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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?

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The part $ij=0:i>j$ is pretty unclear, but I know what you mean. A better way to say it is $\{E^{ij}\mid 1\leq i\leq j\leq n, \}$. – rschwieb Sep 28 '12 at 12:06
up vote 1 down vote accepted

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 detailds to this answer)

Added: to show that this set spans all upper triangular matrices take an upper triangular matrix. if the $(i,j)$ coardinate 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 $n$ ?

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Updated. How do I show that it spans? – CodyBugstein Sep 28 '12 at 2:24
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..? – CodyBugstein Sep 28 '12 at 2:27
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 ? – Belgi Sep 28 '12 at 2:29
Not exactly, but I'm not far from it. Can I show that with scalars I can get any arbitrary UT matrix? – CodyBugstein Sep 28 '12 at 2:32
By definition you need to take a UT matrix and show it can be represented as a linear commbination of your set – Belgi Sep 28 '12 at 2:33

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