4
$\begingroup$

I'm asked to find a basis for $W$, which is a subspace of $M_n(F)$.

$W$ is the subspace containing all upper triangular $n \times n$ matrices.

How do you find this basis? 

My guess is that it's simply a collection of $[n(n+1)]/2$ matrices $E^{ij}$ in which $ij = 0: i>j$.

Would that be correct, and if so, is there a better way to express it? How do I show that it spans?

  • 2
    The part $ij=0:i>j$ is pretty unclear, but I know what you mean. A better way to say it is $\{E^{ij}\mid 1\leq i\leq j\leq n, \}$.2012-09-28

1 Answers 1

1

Your answer is indeed correct.

Showing both that this set spans all upper triangular matrices and showing that it is linear independent should be very easy (but if you are having problems with any feel free to update your question so I could add details to this answer)

Added: To show that this set spans all upper triangular matrices, take an upper triangular matrix. if the $(i,j)$ co-ordinate is $a$ what would you take as the coefficient of $E_{i,j}$ ? Can the other matrices in the sum affect this coordinate ?

Example: $$\begin{pmatrix}a & b\\ 0 & c \end{pmatrix}=a\cdot\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}+b\cdot\begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix}+c\cdot\begin{pmatrix}0 & 0\\ 0 & 1 \end{pmatrix}$$

Can you generelize for $n$ ?

  • 0
    Updated. How do I show that it spans?2012-09-28
  • 0
    I think I got it... I can show that adding two matrices E^ij will still maintain that ij = 0 when i>j since 0 + 0 = 0. Is that concrete enough..?2012-09-28
  • 1
    This only shows this set contained in the space of all upper triangular matrices. you need to show that every upper triangular matrix is a linear combination of those matrices. is it clear to you why ?2012-09-28
  • 0
    Not exactly, but I'm not far from it. Can I show that with scalars I can get any arbitrary UT matrix?2012-09-28
  • 0
    By definition you need to take a UT matrix and show it can be represented as a linear commbination of your set2012-09-28
  • 0
    Is there a quick way to do that?2012-09-28
  • 0
    I added an example. I can't think of any long way to show this...2012-09-28
  • 0
    ok Thanks! I have what to work with now :)2012-09-28
  • 0
    I know this is old, but I'm hoping someone will answer, because I have stumbled across the same problem, and I thought $E_{ij}$ is the matrix having all entries 0, except a 1 at the $(i,j)$ th place, isn't it?2014-09-08
  • 0
    @Diya it's OK. You understand correctly the definition , is there something that bothers you?2014-09-08
  • 0
    Yes, the part that says $ij=0, i>j$ bothers me. That's the definition for the upper triangular matrix, not the basis, isn't it?2014-09-08
  • 0
    @Diya sorry, I miss read your comment. Yes, the definition is that $E_ij$ have exactly one entry which is 1 at coordinate (I, j) and the others are zero. Does everything works for you now?2014-09-08
  • 0
    Yes, perfectly, thank you so much for clearing up my confusion! :)2014-09-08