I'm sorry if this is a silly question. I'm new to the notion of bases and all the examples I've dealt with before have involved sets of vectors containing real numbers. This has led me to assume that bases, by definition, are made up of a number of $n$-tuples.

However, now I've been thinking about a basis for all $n\times n$ matrices and I keep coming back to the idea that the simplest basis would be $n^2$ matrices, each with a single $1$ in a unique position.

Is this a valid basis? Or should I be trying to get column vectors on their own somehow?

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    $\begingroup$ "Vector" just means "element of the vector space." That's their "job title." What they are at home when they aren't working doesn't really matter. If your vector space consists of matrices, then they are "vectors" (which happen to be matrices when they are relaxing at home). If your vector space consists of polynomials, then the polynomials are "vectors" (which happen to be polynomials when they are out on dates). Etc. In your example, the basis consists of vectors (which just happen to be $n\times n$ matrices when they are at home). $\endgroup$ – Arturo Magidin Mar 5 '12 at 16:23

Elements of a basis of a vector space always have to be elements of the vector space in the first place. Hence, if you are looking for a basis of the space of all $n\times n$ matrices, then matrices actually are your vectors and the only choice for what a basis element can be. In fact, the matrices you describe are a valid basis for the space of all $n\times n$ matrices. However, looking at matrices this way (as vectors of the vector space of all $n\times n$ matrices), it might help to realize that they are just tuples with $n^2$ many entries, arranged as a square.

  • $\begingroup$ Fantastic answer - thank you for being so clear! $\endgroup$ – Katherine Rix Mar 5 '12 at 16:23

Yes, you are right. A vector space of matrices of size $n$ is actually, a vector space of dimension $n^2$. In fact, just to spice things up: The vector space of all

  • diagonal,
  • symmetric and
  • triangular matrices of dimension $n\times n$

is actually a subspace of the space of matrices of that size.

As with all subspaces, you can take any linear combination and stay within the space. (Also, null matrix is in all the above three).

Try to calculate the basis for the above 3 special cases: For the diagonal matrix, the basis is a set of $n$ matrices such that the $i^{th}$ basis matrix has $1$ in the $(i,i)$ and $0$ everywhere else. Try to figure out the basis vectors/matrices for symmetric and triangular matrices.

  • $\begingroup$ I went and had a go at this as you suggested. It helped me come to grips with the notion of dimension more. Thanks! $\endgroup$ – Katherine Rix Mar 8 '12 at 20:15

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