# Extending a basis that spans R^3 to R^5

Consider three vectors v=(2,-1,1,5,-3) w=(3,-2,0,0,0) z=(1,1,10,100,0) in R^5. Can the set {v,w,z} be completed to a basis for R^5? If yes, find explicit vectors to complete it.

My attempt: I have row reduced the matrix formed by these vectors down to a matrix 5x3 matrix, where the first 3 rows are the identity matrix for R^3. My question is: I know this matrix spans R^3, but I can also row interchange in any fashion so that it spans any 3 dimensions of R^5. How can I be specific with my choice of vectors in this case? Can I choose any 2 vectors that are not both not in the span of the column space and each other combined with the column space?

• There are an infinite choice of vectors that can extend your basis to $\mathbf {R}^5$. What do you mean by "be specific with my choice of vectors"? They just have to be 2 vectors in $\mathbf {R}^5$ that are not in the span of v, w, and z. – Dapianoman Oct 21 '17 at 0:08
• Your three vectors do not span $\mathbb R^3$. They span a three-dimensional subspace of $\mathbb R^5$. – amd Oct 21 '17 at 0:11
• Right, a subspace of R^3. How can I make sure that my 2 additional vectors are not in this three dimensional subspace? When I row interchange, it leads me to believe that the three dimensions of this subspace could be any combination of 3 out of the 5 dimensions. So how can I avoid choosing vectors in the span of my original 3? – Jungleshrimp Oct 21 '17 at 0:17

So you have your three vectors, $[2,-1,1,5,-3], [3,-2,0,0,0],[1,1,10,100,0]$. You wrote these down as a matrix, which you probably row reduced, as follows: $$\begin{pmatrix} 2 & -1 & 1 & 5 & -3 \\ 3&-2&0&0&0 \\ 1& 1&10&100&0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & -20 & -12 \\ 0&1&0&-30&-18 \\ 0& 0&1&15&3 \end{pmatrix}$$
Therefore, it is clear that the given row vectors span a three dimensional subspace of $\mathbb R^5$. When you want to complete this basis, and make it a basis of $\mathbb R^5$, what you do is make the above matrix upper triangular, as follows:
Note that I added two vectors which happen to be the fourth and fifth standard unit vectors for five dimensions, as two new rows. This matrix has non-zero determinant(why?)., and therefore the rows of this matrix form a basis for $\mathbb R^5$. However, note that the row span of the first three rows, is the same as that of $v,w,z$. Hence, this implies that $v,w,z$ and the fourth and fifth standard basis vectors in $\mathbb R^5$, actually form a basis for $\mathbb R^5$, hence completing the existing linearly independent set $v,w,z$ given to us.
These three vectors, $v,w,z\in\mathbb{R}^5$ do span a $3$-dimensional subspace of $\mathbb{R}^5$ (you already proved this, the right way), say $W$. Given that this subspace is dimensionally "little" with respect to the whole space, you have (mathematical) probability $1$ - choosing randomly other two vectors - to complete $\{v,w,z\}$ to a basis of $\mathbb{R}^5$. This fact justify us in taking $e_4=(0,0,0,1,0)^\top$ and $e_5=(0,0,0,0,1)^\top$ for a try. To demonstrate that $\mathcal{B}=\left(v,w,z,e_4,e_5\right)$ is indeed a base, it suffices to prove its indipendence (because being $\left|\mathcal{B}\right|=5=\dim\mathbb{R}^5$ this would be a maximal indipendent set - so even a generating set for the space). This is a simple problem to solve, take the matrix $M=\left(v|w|z|e_4|e_5\right)$ and show that $\det{M}\neq0$ to prove that $M$ is invertible and her columns indipendent.