the upper triangular matrix problem Given a nonsingular upper-triangular matrix $U$ whose diagonal elements are $u_{ii}$.
Show that:

the diagonal elements of $U^{−1}$ are the reciprocals of the
  diagonal elements of $U$.

I know its inverse $U^{−1}$ is also upper triangular. What about the reciprocals? 
 A: Denote by $u_{ij}$ and $v_{ij}$ the coefficients of $U$ and of $U^{-1}$ respectively. Then, by definition of matrix product,
$$
\sum_{k=1}^n u_{ik}v_{ki}
$$
is the coefficient at place $(i,i)$ in the product, so it is $1$. However, the condition that $U$ and $U^{-1}$ are upper triangular can be expressed by
$$
u_{ij}=0\quad\text{and}\quad v_{ij}=0\quad\text{for }i>j
$$
So we have
$$
1=\sum_{k=1}^n u_{ik}v_{ki}=
\biggl(\sum_{k=1}^{i-1}u_{ik}v_{ki}\biggr)
+u_{ii}v_{ii}+
\biggl(\sum_{k=i+1}^{n}u_{ik}v_{ki}\biggr)
$$
Now, in the first summation we have $k<i$, so $u_{ik}=0$; in the second summation we have $k>i$, so $v_{ki}=0$. Therefore
$$
1=u_{ii}v_{ii}
$$
as required.
A: The answer is straightforward, you can use primary row transformation to find the inverse matrix $U^{-1}$:
$$\begin{pmatrix}
u_{11} & u_{12}&\cdots & \star\\
0&u_{22}&\cdots&\star \\ 
\cdots&\cdots&\ddots&\star
\\ 0&\cdots&0& u_{nn}
\end{pmatrix}\begin{pmatrix}
1 & 0&\cdots & 0\\
0&1&\cdots&0 \\ 
\cdots&\cdots&\ddots&0
\\ 0&\cdots&0& 1
\end{pmatrix}$$
if you use some proper primary row transformation to $U$,then it become identity matrix $I$, the same transformation act on the right identity matrix, it will become $U^{-1}$, and you will easy to find that the diagonal elements is inverse of $u_{ii}$  
