Equivalence of a null determinant and non-null kernel I know that, for a complex square matrix $A$, there exists some non-null vector $v$ such that $Mv=0$ if and only if the determinant, $\det(M)=0$. That is, $$ \exists v: Mv=0 \Leftrightarrow \det(M)=0.$$
What is the simplest, least technical way to prove this theorem?
 A: $M\cdot v$ : linear combination of the columns of the matrix where the coefficient comes from corresponding components of the vectors.
\begin{align}M\cdot v&=\pmatrix{col_1 &col_2&...&col_n\\}\cdot \pmatrix{v_1\\v_2\\...\\v_n}\\&=v_1\cdot col_1+v_2\cdot col_2+\dots+v_n\cdot col_n\\\end{align}
Result: Columns of a sqare matrix $A$ linearly dependent iff $\det(A) =0$. (Use Gaussian elimination, determinant is the product of all pivots.)
$A\cdot v=0$ for some non zero $v\iff $ columns are linearly dependent $\iff \det(A) =0$.
A: The easiest way I know is to use the multiplicativity of the determinant and Cramers rule. More precisely: Let $I_n$ denote the $n\times n$ identity matrix.
For any two matrices $A,B\in M_{n,n}(K)$ it holds that $\det(AB)=\det(A)\det(B)$. So if $Av=0 \implies v=0$, we have that the map $K^n \to K^n, v\mapsto Av$ is injective. Now $K^n$ is of finite dimension over $K$, hence injectivity implies that the map is an isomorphism of $K^n$ (rank-nullity-theorem). Hence there exists an inverse matrix $A^{-1}\in M_{n,n}(K)$. Therefore $1=\det(I_n)=\det(AA^{-1})=\det(A)\det(A^{-1})$, which implies $\det(A)\neq 0$.
Now suppose $\det(A)\neq 0$. Then by Cramers rule (see here for reference)  we have $A \operatorname{adj}(A) = \det(A)I_n$, where $\operatorname{adj}(A)\in M_{n,n}(K)$ denotes the adjoint matrix of $A$. Because $\det(A)\neq 0$ we obtain $A\cdot (\frac{1}{\det(A)}\operatorname{adj}(A))=I_n$ hence $A$ is invertible and thus for any $v\in K^n$ we get $Av=0 \Rightarrow A^{-1}Av=v=0$.
A: By definition, the determinant $\det : M(n, \mathbb{C}) \to \mathbb{C}$ is the unique alternating multilinear map on the space of $n \times n$ complex matrices satisfying $\det(I) = 1$. Here "alternating" means the $\det(A)$ changes sign when you swap two columns of $A$, and multilinear means the determinant is linear in each column of $A$ when you hold the other columns fixed. The fact that such a function $\det$ actually exists can be proven by expanding out the determinant using the properties, and then verifying that the resulting formula (Leibniz formula) satisfies the properties.
The alternating property implies that if $A$ has two identical columns, then $\det(A) = 0$. Multilinearity then implies that if some column of $A$ is a linear combination of the others, then $\det(A) = 0$. Thus if $Av = 0$ for some $v$, then $\det(A) = 0$.
Conversely, if $Av = 0$ only for $v = 0$, then the rank-nullity theorem implies $A$ is invertible. The identity $\det(AB) = \det(A)\det(B)$, which is not too difficult to deduce from the definition of $\det$, then yields
$$1 = \det(I) = \det(A)\det(A^{-1}).$$
Thus $\det(A) \neq 0$.
