Question

Prove that vectors (1,0,0), (0,1,0) and (0,1,1) span R^3 and are linearly independent.

Answer 1

To prove that they span R^3, you have to prove that there exist solutions to the system
\[\begin{cases}
x_1+0x_2+0x_3=b_1\\
0x_1+x_2+0x_3=b_2\\
0x_1+x_2+x_3=b_3
\end{cases}\]
Consider the system Ax=b, where \(b=(b_1,b_2,b_3)\), \(x=(x_1,x_2,x_3)\)and
\[A=\begin{bmatrix}
1 & 0 & 0\\
0 & 1& 0\\
0 & 1&1
\end{bmatrix}\]
By using cofactor expansion along the first row, you can see that A is invertible.
Thus, for all \(b\in\mathbb{R}^3\), there exists a unique solution \(x\) such that Ax=b.
The proof of linear independence is similar.

Answer 2

I was able to prove linear independence by the whole c1v1 c2v2c3v3 = 0 0 0 ideal, but the span was a little difficult. Could I have just said that three vectors that are linearly independent for sure span R^3, because they don't span R^2 or R^1 and there aren't enough vectors to span theoretical R^4? I wasn't sure what explanation to use for a first linear algebra exam, lol.

Answer 3

You just have to prove that if b is some arbitrary vector from R^3, then there exists x1, x2, and x3 such that x1v1+x2v2+x3v3=b.