linear-algebra-done-right-solutions

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Solution to 3D.18 wrongly assumes that V is finite-dimensional

2

In referenced 3.61 vector spaces are set to be finite dimensional. We can't make use assumption for V from 3D.18

4DA

As the example you give, the norm should be 1 and it's permitted.(maybe)

LuNa-shi

https://github.com/celiopassos/linear-algebra-done-right-solutions/blob/4d934545705a679cf534d6d41a8f0a2e3a78f49f/Chapter%2006%20-%20Inner%20Product%20Spaces/6A%20-%20Inner%20Products%20and%20Norms.md?plain=1#L59C54-L59C130 `\sqrt{(4\langle u, v \rangle^2 - 4\langle u, u \rangle\langle v, v \rangle)}` I believe this line is incorrect as `^2 0. An equivalent proof to your is here https://latexonline.cc/compile?git=https://github.com/jubnoske08/linear_algebra&target=chapter_6.tex&command=pdflatex&force=true....

CLARKBENHAM

Exercises 3.B, 22

1

dimnullST

amirli21

In 6.A.29 (b) how to know finite dimensional space V must have orthonormal basis. Despite orthonormal basis, how to deduce Σ|Cjr - Cir| < ε from ||Σ(Cjr - Cir)e|| <...

JTP-123

Should =(2xp+x2p′+p′′)|x=4 be =(2xp+x2p′+p′′')|x=4 ? Nice work BTW! Thanks!

thehoglet

Here I think the following is correct instead of what is written above

OnezeroC