$$
\begin{align*}
&+ s \sum_{h=1}^{n} \left[ (\delta_{(h,k),(j,i)} - \delta_{(h,k),(i,j)}) \delta_{h,l} + \delta_{h,k} (\delta_{(h,l),(j,i)} - \delta_{(h,l),(i,j)}) \right] \\
&+ (c-1) \sum_{h=1}^{n} \left[ (\delta_{(h,l),(i,i)} + \delta_{(h,l),(j,j)}) \delta_{h,l} + \delta_{h,k} (\delta_{(h,l),(i,i)} + \delta_{(h,l),(j,j)}) \right] \\
&+ \sum_{h=1}^{n} \delta_{h,k} \delta_{h,l} \\
&= s^2 (\delta_{(i,k),(i,j)} \delta_{(i,l),(i,j)} + \delta_{(j,k),(j,i)} \delta_{(j,l),(j,i)}) \\
&\quad + (c-1)^2 (\delta_{(i,k),(i,i)} \delta_{(i,l),(i,i)} + \delta_{(j,k),(j,j)} \delta_{(j,l),(j,j)}) \\
&\quad + s(c-1) (-\delta_{(i,k),(i,j)} \delta_{(i,l),(i,i)} - \delta_{(i,k),(i,i)} \delta_{(i,l),(i,j)} + \delta_{(j,k),(j,i)} \delta_{(j,l),(j,j)} + \delta_{(j,k),(j,j)} \delta_{(j,l),(j,i)}) \\
&\quad + s(-\delta_{(i,k),(i,j)} \delta_{i,l} - \delta_{i,k} \delta_{(i,l),(i,j)} + \delta_{(j,k),(j,i)} \delta_{j,l} + \delta_{j,k} \delta_{(j,l),(i,j)}) \\
&\quad + (c-1) (\delta_{(i,l),(i,i)} \delta_{i,l} + \delta_{i,k} \delta_{(i,l),(i,i)} + \delta_{(j,l),(j,j)} \delta_{j,l} + \delta_{j,k} \delta_{(j,l),(j,j)}) \\
&\quad + \delta_{k,l} \\
&= s^2 (\delta_{k,j} \delta_{l,j} + \delta_{k,i} \delta_{l,i}) + (c-1)^2 (\delta_{i,k} \delta_{i,l} + \delta_{k,j} \delta_{l,j}) \\
&\quad + s(c-1) \times 0 + s \times 0 + (c-1) (\delta_{i,l} + \delta_{i,k} \delta_{l,i} + \delta_{j,l} + \delta_{j,k} \delta_{l,j}) + \delta_{k,l} \\
&= (s^2 + (c-1)^2 + c-1) (\delta_{k,j} \delta_{l,j} + \delta_{k,i} \delta_{l,i}) + (c-1) (\delta_{i,l} + \delta_{j,l}) + \delta_{k,l} \\
&= \begin{cases} (-c+1) + (c-1) + 1 & \text{if } k=l=i \text{ or } k=l=j \\ \delta_{k,l} & \text{otherwise.} \end{cases} \\
&= \delta_{k,l} \\
&= [I_n]_{k,l}
\end{align*}
$$

Hence the result.

(ii) For all $k \in \{1, \dots, n\}$, we have :

$$
\begin{align*}
& \left[ G \left( i, j, \frac{x_i}{r}, -\frac{x_j}{r} \right) X \right]_k \\
&= \sum_{h=1}^{n} \left[ G \left( i, j, \frac{x_i}{r}, -\frac{x_j}{r} \right) \right]_{kh} [X]_h \\
&= \sum_{h=1}^{n} \left( -\frac{x_j}{r} (\delta_{(k,h),(j,i)} - \delta_{(k,h),(i,j)}) + \left( \frac{x_i}{r} - 1 \right) (\delta_{(k,h),(i,i)} + \delta_{(k,h),(j,j)}) + \delta_{k,h} \right) x_h \\
&= \left( -\frac{x_j}{r} \delta_{(k,i),(j,i)} + \left( \frac{x_i}{r} - 1 \right) \delta_{(k,i),(i,i)} \right) x_i + \left( -\frac{x_j}{r} \delta_{(k,j),(i,j)} + \left( \frac{x_i}{r} - 1 \right) \delta_{(k,j),(j,j)} \right) x_j + \delta_{k,k} x_k \\
&= \begin{cases} \left( \frac{x_i}{r} - 1 \right) x_i + \frac{x_j}{r} x_j + x_i & \text{if } k=i \\ -\frac{x_j}{r} x_i + \left( \frac{x_i}{r} - 1 \right) x_j + x_j & \text{if } k=j \\ x_k & \text{otherwise.} \end{cases} \\
&= \begin{cases} r & \text{if } k=i \\ 0 & \text{if } k=j \\ x_k & \text{otherwise.} \end{cases}
\end{align*}
$$

Hence the result.

By the lemma, left multiplication by a Givens rotation is a "double transvection" where we obtain an *r* in row *i* and a 0 in row *j*.
By successively multiplying on the left by Givens rotations, we can make the matrix upper triangular, and the product of the Givens rotations will give the matrix Q as we will see in the following subsection.

By multiplying on the right by Givens rotations, we would obtain a RQ decomposition as in section 14.

## D.2 Example

Let's calculate the inverse of:

$$A := \begin{pmatrix} 2 & -2 & 1 \\ 2 & -3 & 2 \\ -1 & 2 & 0 \end{pmatrix}$$

using the QR decomposition and Givens rotations.

Case where $i := 1, j := 2, x_i := 2, x_j := 2$. We have:

$$\begin{aligned} r &:= \sqrt{x_i^2 + x_j^2} \\ &= \sqrt{2^2 + 2^2} \\ &= 2\sqrt{2} \end{aligned}$$

and

$$\begin{aligned} A_1 &:= G\left(i, j, \frac{x_i}{r}, -\frac{x_j}{r}\right) A \\ &= G\left(1, 2, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right) A \\ &= \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & -2 & 1 \\ 2 & -3 & 2 \\ -1 & 2 & 0 \end{pmatrix} \\ &= \begin{pmatrix} 2\sqrt{2} & -\frac{5}{2}\sqrt{2} & \frac{3}{2}\sqrt{2} \\ 0 & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \\ -1 & 2 & 0 \end{pmatrix} \end{aligned}$$

Case where $i := 1, j := 3, x_i := 2\sqrt{2}, x_j := -1$. We have:

$$\begin{aligned} r &:= \sqrt{x_i^2 + x_j^2} \\ &= \sqrt{8 + 1} \\ &= 3 \end{aligned}$$

and

$$\begin{aligned} A_2 &:= G\left(i, j, \frac{x_i}{r}, -\frac{x_j}{r}\right) A_1 \\ &= G\left(1, 3, \frac{2\sqrt{2}}{3}, \frac{1}{3}\right) A_1 \\ &= \begin{pmatrix} 2\sqrt{2}/3 & 0 & -1/3 \\ 0 & 1 & 0 \\ 1/3 & 0 & 2\sqrt{2}/3 \end{pmatrix} \begin{pmatrix} 2\sqrt{2} & -\frac{5}{2}\sqrt{2} & \frac{3}{2}\sqrt{2} \\ 0 & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \\ -1 & 2 & 0 \end{pmatrix} \\ &= \begin{pmatrix} 3 & -4 & 2 \\ 0 & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \\ 0 & \frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \end{pmatrix} \end{aligned}$$
Case where $i := 2, j := 3, x_i := -\sqrt{2}/2, x_j := \sqrt{2}/2$. We have :

$$ 
\begin{aligned}
r &:= \sqrt{x_i^2 + x_j^2} \\
  &= \sqrt{1/2 + 1/2} \\
  &= 1
\end{aligned}
$$

and

$$ 
\begin{aligned}
R &:= A_3 \\
  &:= G\left(i, j, \frac{x_i}{r}, -\frac{x_j}{r}\right) A_2 \\
  &= G\left(2, 3, -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) A_2 \\
  &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 0 & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} 3 & -4 & 2 \\ 0 & -\frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \\ 0 & \frac{1}{2}\sqrt{2} & \frac{1}{2}\sqrt{2} \end{pmatrix} \\
  &= \begin{pmatrix} 3 & -4 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}
\end{aligned}
$$

Let :

$$ Q := G\left(2, 3, -\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right) G\left(1, 3, \frac{2\sqrt{2}}{3}, \frac{1}{3}\right) G\left(1, 2, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right). $$

We have :

$$ 
\begin{aligned}
Q &= \begin{pmatrix} 1 & 0 & 0 \\ 0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \\ 0 & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \end{pmatrix} \begin{pmatrix} 2\sqrt{2}/3 & 0 & -1/3 \\ 0 & 1 & 0 \\ 1/3 & 0 & 2\sqrt{2}/3 \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \\
&= \begin{pmatrix} \frac{2}{3}\sqrt{2} & 0 & -\frac{1}{3} \\ \frac{1}{6}\sqrt{2} & -\frac{1}{2}\sqrt{2} & \frac{2}{3} \\ -\frac{1}{6}\sqrt{2} & -\frac{1}{2}\sqrt{2} & -\frac{2}{3} \end{pmatrix} \begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \\
&= \frac{1}{3} \begin{pmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -2 & -2 \end{pmatrix}
\end{aligned}
$$

It is true (it is unnecessary to show it) :

$$ 
\begin{aligned}
QA &= \frac{1}{3} \begin{pmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -2 & -2 \end{pmatrix} \begin{pmatrix} 2 & -2 & 1 \\ 2 & -3 & 2 \\ -1 & 2 & 0 \end{pmatrix} \\
  &= \begin{pmatrix} 3 & -4 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \\
  &= R
\end{aligned}
$$

Using the formula from proposition A.6.2 for the inverse of an upper triangular matrix, we get :

$$ R^{-1} = \frac{1}{3} \begin{pmatrix} 1 & 4 & 2 \\ 0 & 3 & 0 \\ 0 & 0 & -3 \end{pmatrix} $$
and thus :

$$
\begin{align*}
A^{-1} &= R^{-1}Q \\
&= \frac{1}{9} \begin{pmatrix} 1 & 4 & 2 \\ 0 & 3 & 0 \\ 0 & 0 & -3 \end{pmatrix} \begin{pmatrix} 2 & 2 & -1 \\ 2 & -1 & 2 \\ 1 & -2 & -2 \end{pmatrix} \\
&= \frac{1}{3} \begin{pmatrix} 4 & -2 & 1 \\ 2 & -1 & 2 \\ -1 & 2 & 2 \end{pmatrix}
\end{align*}
$$
Références

[1] Roger A. Horn, Charles R. Johnson, *Matrix analysis*, Cambridge University Press, (1990).

[2] Denis Serre, *Matrices, Theory and Applications*, Springer, (2000).

[3] Wikipedia, *Invertible matrix # Methods of matrix inversion*.