Let $w_1 = (0,1,1)$. Expand {$w_1$} to a basis of $R^3$.

I am reading the book, Applied Linear Algebra and Matrix Analysis. When I was doing the exercise of Section3.5 Exercise 7, I was puzzled at some of it. Here is the problem description:

Let $$w_1 = (0,1,1)$$. Expand {$$w_1$$} to a basis of $$R^3$$.

I don't understand its description well.
I think it wants to get a span set like {$$(0,1,1)$$, $$(1,0,0)$$, $$(0,0,1)$$} which is a basis of $$R^3$$.
And I check the reference answer, which is as followings:

$$(0,1,1)$$, $$(1,0,0)$$, $$(0,1,0)$$ is one choice among many.

I think what I have done is what question wants.
So can anyone tell me am I right or wrong?
Thanks sincerely.

• I think you are right Apr 16, 2019 at 6:02

There is a kind of 'procedure' for dealing with questions of this kind, namely to consider the spanning set $$\left\{ w_1, e_1, e_2, e_3\right\}$$. Consider each vector from left to right. If one of these vectors is in the span of the previous one/s, then throw it out. If not, keep it. So in this case, we start by keeping $$w_1$$. Moving to the next vector, $$e_1$$ is not in the span of $$w_1$$, so we keep it as well. Moving to the next, $$e_2$$ is not in the span of the previous two vectors so we keep it as well. Now, considering the vector $$e_3$$ we see that it is in fact in the span of the previous three vectors, since $$e_3 = w_1 - e_2.$$ So we throw out the vector $$e_3$$ and end up with the basis $$\left\{ w_1, e_1, e_2\right\}$$. This explains the solution in the reference answer. Your solution is also correct, however.
$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}$$ has independent rows. Hence you have found $$3$$ independent vectors in $$\mathbb{R}^3$$, that is it spans $$\mathbb{R}^3$$ and it forms a basis.
You are correct. {$$(0,1,1),(1,0,0),(0,0,1)$$} is a basis of $$\mathbb R^3$$.
Any element $$(a,b,c)$$ in $$\mathbb R^3$$ can be expressed as $$a(1,0,0)+b(0,1,1)+(c-b)(0,0,1).$$
• If your basis is $w_1, w_2, w_3$, the textbook's choice is $w_1, w_2, w_1-w_3$ Apr 16, 2019 at 6:08