Homepage › Solution manuals › Gilbert Strang › Introduction to Linear Algebra › Exercise 4.2.14

Exercise 4.2.14

Answers

The projection of this $b$ onto the column space of $A$ is $b$ itself because $b$ is in that column space. But $P$ is not necessarily $I$ . Here $b = 2 ($ column 1 of $A)$ :

A = [\begin{matrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \end{matrix}] gives P = \frac{1}{21} [\begin{matrix} 5 & 8 & - 4 \\ 8 & 17 & 2 \\ - 4 & 2 & 20 \end{matrix}] and b = P b = p = [\begin{matrix} 0 \\ 2 \\ 4 \end{matrix}] .

mathchakra

2023-09-14 07:31

Exercise 4.2.14

Answers

Comments

Add answer