Question

Show that the sets of upper-triangular and lower-triangular matrices are subspaces of F n,n . Call them U and L respectively. What are their dimensions?

Answer 1

Answer:

dim L = dim U = $\frac{n(n+1)}{2}$

Step-by-step explanation:

We can do it only for the lower-triangular matrices, the case of the upper-triangular matrices is similar. We might caracterice nxn the lower-triangular matrices, as the nxn matrices $A=(a_{ij})$ such that the entry $a_{ij}=0$ if i<j.

Now let $A=(a_{ij})$ and $B=(b_{ij})$ be two lower triangular matrices, now if

$C=A+\lambda B$ for some $\lambda \in \mathbb{F}$

then the entry $c_{ij}$ of C is equal to

$c_{ij}=a_{ij}+\lambda b_{ij}$

Now, if i<j, it must hold that $a_{ij}=0 \quad \text{and} \quad b_{ij}=0$. Therefore, if this is the case we must have that $c_{ij}=0$ and so we get that C is also a lower triangular matrix. This showa that L is closed under sum and scalar multiplcation, hence it is a linear subspace.

To find the dimension, note that all the entries of a lower-triangular matrix over the diagonal must be equal to zero. However, each entry of the matrix under the diagonal and in the diagonal might be any element of $\mathbb{F}$, any entry that can be choosen add up to the dimension of L, we n such elemnts for the first column, (n-1) for the second column, (n-2) for the third column etc.... Therefore,

$\dimL=n+(n-1)+(n-2)+...+2+1=\dfrac{n(n+1)}{2}$