[Math] Is not full rank matrix invertible

determinantlinear algebralinear-transformationsmatricesmatrix-rank


$A$ 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 ?

Best Answer

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.

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