- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.6. $\quad$ Exercises
9. $\quad$ Let $U$ and $V$ be two $n \times n$ upper triangular matrices.
(a) $\quad$ Prove that each of $U + V$ and $UV$ is an upper triangular matrix.
(b) $\quad$ Prove that det $(UV) = (\text{det } U)(\text{det } V).$
(c) $\quad$ If det $U \neq 0$ prove that there is an upper triangular matrix $U^{-1}$ such that $UU^{-1} = I,$ and deduce that det $(U^{-1}) = 1/\text{det } U.$
(a) $\quad$ Prove that each of $U + V$ and $UV$ is an upper triangular matrix.
$\quad$ Proof. $\quad$ If $U$ and $V$ are upper triangular matrices then $u_{ij}$ and $v_{ij}$ are zero if $i \gt j.$ But this means $u_{ij} + v_{ij} = 0$ if $i \gt j.$ Hence, $U+V$ is upper triangular.
$\quad$ Then, by definition, the $ij^{th}$ element of $UV$ is given by the sum \begin{align*} uv_{ij} &= \sum_{k=1}^n u_{ik}v_{kj} \end{align*} Now, assume $i \gt j,$ we will prove that $uv_{ij} = 0.$ Because $U$ and $V$ are upper triangular, $u_{ik} = 0$ if $i \gt k$ and $v_{kj} = 0$ if $k \gt j.$ If we separate the above sum as follows \begin{align*} uv_{ij} &= \sum_{k=1}^n u_{ik}v_{kj} \\ &= \sum_{k=1}^{i - 1} u_{ik}v_{kj} + \sum_{k=i}^{n} u_{ik}v_{kj} \end{align*} we find that $i \gt k$ for every $k$ in the first sum, and $k \gt j$ for every $k$ in the second sum. Hence, $uv_{ij} = 0.$ But because this is true for every pair $(i, j)$ such that $i \gt j,$ this means that $UV$ is upper triangular. $\quad \blacksquare$
(b) $\quad$ Prove that det $(UV) = (\text{det } U)(\text{det } V).$
$\quad$ Proof. $\quad$ From part (a) we know that if $U$ and $V$ are upper triangular matrices, then their product $UV$ is also an upper triangular matrix. Accordingly, the determinant of $UV$ is the product of its diagonal elements:
\begin{align*}
\det (UV) = (uv)_{11} \cdots (uv)_{nn}
\end{align*}
where $(uv)_{ij}$ is the $ij^{th}$ element of $UV.$
$\quad$ To prove that $\det (UV) = \det U \det V,$ it will suffice to show that the $k^{th}$ diagonal element of $UV$ is the product $u_{kk}v_{kk}.$ Let $(uv)_{kk}$ be the $k^{th}$ diagonal element of $UV$ given by the sum $(uv)_{kk} = \sum_{p=1}^n u_{kp}v_{pk}.$ As in part (a), we can separate this sum as follows: \begin{align*} (uv)_{kk} &= \sum_{p=1}^n u_{kp}v_{pk} \\ &= \sum_{p=1}^{k - 1} u_{kp}v_{pk} + (u_{kk}v_{kk}) + \sum_{p=k+1}^{n} u_{kp}v_{pk} \end{align*} But because $U$ and $V$ are upper triangular matrices, $u_{kp} = 0$ when $k \gt p$ and $v_{pk} = 0$ when $p \gt k.$ This means that $u_{kp}v_{pk} = 0$ when $p \neq k,$ giving us \begin{align*} (uv)_{kk} &= u_{kk}v_{kk}. \end{align*} Then, $\det (UV)$ is given by \begin{align*} \det (UV) &= (uv)_{11} \cdots (uv)_{nn} \\ &= u_{11}v_{11} \cdots u_{nn}v_{nn} \\ &= (u_{11}\cdots u_{nn})(v_{11}\cdots v_{nn}) \\ &= \det U \det V. \quad \blacksquare \end{align*}
(c) $\quad$ If det $U \neq 0$ prove that there is an upper triangular matrix $U^{-1}$ such that $UU^{-1} = I,$ and deduce that det $(U^{-1}) = 1/\text{det } U.$
$\quad$ Proof. $\quad$ If $U$ is an upper-triangular matrix with a nonzero determinant, then all its diagonal elements are nonzero. Hence, we can perform the row operations of the Gauss-Jordan elimination process on $U$ until it becomes the $n \times n$ identity matrix. To find $U^{-1},$ we start with the following augmented matrix
\begin{align*}
\begin{bmatrix}
\begin{array}{cccc|cccc}
u_{11} & u_{12} & \cdots & u_{1n} & 1 & 0 & \cdots & 0 \\
0 & u_{22} & \cdots & u_{2n} & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & u_{nn} & 0 & 0 & \cdots & 1
\end{array}
\end{bmatrix},
\end{align*}
First, we divide the $n^{th}$ row by $u_{nn}$ to give
\begin{align*}
\begin{bmatrix}
\begin{array}{cccc|cccc}
u_{11} & u_{12} & \cdots & u_{1n} & 1 & 0 & \cdots & 0 \\
0 & u_{22} & \cdots & u_{2n} & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 & 0 & 0 & \cdots & \frac{1}{u_{nn}}
\end{array}
\end{bmatrix},
\end{align*}
Then, we add multiples of the $n^{th}$ row to rows $1, \dots, n-1$ to make the remaining elements of the $n^{th}$ column zero:
\begin{align*}
\begin{bmatrix}
\begin{array}{cccc|cccc}
u_{11} & u_{12} & \cdots & 0 & 1 & 0 & \cdots & \frac{-u_{1n}}{u_{nn}} \\
0 & u_{22} & \cdots & 0 & 0 & 1 & \cdots & \frac{-u_{2n}}{u_{nn}} \\
\vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 & 0 & 0 & \cdots & \frac{1}{u_{nn}}
\end{array}
\end{bmatrix},
\end{align*}
Repeating this process for the remaining $n - 1$ rows of $U,$ we get the $n \times n$ identity matrix to the left of the vertical line, with the upper triangular matrix $U^{-1}$ to the right of the vertical line. But as a consequence of this process, the diagonal elements of $U^{-1}$ are
\begin{align*}
\frac{1}{u_{11}}, \frac{1}{u_{22}}, \dots, \frac{1}{u_{nn}}
\end{align*}
And because $U^{-1}$ is an upper triangular matrix, its determinant is the product of these elements. Hence, we have
\begin{align*}
\det U^{-1} &= \frac{1}{u_{11}}\frac{1}{u_{22}}\dots\frac{1}{u_{nn}} \\
&= \frac{1}{u_{11}u_{22} \dots u_{nn}} \\
&= \frac{1}{\det U}. \quad \blacksquare
\end{align*}