Theorem: Construction of Basis for Vector Space from Basis for Vector Subspace and Basis of Quotient

Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $Y$ be a vector subspace of $X$.
Let $X/Y$ be the quotient vector space of $X$ modulo $Y$.
Let $q : X \to X/Y$ be the quotient mapping. 
Let $\BB_1 = \set {y_\alpha : \alpha \in I_1}$ be a basis for $Y$, where $I_1$ is an index set with cardinality $\dim Y$. 
Let $\widetilde {\BB_2} = \set {\map q {z_\beta} : \beta \in I_2}$ be a basis for $\map \dim {X/Y}$, where $I_2$ is an index set with cardinality $\map \dim {X/Y}$.
Let $\BB_2 = \set {z_\beta : \beta \in I_2}$.
Then $\BB_1 \cup \BB_2$ is a basis of $X$.
Further, $\BB_1 \cap \BB_2 = \O$.

Proof:
For $\beta, \beta' \in I_2$ with $\beta \ne \beta'$, we have $\map q {z_\beta} \ne \map q {z_{\beta'} }$ and hence $z_\beta \ne z_{\beta'}$.
We first establish that $\BB_1$ and $\BB_2$ are disjoint.
From Kernel of Quotient Mapping, we have that $\map q {y_\alpha} = {\mathbf 0}_{X/Y}$ for each $\alpha \in I_1$. 
Since $\widetilde {\BB_2} = \set {\map q {z_\beta} : \beta \in I_2}$ is a basis for $\map \dim {X/Y}$, it is in particular a linearly independent set and so:
:$\map q {z_\beta} \ne {\mathbf 0}_{X/Y}$ for each $\beta \in I_2$. 
Hence we must have $\BB_1 \cap \BB_2 = \O$.
We now show that $\BB_1 \cup \BB_2$ is a linearly independent set.
Let $\lambda_1, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in \GF$, $\alpha_1, \ldots, \alpha_n \in I_1$ and $\beta_1, \ldots, \beta_m \in I_2$ be such that:
:$\ds \sum_{j \mathop = 1}^n \lambda_j y_{\beta_j} + \sum_{j \mathop = 1}^m \mu_j z_{\beta_j} = {\mathbf 0}_X$
From Kernel of Quotient Mapping, we have:
:$\map q {y_{\beta_j} } = {\mathbf 0}_{X/Y}$ for each $1 \le j \le n$.
Hence from Quotient Mapping is Linear Transformation, we have:
:$\ds \sum_{j \mathop = 1}^m \mu_j \map q {z_{\beta_j} } = {\mathbf 0}_{X/Y}$
Since $\widetilde {\BB_2}$ is linearly independent, we have:
:$\mu_1 = \mu_2 = \ldots = \mu_m = {\mathbf 0}_K$
Hence we have:
:$\ds \sum_{j \mathop = 1}^n \lambda_j y_{\beta_j} = {\mathbf 0}_X$
Since $\BB_1$ is linearly independent, we have:
:$\lambda_1 = \lambda_2 = \ldots = \lambda_n = {\mathbf 0}_K$
So $\BB_1 \cup \BB_2$ is linearly independent.
We finally show that $\BB_1 \cup \BB_2$ spans $X/Y$.
Finally, take $x \in X$ so that $\map q x \in X/Y$.
Let $\beta_1, \ldots, \beta_n \in I_2$ and $\mu_1, \ldots, \mu_n \in K$ be such that:
:$\ds \map q x = \sum_{j \mathop = 1}^n \mu_j \map q {z_{\beta_j} }$
Hence by Quotient Mapping is Linear Transformation, we have:
:$\ds \map q {x - \sum_{j \mathop = 1}^n \mu_j z_{\beta_j} } = {\mathbf 0}_{X/Y}$
From Kernel of Quotient Mapping, we have:
:$\ds x - \sum_{j \mathop = 1}^n \mu_j z_{\beta_j} \in Y$
Then there exists $\alpha_1, \ldots, \alpha_m \in I_1$ and $\lambda_1, \ldots, \lambda_m \in K$ such that:
:$\ds x - \sum_{j \mathop = 1}^n \mu_j z_{\beta_j} = \sum_{j \mathop = 1}^m \alpha_j y_{\alpha_j}$
so that:
:$\ds x = \sum_{j \mathop = 1}^n \mu_j z_{\beta_j} + \sum_{j \mathop = 1}^m \alpha_j y_{\alpha_j}$
Hence $x \in \map \span {\BB_1 \cup \BB_2}$.
Hence $\BB_1 \cup \BB_2$ is a linearly independent spanning set of $X$.
{{qed}}
Category:Quotient Vector Spaces
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