MHB Extending Vectors to a Basis of R^4: Why Notation Matters

[Yankel]

Hello all

I am trying to solve this problem:

Extend the following vectors to a basis of R^4.

\[u_{1}=\left ( \begin{matrix} 1\\1 \\1 \\1 \end{matrix} \right )\]

and

\[u_{2}=\left ( \begin{matrix} 2\\2 \\3 \\4 \end{matrix} \right )\]

What I did, I put these vectors as columns of a matrix, and surprisingly I found that they are already linear independent. In the book where I took it from, they put the vectors as rows, and they were dependent. I don't understand. I always put vectors as columns when I want to check for dependency, span, or linear combination. How come this time it has to be as rows ? Thank you.

 

[Nono713]

Are you sure the book is using the same notation as you, i.e. multiplying vectors on the right rather than the left? If it is using the "opposite" notation then things will be changed around, including row/column stuff. Your question isn't very clear.

Anyway, your two vectors are certainly linearly independent, but they aren't a basis of R^4, for they do not span R^4 (their span is not R^4 but some hyperplane of dimension 2). Remember that for a set of vectors to be a basis of a vector space they need to be linearly independent as well as span the vector space. Thus a basis of R^4 necessarily is a set of exactly four vectors, since R^4 has dimension 4.

To "extend" this set of vectors to a basis of R^4 you need to find another vector that cannot be expressed as a linear combination of the two you have, and then you need to find one more vector that cannot be expressed as a linear combination of the previous three. At that point you will find that every vector you choose can be written as a linear combination of the four vectors you have, and that will be your basis. Needless to say, there is more than one possible basis.​