\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?
Best Answer
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).