nullity of a square matrix $(n\times n)$ A is equal to 0.

So this means that for any vector $(n\times 1)$, $Av=0$ has only $1$ solution, which is the trivial one: $v=0$.

And we know that this is equivalent to say that the determinant of A is nonzero.

And finally, because the determinant of A is nonzero, it means it is invertible

This completes the proof of equivalence as Dave already posted the $(\Longrightarrow)$.