Solution to Exercise 6.5.29 from „Linear Algebra“

We have the formula

∥ v_{k} ∥ u_{k} = v_{k} = w_{k} - \sum_{j = 1}^{k} \frac{⟨ w_{k}, v_{j} ⟩}{∥ v_{j} ∥^{2}} v_{j}

= w_{k} - \sum_{j = 1}^{k} \frac{⟨ w_{k}, v_{j} ⟩}{∥ v_{j} ∥} u_{j} = w_{k} - \sum_{j = 1}^{k} ⟨ w_{k}, u_{j} ⟩ u_{j} .

It directly comes from the formula above and some computation.

We have

w_{1} = (1, 1, 0), w_{2} = (2, 0, 1), w_{3} = (2, 2, 1)

and

v_{1} = (1, 1, 0), v_{2} = (1, - 1, 1), v_{3} = (- \frac{1}{3}, \frac{1}{3}, \frac{2}{3})

by doing the Gram-Schmidt process. This means that we have

(\begin{matrix} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 & - \frac{1}{3} \\ 1 & - 1 & \frac{1}{3} \\ 0 & 1 & \frac{2}{3} \end{matrix}) (\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix}) .

Then we may also compute

∥ v_{1} ∥ = \sqrt{2}, ∥ v_{2} ∥ = \sqrt{3}, ∥ v_{3} ∥ = \frac{\sqrt{2}}{\sqrt{3}} .

Now we have

(\begin{matrix} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix}) = (\begin{matrix} 1 & 1 & - \frac{1}{3} \\ 1 & - 1 & \frac{1}{3} \\ 0 & 1 & \frac{2}{3} \end{matrix}) (\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix})

= (\begin{matrix} 1 & 1 & - \frac{1}{3} \\ 1 & - 1 & \frac{1}{3} \\ 0 & 1 & \frac{2}{3} \end{matrix}) {(\begin{matrix} \sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix})}^{- 1} (\begin{matrix} \sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix}) (\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & \frac{1}{3} \\ 0 & 0 & 1 \end{matrix})

= (\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & - \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix}) (\begin{matrix} \sqrt{2} & \sqrt{2} & 2^{\frac{3}{2}} \\ 0 & \sqrt{3} & \frac{1}{\sqrt{3}} \\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix}) .

Here the we have

Q = (\begin{matrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} & - \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} \\ 0 & \frac{1}{\sqrt{3}} & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix})

and

R = (\begin{matrix} \sqrt{2} & \sqrt{2} & 2^{\frac{3}{2}} \\ 0 & \sqrt{3} & \frac{1}{\sqrt{3}} \\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix}) .

First that

Q_{1}

,

Q_{2}

and

R_{1}

,

R_{2}

are invertible otherwise

A

cannot be invertible. Also, since

Q_{1}

,

Q_{2}

is unitary, we have

Q_{1}^{∗} = Q_{1}^{- 1}

and

Q_{2}^{∗} = Q_{2}^{- 1}

. Now we may observe that

Q_{1} Q_{2}^{∗} = R_{2} R_{1}^{- 1}

is an unitary matrix. But

R_{2} R_{1}^{- 1}

is upper triangular since

R_{2}

and the inverse of an upper triangular matrix

R_{1}

are triangular matrices. So

D = R_{2} R_{1}^{- 1}

is both upper triangular and unitary. It could only be a unitary diagonal matrix.

Denote

b

by

(\begin{matrix} 1 \\ 11 \\ - 1 \end{matrix})

. Now we have

A = QR = b

. Since

Q

is unitary, we have

R = Q^{∗} b

. Now we have

(\begin{matrix} \sqrt{2} & \sqrt{2} & 2^{\frac{3}{2}} \\ 0 & \sqrt{3} & \frac{1}{\sqrt{3}} \\ 0 & 0 & \frac{\sqrt{2}}{\sqrt{3}} \end{matrix}) = R = Q^{∗} b = (\begin{matrix} 3 \cdot 2^{\frac{3}{2}} \\ - \frac{11}{\sqrt{3}} \\ \frac{2^{\frac{5}{2}}}{\sqrt{3}} \end{matrix}) .

Then we may solve it to get the answer $x = 3$ , $y = - 5$ , and $z = 4$ .

Exercise 6.5.29