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I'm in the process of reading my first Linear Algebra textbook, and was just wondering...Is the standard basis of a vector space in n dimensions equivalent to the row space of the n x n identity matrix?

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You seem to be a bit confused there. The row space of a matrix is a vector subspace. The row space of the identity matrix is just the whole space. A basis on the other hand is a (usually ordered) set of elements of the vector space. What is true is that the standard basis is given by the rows of the identity matrix, but that's not a terribly exciting statement. – Alex B. Jan 27 '11 at 6:57
Yeah, I phrased the question wrong; that is what I wanted to know though. As I said, I'm in the process of learning Linear Algebra for the first time and just wanted to make sure that I understood the idea of a "standard basis". Thanks! – Dan M. Katz Jan 27 '11 at 16:13
up vote 3 down vote accepted

I assume you're talking about $\mathbb{R}^n$? First off, the standard basis $\{e_1,e_2,\dots,e_n\}$ is a linearly independent set of $n$ vectors, which spans the vector space.

However, the row space of $I_n$ is the set of all linear combinations of the row vectors, which gives a subspace of $n$-dimensional space, which is just the whole space in this case.

The key difference is that the standard basis is just a set of basis vectors, but the row space of $I_n$ is a vector space, not a standard basis, which are two different objects.

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I understand that they are two different objects, but I was inquiring as to whether the vectors that they contain are equivalent...I suppose I should have phrased this better. – Dan M. Katz Jan 27 '11 at 14:06
@user6300: The span of the standard basis is the same as the rowspace of the identity matrix. In fact, the rows of the identity are the standard basis for $\mathbb{R}^n$, if you view the latter as made up of row vectors (sometimes we prefer to view it as made up of column vectors). – Arturo Magidin Jan 27 '11 at 14:53
@user6300 Ah, my mistake that I didn't answer your intended question. Thankfully Arturo has better explained it to that end. – yunone Jan 27 '11 at 20:19

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