1. Sea ${\displaystyle A:={\begin{pmatrix}-1&1&0\\1&1&-2\\-2&1&1\end{pmatrix}}}$ . Entonces:
   ${\displaystyle \operatorname {Col} (A)=\langle (-1,1,-2),(1,1,1),(0,-2,1)\rangle =\langle (-1,1,-2),(1,1,1)\rangle }$ ;
   ${\displaystyle \operatorname {Fil} (A)=\langle (-1,1,0),(1,1,-2),(-2,1,1)\rangle =\langle (-1,1,0),(1,1,-2)\rangle }$ ;
   ${\displaystyle \operatorname {Nul} (A)=\langle (1,1,1)\rangle }$ .
   La matriz no tiene por qué ser cuadrada; veamos otro ejemplo:
2. Sea ${\displaystyle B:={\begin{pmatrix}-1&2\\3&-6\\-2&4\end{pmatrix}}}$ . Entonces:
   ${\displaystyle \operatorname {Col} (B)=\langle (-1,3,-2),(2,-6,4)\rangle =\langle (-1,3,-2)\rangle }$ ;
   ${\displaystyle \operatorname {Fil} (B)=\langle (-1,2),(3,-6),(-2,4)\rangle =\langle (-1,2)\rangle }$ ;
   ${\displaystyle \operatorname {Nul} (B)=\langle (2,1)\rangle }$ .

Para las relaciones de ortogonalidades entre conjuntos, siempre se considera el producto escalar estándar de ${\displaystyle K^{m}}$  o ${\displaystyle K^{n}}$ :

• ${\displaystyle \operatorname {Col} (A^{T})=\operatorname {Fil} (A)}$ 

• ${\displaystyle \operatorname {Fil} (A^{T})=\operatorname {Col} (A)}$ 

• ${\displaystyle \operatorname {Nul} (A)\perp \operatorname {Fil} (A)}$ 

• ${\displaystyle \operatorname {Col} (A)\perp \operatorname {Nul} (A^{T})}$ 

• ${\displaystyle \dim(\operatorname {Col} (A))=\dim(\operatorname {Col} (A^{T}))=\operatorname {rg} (A)=\operatorname {rg} (A^{T})}$ .

• Si ${\displaystyle A=(a_{ij})\in {\mathcal {M}}_{n\times n}(K)}$  y además las columnas de ${\displaystyle A}$ , ${\displaystyle \{(a_{11},\ldots ,a_{n1}),\ldots ,(a_{1n},\ldots ,a_{nn})\}}$ , forman un conjunto linealmente independiente de ${\displaystyle K^{n}}$ , entonces ${\displaystyle \det(A)\neq 0}$ , o sea, la matriz es invertible.

• Si ${\displaystyle A=(a_{ij})\in {\mathcal {M}}_{n\times n}(K)}$  y además ${\displaystyle \operatorname {Nul} (A)\neq \{0\}}$ , entonces ${\displaystyle \det(A)=0}$ , o sea, la matriz no es invertible.

• ${\displaystyle \dim(\operatorname {Col} (A))+\dim(\operatorname {Nul} (A))=n}$ .

• Sean ${\displaystyle A\in {\mathcal {M}}_{n\times m}(K)}$  y ${\displaystyle B\in {\mathcal {M}}_{p\times n}(K)}$ . Si ${\displaystyle x\in \operatorname {Col} (BA)}$ , entonces existe ${\displaystyle y\in K^{m}}$  tal que ${\displaystyle BAy=x}$ . Si tomamos ${\displaystyle w=Ay}$ , entonces ${\displaystyle Bw=x}$ , así que ${\displaystyle x\in \operatorname {Col} (B)}$ . Por lo tanto, ${\displaystyle \operatorname {Col} (BA)\subseteq \operatorname {Col} (B)}$ . Además ${\displaystyle \operatorname {Col} (BA)=\operatorname {Col} (B)}$  si y solo si ${\displaystyle \operatorname {rg} (A)=n}$ .

• Sean ${\displaystyle A\in {\mathcal {M}}_{n\times m}(K)}$  y ${\displaystyle B\in {\mathcal {M}}_{p\times n}(K)}$  —en particular, ${\displaystyle BA\in {\mathcal {M}}_{p\times m}(K)}$ —. Entonces si ${\displaystyle Ax=0}$ , también se tiene que ${\displaystyle BAx=0}$ . Así, ${\displaystyle \operatorname {Nul} (A)\subseteq \operatorname {Nul} (BA)}$ , y ocurre que ${\displaystyle \operatorname {Nul} (A)=\operatorname {Nul} (BA)}$  si y solo si ${\displaystyle \operatorname {rg} (B)=n}$ .

• Supongamos que ${\displaystyle K=\mathbb {R} }$  y sea ${\displaystyle A\in {\mathcal {M}}_{m\times n}(\mathbb {R} )}$ . Veamos que ${\displaystyle \operatorname {Nul} (A)=\operatorname {Nul} (A^{T}A)}$ . Sea ${\displaystyle x\in \operatorname {Nul} (A)}$ , entonces ${\displaystyle Ax=0}$ , por lo que ${\displaystyle A^{T}Ax=0}$ . Por otro lado, si ${\displaystyle x\in \operatorname {Nul} (A^{T}A)}$ , tenemos que ${\displaystyle A^{T}Ax=0}$ , por lo tanto ${\displaystyle x^{T}A^{T}Ax=0}$ . Como ${\displaystyle x^{T}A^{T}Ax=(Ax)^{T}Ax=\langle Ax,Ax\rangle }$ , donde ${\displaystyle \langle \cdot ,\cdot \rangle }$  denota el producto escalar estándar de ${\displaystyle \mathbb {R} ^{m}}$ , necesariamente ${\displaystyle Ax=0}$ , luego, ${\displaystyle x\in \operatorname {Nul} (A)}$ .

1. Matriz
2. Determinante de una matriz
3. Producto escalar estándar