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\begin{bmatrix} 1 & -1 &0 \\ 0& 1 &2 \\ 2& 1 & -3\\ 1 &-3 & 4 \end{bmatrix}

Determine if the set of column vectors of the matrix above form a basis.

The column vectors are $[1,0,2,1],[-1,1,1,-3],[0,2,-3,4]$

After forming a series of row operations, I get

\begin{bmatrix} 1 & 0 &0 \\ 0& 1 &0 \\ 0& 0 & 1\\ 0 &0 &0 \end{bmatrix}

Since all the columns have pivots, that means that all three columns form a basis, right?

So the vectors $[1,0,2,1],[-1,1,1,-3],[0,2,-3,4]$ form a basis. But what dimension? I know that these vectors must span an entire dimension, and it can't span $\mathbb{R}^4$ since there are only three vectors. So it must span $\mathbb{R}^3$. So my answer would be that $[1,0,2,1],[-1,1,1,-3],[0,2,-3,4]$ are a basis that span $\mathbb{R}^3$.

Is this right?

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    By definition the dimension of a space is the cardinality of a basis. Since you have 3 vectors which are linearly independent, they span a 3-dimensional space. They **do not** span $\Bbb R^3$, remember $\Bbb R^3$ is sets of 3-tuples of real numbers, none of your vectors are 3-tuples. – 2017-02-12
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    So then what dimension are they a basis of? – 2017-02-12
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    I already answer that in my last comment. I think you're assuming all three dimensional spaces are just automatically $\Bbb R^3$, which is false. (They are what is called "isomorphic" but that's another story). – 2017-02-12
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    @AdamHughes: But they _are_ a basis for _the subspace they span_, as stated in the question title. (This is just a roundabout way of saying they are linearly independent). – 2017-02-12
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    So can I say that these vectors are a basis for the 3 dimensional space they span. It that correct? – 2017-02-12
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    @HenningMakholm yeah, I don't dispute that. I was just giving the op a heads up that this thing is not the thing which is called $\Bbb R^3$. – 2017-02-12

2 Answers 2

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The Gaussian elimination that you performed showed only one thing: that the three original vectors are linearly independent

Because they are linearly independent, they form a basis for the space that they span. That space is not all of $\mathbb{R}^4$, but is a three dimensional subspace of $\mathbb{R}^4$. For that subspace, they form a basis.

But they are certainly not a basis for all of $\mathbb{R}^4$. All of $\mathbb{R}^4$ is $4$-dimensional: all of its bases have exactly $4$ (linearly independent) vectors.

Finally, that subspace is not $\mathbb{R}^3$. It's just a $3$-dimensional subspace of $\mathbb{R}^4$. The symbol $\mathbb{R}^3$ refers to "triplets of numbers". Here, we're are not dealing with triplets of numbers. We are still dealing with quadruplets of numbers (which, to repeat, all live in a $3$-dimensional subspace of $\mathbb{R}^4$).

Here's a link to the appropriate place in our Linear Algebra course: http://lem.ma/Dh (although you may need to start a little bit earlier to get used to our approach).

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    Are you a company? Thanks for the answer – 2017-02-13
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    I'm personally not a company, I'm a teacher of mathematics. But I've started a learning company called Lemma. And when I come here, I look for questions that I have already answered in detail on lem.ma and so the answers I provide have pretty good *backup*. – 2017-02-13
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We're in $R^{4}$. Do the columns form a basis? The key-phrase here is a basis for what space within $R^4$?

All the bases for $R^4$ have 4 vectors, so they're not a basis for $R^4$ because the number of vectors in any basis for a space is equal to its dimension; and the dimension of $R^4$ is 4 and those columns are only 3 in number.

More specifically, we know they're not a basis for $R^4$ because they fail to span $R^4$ (instead of failing to be linearly independent, or both) because back-solving the echelon matrix you obtained yields only the trivial solution (x1=x2=x3=0). This indicates linear independence, and since we know it's not a basis, we're forced to conclude the columns do not span $R^4$ to avoid a logical contradiction.

However, the columns are linearly independent, and do span their span; therefore they are a basis for the subspace they span (Sounds nearly tautological, doesn't it?).

Thus, the short answer is yes, they are a basis in $R^4$, but not for $R^4$.

Adam V. Nease