Answer:
dim L = dim U = [tex]\frac{n(n+1)}{2}[/tex]
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 [tex]A=(a_{ij})[/tex] such that the entry [tex]a_{ij}=0[/tex] if i<j.
Now let [tex]A=(a_{ij})[/tex] and [tex]B=(b_{ij})[/tex] be two lower triangular matrices, now if
[tex]C=A+\lambda B[/tex] for some [tex]\lambda \in \mathbb{F}[/tex]
then the entry [tex]c_{ij}[/tex] of C is equal to
[tex]c_{ij}=a_{ij}+\lambda b_{ij}[/tex]
Now, if i<j, it must hold that [tex]a_{ij}=0 \quad \text{and} \quad b_{ij}=0[/tex]. Therefore, if this is the case we must have that [tex]c_{ij}=0[/tex] 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 [tex]\mathbb{F}[/tex], 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,
[tex]\dimL=n+(n-1)+(n-2)+...+2+1=\dfrac{n(n+1)}{2}[/tex]
