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Rank-nullity theorem Let V, W be vector spaces, when V is finite dimensional. Let T:V$ \to $W be a linear transformation. Then Rank(T)+Nullity(T)=dimV

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$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

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Steinitz exchange lemma If $ \{v_1,\ldots,v_m\} $ is a set of linearly independent vectors in vector space V, and $ \{w_1,\ldots,w_n\} $ spans V, then:

Possibly after reordering the $ w_i $, $ \{v_1,\ldots,v_m,w_{m+1},\ldots,w_n\} $ spans V.
$ m\,\leq\,n $

$ H_m $: Steinitz exchange lemma holds for all non-negative integers m.
$ H_0 $: Steinitz exchange lemma holds when $ m=0 $ $ \emptyset $ is a set of linearly independent vectors in vector space V. Based on the conditions given $ \{w_1,\ldots,w_n\} $ spans V. And clearly, $ m=0\leq n $
$ H_k $: Steinitz exchange lemma holds for some non-negative integers $ k $
$ H_{k+1} $: Steinitz exchange lemma holds for $ (k+1) $. Based on $ H_k $, $ \{v_1,\ldots,v_k,w_{k+1},\ldots,w_n\} $ spans V.
\[ v_{k+1}\in \textbf{V}\Leftrightarrow v_{k+1}=\sum_{i=1}^k{\lambda_i v_i}+\sum_{i=k+1}^n{\lambda_i w_i} \]
Since $ v_1,\ldots,v_{k+1} $ are linearly independent,

\begin{align*} \Rightarrow &v_{k+1}-\sum_{i=1}^k{\lambda_i v_i}\neq0\\ \Rightarrow &v_{k+1}\neq\sum_{i=1}^k{\lambda_i v_i}\\ \Rightarrow &\text{At least one of }\lambda_{k+1},\ldots\lambda_n~\text{is non-zero} \end{align*}
Rearrange $ w_i $ so that $ \lambda_{k+1}\neq0 $

\begin{align*} \Rightarrow &v_{k+1}=\sum_{i=1}^k{\lambda_i v_i}+\lambda_{k+1}w_{k+1}+\sum_{i=k+2}^n{\lambda_i w_i}\\ \Leftrightarrow &w_{k+1}=\frac{1}{-\lambda_{k+1}}(\sum_{i=1}^k{\lambda_i v_i}-v_{k+1}+\sum_{i=k+2}^n{\lambda_i w_i}) \end{align*}
$ \forall v\in $V, $ \exists $ a linear combination of$ ~\{v_1,\ldots,v_k,w_{k+1},\ldots,w_n\}=v $ Now we substitute $ w_{k+1}=\frac{1}{-\lambda_{k+1}}(\sum_{i=1}^k{\lambda_i v_i}-v_{k+1}+\sum_{i=k+2}^n{\lambda_i w_i}) $ into the linear combination. $ \Rightarrow\forall v\in $V, $ \exists $ a linear combination of$ ~\{v_1,\ldots,v_k,v_{k+1},w_{k+2},\ldots,w_n\}=v $ Therefore $ H_{k+1} $ holds, if $ H_k $ is true.
Based on M.I., $ H_m $ is true.