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I'm asking in regards to Question 8 here: http://www.math.ucla.edu/~jheez/hw2solutions.pdf

The answer given says that you can stop searching once you've identified four vectors that are a basis to this $\mathbb{R}^5$ space, since the last [unknown] one is the negative sum of all the others.

I don't understand how they arrived at that conclusion. How can you know forsure? And what is the rule in general? What is it called?

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A good explanation probably depends on how you've defined the term "basis." –  user17794 Sep 27 '12 at 19:54

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"you can stop searching once you've identified four vectors that are a basis to this R5 space, since the last [unknown] one is the negative sum of all the others".

That's not the idea.

What's true is that within each vector of this subspace, its last, fifth coordinate is determined by the first four coordinates of that vector, as the negative sum of the first four.

Given four independent vectors in the subspace, they generate that space without adding any last unknown vector, because the dimension is $4$ for the subspace, and this is why one can stop at four. If you did introduce a fifth vector that is the negative sum of the first four it would not "complete" or make them any more of a basis, for either the subspace or the whole space, than they were by themselves. The (negative) sum is, by definition, in the subspace spanned by the four vectors so cannot add any more material that enlarges that subspace. So the idea cannot be to take four vectors and add a fifth, it is about the relationship between the coordinates of individual vectors (in the given subspace, which is defined as vectors with a coordinate sum of zero).

"why is the dimension less than 5?"

Finite dimensional vector spaces have a well-defined dimension so that any subspace has a unique dimension regardless of the specific manner in which it was defined (say, as the space of solutions to some linear equations, or the subspace generated by several given vectors, or something else). This is not a logically obvious statement, it is something proved in a linear algebra book and a basic reason the theory works.

Here the subspace was defined by imposing one equation, so its dimension will be one less than the dimension of the entire space: $4 = 5-1$. The dimension calculated in terms of number of independent vectors that can all fit inside the space has to equal this and the problem is about finding a particular set of four vectors that illustrates that principle.

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To summarize what you're saying (tell me if I got this): This vector space only requires 4 vectors since there are only four independent components in each vector and not five because the 'fifth' component is always a combination of the other 4. –  Imray Sep 28 '12 at 2:08
    
I just reread the question and realized I missed that part! I was wondering how you noticed that by looking at all the vectors! –  Imray Sep 28 '12 at 2:11
    
Your first comment is correct, but to talk formally about components of vectors being independent one can write down the whole set of vectors, usually as a rectangular matrix where the rows are the vectors, and then the columns of the matrix are vectors that represent the components. –  zyx Oct 2 '12 at 22:48
    
... I don't follow.. If you have independent vectors, you can write them as rows in a rectangular matrix and then the columns represent components of what? –  Imray Oct 4 '12 at 17:30
    
Of the rows. In a rectangular matrix you can read the rows or the columns as vectors. Saying that the first four, but not the first five, components in each (row) vector are independent is only loosely correct. If you try to pin down precisely what it means, it is really saying that the first four column vectors are independent but the set of all five column vectors is not. –  zyx Oct 5 '12 at 2:03

The subspace $W$ descriped in that problem is a proper subsapce of $\mathbb R^5$, hence has smaller dimension, i.e. $\dim W<5$. Once you have found four linearly independant vectors, you have thus established that $\dim W\ge 4$, hence $\dim W=4$ and this implies that your linear indepent system is already a basis.

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Why is the dimension less than 5?? –  Imray Sep 27 '12 at 20:03
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If $W\subset V$ are finite-dimensional vector spaces, then $\dim W\le \dim V$ (because you can produce a basis of $V$ by possibly adding vectors to a basis of $W$ - a basis of $W$ is at least an independent family of $V$) and equality holds only if $W=V$. Here $W\ne V=\mathbb R^5$ because you can easily find a vector not in $W$, and of course $\dim \mathbb R^5 = 5$. –  Hagen von Eitzen Sep 27 '12 at 20:08

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