Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
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
add comment

1 Answer

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.

share|improve this answer
    
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
1  
@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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.