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This question already has an answer here:

My textbook says that its true but I can't find a proof of this on the Internet.

"If A is an $n \times n$ matrix with linearly independent columns, then the columns of A span $\mathbb{R}^{n}$."

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marked as duplicate by user147263, Dirk, Community Nov 25 '15 at 22:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It is a fundamental fact. $n$ linearly independent vectors in an $n-$ dimensional space form a basis. $\endgroup$ – David Peterson Nov 25 '15 at 19:35
  • $\begingroup$ Can you elaborate why it is the case? Or link to a proof? $\endgroup$ – Jerry Dean Nov 25 '15 at 19:37
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The dimension of a finite dimensional vector space is the maximal number of linearly independent vectors, and it is also the minimal number of a system of generators. A maximal system of linearly independent vectors is a basis. A minimal system of generators is a basis.

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  • $\begingroup$ But why do any two maximal systems of linearly independent vectors have to have the same number of elements? This isn't obvious at all, and is what the question is asking. $\endgroup$ – Eric Wofsey Nov 25 '15 at 22:46
  • $\begingroup$ It all depends on what the O.P. is supposed to know. It was not apparent to me the O.P. wanted to know why two bases have the same cardinality. $\endgroup$ – Bernard Nov 25 '15 at 23:18
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Notice that the columns of the matrix (lets call it $A$) are $n$ linearly independent vectors. On the other hand, any basis of $\mathbb{R}^n$ requires $n$ linearly independent vectors. You already have them: they are the vectors of $A$.

Formally, let $x \in \mathbb{R}^n$ then \begin{equation} x = \sum\limits_{i=1}^{n} \alpha_i v_i \end{equation} where $\alpha_i$ denote scalars and $v_i$ denote linearly independent vectors. In particular there is a change of coordinates that transforms $\alpha_i v_i \to \beta_i A_i$ where $A_i$ are the columns of your matrix. And thus you can write any $x$ as: \begin{equation} x = \sum\limits_{i=1}^{n} \beta_i A_i \end{equation}

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The idea is that, if this matrix is called $A$, and the columns are linearly independent, then $A$ is invertible. This means given any system $$Ax=b$$ we can solve for $x$ simply by multiplying by $A^{-1}$, so $$x=A^{-1}b$$

Since $b$ could have been any vector in $\mathbb{R}^n$, the columns span $\mathbb{R}^n$, because $Ax$ is just a linear combination of the columns of $A$.

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  • $\begingroup$ But why is $A$ invertible? This is just a paraphrase of the question itself, not anything resembling a proof. $\endgroup$ – Eric Wofsey Nov 25 '15 at 22:43

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