Is it acceptable to use reduced row echelon to show basis?

Question

I am asked to show that {a, b, c} forms a basis for $\Bbb R^3$. I'm just wondering if it is acceptable to use reduced row echelon to show it since it is not shown that way in the marking scheme?

$$a= \begin{pmatrix} 2\\ -1\\ 1 \\ \end{pmatrix}b= \begin{pmatrix} 1\\ 1\\ 1 \\ \end{pmatrix} c= \begin{pmatrix} 0\\ 1\\ -1 \\ \end{pmatrix}$$

Here's how I did it:

For the vectors to form a basis, they must be linearly independent such that $$a_1\begin{pmatrix} 2\\ -1\\ 1 \\ \end{pmatrix}+a_2\begin{pmatrix} 1\\ 1\\ 1 \\ \end{pmatrix}+a_3\begin{pmatrix} 0\\ 1\\ -1 \\ \end{pmatrix}=\begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix}$$

So, I did reduced row echelon and I got.. $$\begin{bmatrix} 2 & 1 & 0 \\-1 & 1 & 1\\ 1 & 1 & -1 \\ \end{bmatrix} \rightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \\ \end{bmatrix}$$ After that, I solve for $a_1, a_2$ and $a_3$ $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \\ \end{bmatrix}\cdot \begin{Bmatrix} a_1\\ a_2\\ a_3 \\ \end{Bmatrix}=0$$

and I got $a_1=a_2=a_3=0$. Then, I conclude that {a,b,c} forms a basis.

Answer 1

Your proof is completely correct.

As a matter of style and comprehension, I'd change two very small (purely cosmetic) things before turning it in.

1. When you write "So, I did reduced row echelon and I got", I'd specifically mention that you are suppressing the column of zeroes in your augmented matrix. So I'd state "I used elementary row operations on this augmented matrix (with the column of zeroes suppressed) until getting RREF"
2. Once you'd gotten the RREF, I'd simply state "Converting this into a system of equations, we can see that $$\begin{cases} a_1 = 0 \\ a_2 = 0 \\ a_3 = 0\end{cases}$$ I would do this because $$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \\ \end{bmatrix}\cdot \begin{Bmatrix} a_1\\ a_2\\ a_3 \\ \end{Bmatrix}=0$$ is somewhat nonstandard notation. And my way gets to the point quicker.

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Edit: As hardmath pointed out to me in the comments, you should also mention at the end that because the dimension of $\Bbb R^3$ is $3$ (presumably this was proven at some point in your course?), any three linearly independent vectors will form a basis.

Answer 2

There are many ways to skin a cat in linear algebra.

Strictly speaking you showed the columns $a,b,c$ are linearly independent. It remains to say something about their spanning $\mathbb{R}^3$ to complete the proof that these form a basis.

At some point in our linear algebra studies we learn that every linearly independent set can be extended to a basis, and that all bases of vector spaces have the same size, so in a finite dimensional setting, knowing that $\mathbb{R}^3$ is three-dimensional allows us to assert $\{a,b,c\}$ is a basis by counting how many linear independent vectors we have.

Your calculation could also serve to demonstrate that the determinant is nonzero, since we know how elementary row operations affect the determinant. Since the end result of your row reduction was the identity matrix (nonzero determinant), you can argue that the original determinant was also nonzero.

Perhaps this is simply a matter of not having said quite enough of the right words to reassure the grader that you know the definition of basis and/or the specific equivalences (there are many) in terms of ranks of matrices.

My point was that the row reduction of a matrix preserves the row space. Thus since your steps produced rows that form a standard basis of $\mathbb{R}^3$, the original rows were also a basis. But the question you were asked was about the columns of that matrix, so perhaps a few extra words about equality of row and column rank would have sufficed. Or perhaps you haven't covered that topic yet.