1

So I have been reading countless posts on extending a matrix basis, however I still am unable to grasp it and apply what I've read to my problem.

w1 = [1,0,-2,3] w2 = [1,-2,3,-1] w3 = [1,-8,4,0]

W = sp(w1, w2, w3)

With the above provided, I was asked to find a basis for W, which I solved to be:

{$[1,0,-2,3]^T$, $[1,-2,3,-1]^T$, $[1,-8,4,0]^T$}

I got this answer by reducing the matrix formed by w1, w2, and w3, and checking for linear independence, then selecting the pivot columns.

Anyways, the question I am stuck on is how do I now extend this basis for $R^4$? As I mentioned previously, I have read multiple previous questions on extending a basis. However I am still confused and unable to apply what I read previously to this question.

Any help is appreciated!

Marc van Leeuwen
  • 115,048
Matthew V.
  • 113
  • The goal is to find a column vector that is not in $\textrm{span}{w_1,w_2,w_3}$ (provided $w_1,w_2,w_3$ are linearly independent). – Justin Benfield Apr 13 '16 at 05:02
  • One important point to realise is that adding the right amount of vectors (and you need only one here) almost always works; you just need to avoid the rare situation of linear dependence between the given vectors and the vectors you add. And if you are out of inspiration for fresh vectors, just take the standard basis (or any basis, but why not go for the simplest) and take any vector(s) that pass the linear independence test (such a subset always exists). – Marc van Leeuwen Apr 13 '16 at 05:13

2 Answers2

1

There's an algorithmic way to do this. Start by putting your vectors into the columns of a matrix $$ \left[\begin{array}{rrr} 1 & 1 & 1 \\ 0 & -2 & -8 \\ -2 & 3 & 4 \\ 3 & -1 & 0 \end{array}\right] $$ Then augment this matrix with the $4\times 4$ identity matrix $$ A= \left[\begin{array}{rrr|rrrr} 1 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & -2 & -8 & 0 & 1 & 0 & 0 \\ -2 & 3 & 4 & 0 & 0 & 1 & 0 \\ 3 & -1 & 0 & 0 & 0 & 0 & 1 \end{array}\right] $$ Now, row reduce this matrix to find its reduced row-echelon form. In our case we have $$ \DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrr|rrrr} 1 & 0 & 0 & 0 & \frac{1}{8} & \frac{1}{4} & \frac{1}{2} \\ 0 & 1 & 0 & 0 & \frac{3}{8} & \frac{3}{4} & \frac{1}{2} \\ 0 & 0 & 1 & 0 & -\frac{7}{32} & -\frac{3}{16} & -\frac{1}{8} \\ 0 & 0 & 0 & 1 & -\frac{9}{32} & -\frac{13}{16} & -\frac{7}{8} \end{array}\right] $$ The pivot columns are the first four columns. This implies that the first four columns of $A$ form a basis for $\Bbb R^4$. Of course, the first three of these columns are your vectors by design.

Brian Fitzpatrick
  • 26,559
0

An alternative way to Brian's is to determine the solution space to $W$, i.e. solve for some vector $\mathbf{x}$ $$ \left(\begin{array}{ccc|c} \vdots & \vdots & \vdots & \vdots \\ \mathbf{w}_1 & \mathbf{w}_2 & \mathbf{w}_3 & \mathbf{x}\\ \vdots & \vdots & \vdots & \vdots \\ \end{array}\right) $$ for which at least the bottom row in the w-matrix will be zero on the LHS, and non-zero on the RHS. If solution exists, then the RHS of this zero row must equal to zero, so you will get a (or more) linear equation(s) in terms of $x$'s. All vectors in $W$ must satisfy these equations (try with the $\mathbf{w}$'s). So pick a vector $\mathbf{u}$, which does not satisfy it $\implies \mathbf{u} \notin W$, which is obviously linearly independent of the others.