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If the RREF of a matrix is the identity matrix, would the standard basis be a basis for its column space? And would the theorem that says a basis for the column space is the corresponding columns with the leading ones still hold? Thanks!

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  • $\begingroup$ Yes, if the RREF is the identity matrix of size $n \times n$, then the matrix has full rank (rank $n$), so the column space is all of $\mathbb{R}^n$, and one such basis is the standard basis. $\endgroup$ – angryavian Jan 9 '16 at 3:43
  • $\begingroup$ @angryavian: Thanks for replying! Can you explain more why when the matrix has full rank, the column space is all of R^n? $\endgroup$ – secretsO_o Jan 10 '16 at 5:32
  • $\begingroup$ I don't know what definition of rank you are using, but the rank of a matrix is the dimension of the column space. Since your matrix has columns of length $n$, the column space is a subspace of $\mathbb{R}^n$. The only $n$-dimensional subspace of $\mathbb{R}^n$ is $\mathbb{R}^n$ itself. $\endgroup$ – angryavian Jan 10 '16 at 17:21
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If the RREF of your matrix $A$ is an identity matrix, then we're dealing with an $n \times n$ matrix.

A basis for the column space of $A$ is the set of columns of $A$ corresponding to the columns with leading 1s in the RREF of $A$.

In your case, every column in the RREF of $A$ has a leading 1, so the $n$ column vectors of $A$ form a basis, and the column space has dimension $n$, i.e. it's all of $\mathbb{R}^n$.

Since the column space of $A$ is $\mathbb{R}^n$, any basis of $\mathbb{R}^n$ is also a basis for the column space. So yes, the standard basis of $\mathbb{R}^n$ is a basis for the column space of $A$.

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