Basis compatible

Definition

$V$ is a finite-dimension vector space, $\mathcal{B}$ is a basis for $V$, for any subspace $A\preceq V$, we call $\mathcal{B}$ is compatible with $A$ iff $\mathcal{B} \cap A$ is a basis of $A$.

Examples

• $\mathcal{B}=\left\{ (1,0,0),(0,1,0),(0,0,1) \right\}$ is a basis of $\mathbb{R}^3$, the $x-y$ plane is the subspace $A$, $\mathcal{B} \cap A=\left\{ (1,0,0),(0,1,0) \right\}$ is a basis of $A$​, so compatible.
• $\mathcal{B}=\left\{ (1,-1,0),(0,0,1),(0,1,-1) \right\}$ is a basis of $\mathbb{R}^3$, $A=\left\{ (x_1,x_2,x_3)|x_1+x_2+x_3=0 \right\}$, $\mathcal{B} \cap A=\left\{ (1,-1,0),(0,1,-1) \right\}$ is a basis of $A$, compatible.
• $\mathcal{B}=\left\{ (1,0),(0,1) \right\}$ is a basis of $\mathbb{R}^2$ , $A=\left\{ (x,y)|y=x \right\}$, $\mathcal{B} \cap A = \emptyset$, which implies that $\mathcal{B}$ is not compatible with $A$.

Property

For any subspace $A_1,A_2 \preceq V$, there is a basis $\mathcal{B}$ of $V$ which is both compatible with $A_1$ and $A_2$.

proof:

As already known that any set of linearly independent vectors in $V$ can be expanded to a basis of $V$, so

$\mathcal{B}_{A_1 \cap A_2}=\left\{ x_1,...,x_r \right\}$, $\mathcal{B}_{A_1}=\left\{ x_1,...,x_r,y_1,...,y_s \right\}$, $\mathcal{B}_{A_2}=\left\{ x_1,...,x_r ,z_1,...,z_t\right\}$.​

We just need to prove $\left\{ x_1,...,x_r, y_1...,y_s,z_1,...,z_t\right\}$ is a linearly independent set.

Ausume that $\displaystyle{\sum_{i=1}^r \alpha_i x_i + \sum_{j=1}^s \beta_j y_j + \sum_{k=1}^t \gamma_k z_k = 0}$ and define $\displaystyle{\alpha = \sum_{i=1}^r \alpha_i x_i}$ $, \displaystyle{\beta=\sum_{j=1}^s \beta_j y_j}$ $,\displaystyle{\gamma= \sum_{k=1}^t \gamma_k z_k}$​.

Consider $\xi = \alpha + \beta = -\gamma$,

$\xi = \alpha + \beta \leftrightarrow x \in A_1$, $\xi=-\gamma \leftrightarrow x \in A_2$, so $\xi \in A_1 \cap A_2$, $\displaystyle{\xi = \sum_{i=1}^r \lambda_i x_i = -\sum_{k=1}^t \gamma_k z_k}$

$\mathcal{B}_{A_2}$ is linearly independent, so $\lambda_1,..., \lambda_r,\gamma_1,..., \gamma_t=0$ and $\alpha + \beta = 0$.

$\displaystyle{\sum_{i=1}^r \alpha_i x_i + \sum_{j=1}^s \beta_j y_j = 0} \rightarrow \alpha_i, \beta_j = 0$ . $\blacksquare$

Why can't generalize

The property above can't be promoted to the situation of $3$ subspaces.

Consider a basis of $\mathbb{R}_4:\left\{ a,b,c,d \right\}$ and $3$ subspaces with basis $\left\{a,b\right\}$, $\left\{c,d\right\}$, $\left\{a+c\right\}$, you'll find it's a counterexample!