An Algebraic Geometry view of singular matrix (The graph is for real number!)

Traditionally, we view ${GL}_n(\mathbb{C})$${GL}_n(\mathbb C)$ is automorphism of ${\mathbb{C}}^n$$\mathbb C^n$ in category of ${\mathrm{Vect}}_{\mathbb{C}}$$\mathrm{Vect}_{\mathbb C}$.

But actually, we can view it as an automorphism of a variety.

It just uses abstract nonsense to rewrite something well-known, just like the openest category, but on the other hand, it can be seen as a new view of old things. I do not know if it is powerful or not.

Notation $X^{n\times n}$$X^{n\times n}$ form a $n\times n$$n\times n$ matrix, each entry $a_{ij}=X_{ij}$$a_{ij}=X_{ij}$ is variable.

You can view $X^{n\times n}:{\mathbb{C}}^{n\times n}\to {\mathbb{C}}^{n\times n}$$X^{n\times n}:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ is just the identity function of affine space ${\mathbb{C}}^{n\times n}$$\mathbb C^{n\times n}$.

Then, you can consider the action ${\mathrm{End}}_{{\mathrm{Vect}}_{\mathbb{C}}}({\mathbb{C}}^{n\times n})\times X^{n\times n}$$\mathrm{End}_{\mathrm{Vect}_{\C}}(\mathbb C^{n\times n})\times X^{n\times n}$, that is $A{X}^{n\times n},$$AX^{n\times n},$ where $A$$A$ is a $n\times n$$n\times n$ matrix.

For convenience, I will denote $X^{n\times n}$$X^{n\times n}$ by $X$$X$, since the result is not dependent on the dimension (finite).

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Then, you can consider a map from ${\mathrm{End}}_{{\mathrm{Vect}}_{\mathbb{C}}}({\mathbb{C}}^{n\times n})\times X^{n\times n}⟶\mathbb{C}[X_{11},...,X_{n,n}]$$\mathrm{End}_{\mathrm{Vect}_{\C}}(\mathbb C^{n\times n})\times X^{n\times n}\longrightarrow\mathbb C[X_{11},...,X_{n,n}]$, that is, the determinant.

The solution of $detX=0$$\det X=0$ , or the variety $V(detX)$$V(\det X)$ is just all the singular matrix (not invertible matrix).

The obvious but interesting thing is $V(detGX)=V(detX)$$V(\det GX)=V(\det X)$ when $G\in GL(\mathbb{C})$$G\in GL(\mathbb{C})$. Since $det(GX)=detG\cdot detX$$\det(GX)=\det G\cdot\det X$.

Similarly $V(detFX)={\mathrm{End}}_{{\mathrm{Vect}}_{\mathbb{C}}}({\mathbb{C}}^{n\times n})$$V(\det FX)=\mathrm{End}_{\mathrm{Vect}_{\C}}(\mathbb C^{n\times n})$, since $det(FX)=detF\cdot detX=0$$\det(FX)=\det F\cdot\det X=0$. $F$$F$ is a singular matrix.

By the way, it looks like a Boolean Value, since ${\mathrm{End}}_{{\mathrm{Vect}}_{\mathbb{C}}}({\mathbb{C}}^{n\times n})$$\mathrm{End}_{\mathrm{Vect}_{\C}}(\mathbb C^{n\times n})$ equal to the disjoint union of the set of invertible and singular matrices.

When we consider the endomorphism act on the Variety, it maps all the $G$$G$ to ${\mathrm{id}}_{detX}$$\mathrm{id}_{\det X}$ or $1$$1$, and maps all the $F$$F$ to $0$$0$.

Actually, it is a continuous function From $f:{\mathrm{End}}_{{\mathrm{Vect}}_{\mathbb{C}}}({\mathbb{C}}^{n\times n})⟶{\mathbb{F}}_2$$f:\mathrm{End}_{\mathrm{Vect}_{\C}}(\mathbb C^{n\times n})\longrightarrow\mathbb F_2$. We define the topology on ${\mathbb{F}}_2$$\mathbb F_2$ as $\tau :=\{\mathrm{\varnothing },\{1\},{\mathbb{F}}_2\}$$\tau:=\{\empty,\{1\},\mathbb F_2\}$.

Since $detX$$\det X$ is a polynomial function; thus, it is continuous $g:{\mathbb{C}}^{n\times n}\to \mathbb{C}$$g:\mathbb C^{n\times n}\to\mathbb C$, then you quotient the relation $h:\mathbb{C}\to \mathbb{C}{P}^0$$h:\C \to\mathbb CP^{0}$.

Then $f=h\circ g$$f=h\circ g$ . $f^{-1}(0)=\{F\in M_{n\times n}(\mathbb{C})|detF=0\}$$f^{-1}(0)=\{F\in M_{n\times n}(\mathbb C)|\det F=0\}$ is closed. $f^{-1}(1)=G{L}_n(\mathbb{C})$$f^{-1}(1)=GL_n(\mathbb C)$ is open.

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There exists some much more interesting point of view.

1.Observe that $\mathrm{deg}(det{X}^{n\times n})=n$$\deg(\det X^{n\times n})=n$, and the coordinate ring of $V(det{X}^{n\times n})$$V(\det X^{n\times n})$, $\mathbb{C}[X_{11},...,X_{n,n}]/(det{X}^{n\times n})$$\mathbb C[X_{11},...,X_{n,n}]/(\det X^{n\times n})$.

2.You can consider the geometry of singular matrix.

For example, consider $2\times 2$$2\times 2$ matrix $\left(\begin{array}{cc}{X}_{11}& X_{12}\\ X_{21}& X_{22}\end{array}\right)$$\begin{pmatrix}X_{11} &X_{12}\\X_{21}&X_{22}\end{pmatrix}$What is the geometry of $X_{11}{X}_{22}-X_{21}{X}_{12}=0$$X_{11}X_{22}-X_{21}X_{12}=0$?

If you fix $X_{11}=1$$X_{11}=1$ then, you can visualise the surface $X_{22}=X_{21}{X}_{12}$$X_{22}=X_{21}X_{12}$, or $y=xz$$y=xz$

We might need to have a projective view of it Since if $\left(\begin{array}{cc}{X}_{11}& X_{12}\\ X_{21}& X_{22}\end{array}\right)$$\begin{pmatrix}X_{11} &X_{12}\\X_{21}&X_{22}\end{pmatrix}$ is the solution of $detX$$\det X$,

then so as $det(\lambda X)$$\det(\lambda X)$ or more general $detGX$$\det GX$.

We can consider $\begin{array}{c}=X_{11}\left(\begin{array}{cc}1& X_{12}/X_{11}\\ X_{21}/X_{11}& X_{22}/X_{11}\end{array}\right)\end{array}$$=X_{11} \begin{pmatrix}1 &X_{12}/X_{11}\\X_{21}/X_{11}&X_{22}/X_{11}\end{pmatrix}$

Each point on the surface is a family of solutions of $\begin{array}{c}det\left(\begin{array}{cc}1& Y_{12}\\ Y_{21}& Y_{22}\end{array}\right)=0\end{array}$$\det\begin{pmatrix}1 &Y_{12}\\Y_{21}&Y_{22}\end{pmatrix}=0$, that is the singular matrix.

The graph is as follows.

 

TT