Intuition behind a matrix being invertible iff its determinant is non-zero

Question

Question

I have been wondering about this question since I was in school. How can one number tell so much about the whole matrix being invertible or not?

I know the proof of this statement now. But I would like to know the intuition behind this result and why this result is actually true.

---

My Proof

If $A$ is invertible, then $$ 1 = \det(I) = \det(AA^{-1}) = \det(A)\cdot\det(A^{-1})$$ whence $\det(A) \neq 0$. Conversely, if $\det(A) \neq 0$, we have $$ A adj(A) = adj(A)A = \det(A)I$$ whence $A$ is invertible. $adj(A)$ is the adjugate matrix of $A$. $$ adj(A)_{ji} = (-1)^{i+j}\det(A_{ij})$$ where $A_{ij}$ is the matrix obtained from $A$ by deleting $ith$ row and $jth$ column.

Any other insightful proofs are also welcome.

Answer 1

Another classical way is more understandable: note that a determinant is not changed if we add one row to other and one column to another. Thus we obtain a diagonal matrix $B$. This matrix differs from $A$ by matrix-multipliers which correspond to elementary transformations and are invertible. So $A$ is invertible iff $B$ is invertible iff $\det(B) \neq 0$ iff $\det(A) \neq 0$.

Answer 2

The absolute value of the determinant of a matrix is the volume of the parallelepiped spanned by the column vectors of that matrix.

Michael

Answer 3

Here's an explanation for three dimensional space ($3 \times 3$ matrices). That's the space I live in, so it's the one in which my intuition works best :-).

Suppose we have a $3 \times 3$ matrix $\mathbf{M}$. Let's think about the mapping $\mathbf{y} = f(\mathbf{x}) = \mathbf{M}\mathbf{x}$. The matrix $\mathbf{M}$ is invertible iff this mapping is invertible. In that case, given $\mathbf{y}$, we can compute the corresponding $\mathbf{x}$ as $\mathbf{x} = \mathbf{M}^{-1}\mathbf{y}$.

Let $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ be 3D vectors that form the columns of $\mathbf{M}$. We know that $\det{\mathbf{M}} = \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$, which is the volume of the parallelipiped having $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ as its edges.

Now let's consider the effect of the mapping $f$ on the "basic cube" whose edges are the three axis vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$. You can check that $f(\mathbf{i}) = \mathbf{u}$, $f(\mathbf{j}) = \mathbf{v}$, and $f(\mathbf{k}) = \mathbf{w}$. So the mapping $f$ deforms (shears, scales) the basic cube, turning it into the parallelipiped with sides $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$.

Since the determinant of $\mathbf{M}$ gives the volume of this parallelipiped, it measures the "volume scaling" effect of the mapping $f$. In particular, if $\det{\mathbf{M}} = 0$, this means that the mapping $f$ squashes the basic cube into something flat, with zero volume, like a planar shape, or maybe even a line. A "squash-to-flat" deformation like this can't possibly be invertible because it's not one-to-one --- several points of the cube will get "squashed" onto the same point of the deformed shape. So, the mapping $f$ (or the matrix $\mathbf{M}$) is invertible if and only if it has no squash-to-flat effect, which is the case if and only if the determinant is non-zero.