Determine whether the set of vectors is a basis for the subspace of $\mathbb{R}^n$ that the vectors span

Question

\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?

Answer 1

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).

Answer 2

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