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?

  • $\begingroup$ Please tell us where you are stuck. $\endgroup$ – Michael Albanese Dec 28 '14 at 13:31
  • $\begingroup$ I'm sorry, I don't understand what you mean. $\endgroup$ – Michael Albanese Dec 28 '14 at 13:41
  • $\begingroup$ I don't understand what you're asking. Do you know what reciprocals are? $\endgroup$ – Michael Albanese Dec 28 '14 at 13:46

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.


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 $u_{ii}$


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