Inverse and singularity relation reason

Question

Why can't a square matrix whose determinant is zero or in other words has linearly dependent rows/columns be invertible?

Answer 1

We'll assume that $\det(A) = 0$ iff $A$ has linearly independent columns.

Saying that $A$ has linearly independent columns is precisely the same as stating that there is a non-zero column vector $x$ for which $Ax = 0$.

Now, suppose for the sake of contradiction that this matrix $A$ had an inverse $A^{-1}$. We could then take the equation $Ax = 0$ on both sides by $A^{-1}$ conclude that $A(A^{-1}x)$, which is $x$, is equal to $A^{-1}(0)$, which is $0$. But we said that $x \neq 0$, so this is a contradiction.

An interesting observation here is that $A$ has linearly independent rows if and only if it has linearly independent columns. One way to see this is to show that $\det(A) = \det(A^T)$. Another is to consider the matrix $A^TA$ as an intermediary.

Answer 2

Let's consider a matrix $A=(v_1,...,v_n) \in K^{n \times n}$ (that means $v_1,...,v_n \in K^n$ are the columns) with $det(A)=0$. As you mentioned, we then know that the columns of the matrix are linearly dependent. So there exist $a_1,...,a_n \in K$ (at least one of them is $\neq 0$) such that $a_1v_1+..+a_nv_n = (0,...,0)^T$.

Of course $A \cdot (0,...,0)^T = 0 \cdot v_1 + ... + 0 \cdot v_n = (0,...,0)^T$. Each square marix has this property. But on the other hand, $A \cdot (a_1,...,a_n)^T = a_1v_1+..+a_nv_n = (0,...,0)^T$ by Definition of the $a_i$. So both $(0,...,0)^T$ and $(a_1,...,a_n)^T$ are mapped to $(0,...,0)^T$ by $A$. Note that $(a_1,...,a_n) \neq (0,...,0)$.

Now, if an inverse matrix $A^{-1}$ existed, then we had $A^{-1} \cdot (0,...,0)^T = (0,...,0)^T$ as well as $A^{-1} \cdot (0,...,0)^T = (a_1,...,a_n)^T$. (You just have to multiply the equations of the last passage by $A^{-1}$ from the left side to see this. Use $A^{-1} \cdot A = \mathbb{I}_n$.) But that is not possible because this operation is well-defined, so $A$ cannot be invertible.