Problem of the week - Is the spanning set a basis?

Vector spaces are one of the key subjects of linear algebra, and their theory has found application in mathematics, engineering, physics, chemistry, biology, the social sciences, and other areas. The theory, basically, consists in generalizing the familiar ideas of geometrical vectors of calculus to vectors of any size, but it provides an abstract, coordinate-free way of dealing with geometrical and physical objects such as tensors. The beauty of vector spaces theory can be found in every problem, where many of them, are just the appropriate linear-algebraic notion of very well known problems like solving systems of linear equations.

The problem of this week is related to spanning sets and bases, two key concepts from vector spaces you should master. The solution, as usual in most of Linear Algebra problems, uses basic concepts from matrices and their operations.

Find the values of 'a' such that the spanning set S is not a basis, and in such a cases, extract one basis from S.

S = {(6 -1 2 a); (-6 -2 -5 -4); (-2 -1 -2 a); (-4 a -3 -3)}

The spanning set will not be a basis when the vectors are linearly dependent. As there are 4 vectors, this problem boils down to find when the rank of the matrix whose columns are the given vectors is less than 4.

┌                ┐
│  6  -6  -2  -4 │
│ -1  -2  -1   a │
│  2  -5  -2  -3 │
│  a  -4   a  -3 │
└                ┘

Let's transform the matrix to row echelon form.

r1 <———> r2
┌                ┐
│ -1  -2  -1   a │
│  6  -6  -2  -4 │
│  2  -5  -2  -3 │
│  a  -4   a  -3 │
└                ┘
r2 <———> r2 + 6•r1
r3 <———> r3 + 2•r1
r4 <———> r4 + a•r1
┌                    ┐
│ -1   -2  -1      a │
│  0  -18  -8  -4+6a │
│  0   -9  -4     -7 │
│  0    0   0  -3+a² │
└                    ┘
r2 <———> r3
┌                    ┐
│ -1   -2  -1      a │
│  0   -9  -4     -7 │
│  0  -18  -8  -4+6a │
│  0    0   0  -3+a² │
└                    ┘
r3 <———> r3 - 2•r2
┌                   ┐
│ -1  -2  -1     -2 │
│  0  -9  -4     -7 │
│  0   0   0  10+6a │
│  0   0   0  -3+a² │
└                   ┘

At this point we can see that if a=-10/6 the third row is null, but the last one is not. The opposite occurs when a = ±√3. So, whatever be the value of 'a', the spanning set is never a basis, and the dimension of the spanned subspace is always 3.

To extract a basis from the spanning set, just check the echelon form and output the vectors which correspond to columns with pivots, it this case, the first, second and fourth column: {(6 -1 2 a); (-6 -2 -5 -4); (-4 a -3 -3)} where 'a' can take any value.