Homepage › Solution manuals › Gilbert Strang › Linear Algebra and Learning from Data › Exercise 2.2.3

Exercise 2.2.3

Answers

$A = \sum σ_{i} u_{i} v_{i}^{T}$ , $A^{+} = \sum \frac{v_{i} u_{i}^{T}}{σ_{i}}$ .

So we have

\begin{array}{l} A^{+} A & = \sum_{i} \frac{v_{i} u_{i}^{T}}{σ_{i}} \sum_{j} σ_{j} u_{j} v_{j}^{T} \\ = \sum_{i} \sum_{j} \frac{v_{i} u_{i}^{T}}{σ_{i}} σ_{j} u_{j} v_{j}^{T} \\ = \sum_{i} \frac{v_{i} u_{i}^{T}}{σ_{i}} σ_{i} u_{i} v_{i}^{T} \\ = \sum_{i} v_{i} v_{i}^{T} \\ = A^{+} A \end{array}

Now ${(A^{+} A)}^{2} = A^{+} A A^{+} A = A^{+} \sum_{i} u_{i} u_{i}^{T} A = A^{+} \sum_{i} u_{i} u_{i}^{T} \sum_{j} σ_{j} u_{j} v_{j}^{T} = A^{+} \sum_{i} σ_{i} u_{i} v_{i}^{T} = \sum_{j} \frac{v_{j} u_{j}^{T}}{σ_{j}} \sum_{i} σ_{i} u_{i} v_{i}^{T} = \sum_{j} v_{j} v_{j}^{T} = A^{+} A$

niuers

2020-03-20 00:00

Exercise 2.2.3

Answers

Comments

Add answer