# Is not full rank matrix invertible?

## Problem

$$A$$ is a $$4 \times 4$$ matrix. It is known that $$\text{rank}(A)=3$$. Is matrix A invertible ?

## Attempt to solve

$$\text{rank(A)}=3 \implies \det(A)=0$$ which implies matrix is $$\textbf{not}$$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $$0$$ and that some information is lost in this linear transformation.

Is my intuition behind this correct ?

• Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together. – Doug M Dec 14 '18 at 16:14

Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.

The following properties are equivalent for a square matrix $$A$$:

• $$A$$ has full rank
• $$A$$ is invertible
• the determinant of $$A$$ is non-zero

There are more, but the first two are sufficient to immediately draw the desired conclusion.

This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $$f:U \to V$$ we have $$\mbox{rank}+\mbox{nullity} = \dim U$$ where the nullity is the dimension of the kernel, $$\ker f$$.

A four-by-four matrix represents a linear map $$f: U \to V$$ where $$\dim U = \dim V = 4$$. If the rank is three then $$3+\mbox{nullity}=4$$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.

$$rank(A) = 3 \Rightarrow det(A) = 0$$ needs to be proven, it is right though. basically, you can say that: $$rank(A) < dim(A) \Rightarrow det(A) = 0$$ but it still needs to be proven. an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.

If $$A$$ and $$B$$ are matrices for which $$AB$$ makes sense, then $$\operatorname{rank}(AB)\le\min\{\operatorname{rank}(A),\operatorname{rank}(B)\}$$ In particular, for every $$B$$, $$\operatorname{rank}(AB)\le\operatorname{rank}(A)=3$$. Can now $$AB=I$$?