Singular value decomposition

Problem

Let

α = {v_{1}, v_{2}, v_{3}}

with

v_{1} = \frac{1}{\sqrt{3}} [\begin{matrix} 1 \\ 1 \\ 1 \end{matrix}], v_{2} = \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ - 1 \\ 0 \end{matrix}], v_{3} = \frac{1}{\sqrt{6}} [\begin{matrix} 1 \\ 1 \\ - 2 \end{matrix}] .

Let

β = {u_{1}, u_{2}, u_{3}, u_{4}}

with

u_{1} = \frac{1}{\sqrt{2}} [\begin{matrix} 1 \\ 1 \\ 0 \\ 0 \end{matrix}], u_{2} = \frac{1}{\sqrt{2}} [\begin{matrix} - 1 \\ 1 \\ 0 \\ 0 \end{matrix}], u_{3} = \frac{1}{\sqrt{2}} [\begin{matrix} 0 \\ 0 \\ 1 \\ 1 \end{matrix}], u_{4} = \frac{1}{\sqrt{2}} [\begin{matrix} 0 \\ 0 \\ - 1 \\ 1 \end{matrix}] .

Let

A

be a

4 \times 3

matrix such that

[f_{A}]_{α}^{β} = [\begin{matrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] .

For

x = [\begin{matrix} 0 \\ 1 \\ 2 \end{matrix}],

find

A x

.

Thought

Recall that

[f_{A}]_{α}^{β} [x]_{α} = [A x]_{β}

. Therefore, we need to transform

x

into

[x]_{α}

and then transform

[A x]_{β}

into

A x

to get the answer.

Sample answer

Note that

α

is an orthonormal basis, so we have

x = c_{1} v_{1} + c_{2} v_{2} + c_{3} v_{3}

with

c_{i} = ⟨ x, v_{i} ⟩

for

i = 1, 2, 3

. By direct computation, we have

x = \sqrt{3} v_{1} - \frac{1}{\sqrt{2}} v_{2} - \frac{3}{\sqrt{6}} v_{3} and [x]_{α} = [\begin{matrix} \sqrt{3} \\ - \frac{1}{\sqrt{2}} \\ - \frac{3}{\sqrt{6}} \end{matrix}] .

Now we may compute

[A x]_{β} = [f_{A}]_{α}^{β} [x]_{α} = [\begin{matrix} 3 \sqrt{3} \\ - \sqrt{2} \\ 0 \\ 0 \end{matrix}] .

This means

\begin{aligned} A x & = 3 \sqrt{3} u_{1} - \sqrt{2} u_{2} + 0 u_{3} + 0 u_{4} \\ = \frac{3 \sqrt{3}}{\sqrt{2}} [\begin{array}{c} 1 \\ 1 \\ 0 \\ 0 \end{array}] - [\begin{array}{c} - 1 \\ 1 \\ 0 \\ 0 \end{array}] = [\begin{array}{c} \frac{3 \sqrt{3}}{\sqrt{2}} + 1 \\ \frac{3 \sqrt{3}}{\sqrt{2}} - 1 \\ 0 \\ 0 \end{array}] . \end{aligned}

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