The definition of determinat

Main reference: Linear Algebra Done Right by Sheldon Axler.

Definition

For vector space $𝑉$ with base field $𝕜$ and a fixed basis $({𝑒}_{𝑖})$, we use ${\mathrm{A}\mathrm{l}\mathrm{t}}_{𝑛}(𝑉)$ to denote its all alternating n-froms. It’s a well known fact that this is also a vector space with dimension $1$.

For any endomorphism $𝑇$ on $𝑉$, i.e., a matrix, we define a endomorphism of ${\mathrm{A}\mathrm{l}\mathrm{t}}_{𝑛}(𝑉)$ as $𝛼\mapsto {𝑇}^{\ast }𝛼$, the latter applied to a vector $({𝑣}_{𝑖})$ in $𝑉$ as ${𝑇}^{\ast }𝛼({𝑣}_{𝑖})=𝛼(𝑇{𝑣}_{𝑖})$.

We thus obtain a endomorphism ${𝑇}^{\ast }(-)$ on ${\mathrm{A}\mathrm{l}\mathrm{t}}_{𝑛}(𝑉)$, since ${\mathrm{A}\mathrm{l}\mathrm{t}}_{𝑛}(𝑉)\cong 𝕜$, ${𝑇}^{\ast }$ must be identified with a scalar multiplication $𝑐\cdot -$, and the constant $𝑐\in 𝕜$ is unique.

Finally, we define $\mathrm{d}\mathrm{e}\mathrm{t}𝑇$ as the unique constant $𝑐$ in $𝕜$.

Some quick facts

• The determinat of identity matrix is what you want: $\mathrm{d}\mathrm{e}\mathrm{t}(\mathrm{i}\mathrm{d})=1$ by definition.
• Determinat plays well with multiplication: $\mathrm{d}\mathrm{e}\mathrm{t}(𝐴𝐵)=\mathrm{d}\mathrm{e}\mathrm{t}𝐴\cdot \mathrm{d}\mathrm{e}\mathrm{t}𝐵$ for all morphisms $𝐴$ and $𝐵$. Again this is a immediately result by definition.
• Definition plays well with scalar: $\mathrm{d}\mathrm{e}\mathrm{t}(𝑐𝐴)={𝑐}^{𝑛}\mathrm{d}\mathrm{e}\mathrm{t}𝐴$, by definition.

• Determinat encodes the invertibility of morphisms: For all morphisms $𝐴$, $𝐴$ invertible if and only if $\mathrm{d}\mathrm{e}\mathrm{t}𝐴\ne 0$.

  • Proof: If $\mathrm{d}\mathrm{e}\mathrm{t}𝐴\ne 0$, we need to prove that $𝐴$ is an isomorphism. Thanks to the rank theorem of vector spaes, we only need to prove $𝐴$ is injective, which is equivalent to that kernel is trivial. For all non-zero $𝑣\in 𝑉$, we can extend it into a basis $(𝑣,{𝑒}_{𝑖})$, choose an alternating n-form $𝛼\ne 0$, we have $𝛼(𝑣,{𝑒}_{𝑖})\ne 0$, so $𝛼(𝑇𝑣,𝑇{𝑒}_{𝑖})={𝑇}^{\ast }𝛼(𝑣,{𝑒}_{𝑖})=\mathrm{d}\mathrm{e}\mathrm{t}𝑇\cdot 𝛼(𝑣,{𝑒}_{𝑖})\ne 0$. Note that for for vectors $({𝑤}_{𝑖})$, $𝛼({𝑤}_{𝑖})\ne 0$ if and only if these vectors independant. So, we have that $𝑇𝑣\ne 0$. Since $𝑣$ is arbitrary we deduce that $𝑇$ injective.

    The other direction is simple, due to determinat plays well with multiplication.