Question

I need help.

Answer 1

In the 15th line of the proof, how are those e1, e2 coming?

Answer 2

@Directrix

Answer 3

@HolsterEmission

Answer 4

@HarryCraig any help?

Answer 5

@mathmate

Answer 6

@ganeshie8

Answer 7

@welshfella

Answer 8

Sorry I've forgotten this stuff.

Answer 9

@Loser66

Answer 10

Hey, graduate level student!! how can I help you while my level is lower than yours?

Answer 11

@amorfide

Answer 12

i don't think it holds any significance, and is just shorter to write e1|e2, instead of the col1, col2, col3, so you can write a general form easier

Answer 13

@amorfide How did they got that line, I want to know.

Answer 14

@mhchen

Answer 15

@3mar

Answer 16

Well, I am here.
How could I help you?

Answer 17

What is your question?

Answer 18

OK. See the pdf.. In line 15 , how they calculated that, I want to know

Answer 19

P−1AP = [P−1col1(P)| · · · |P−1colk(P)|P−1B] = [e1|e2| · · · |ek|P−1B]
these e1, e2, how they obtained

Answer 20

By the way, I looked at it as you just share it but I did not have any idea!

Answer 21

@imqwerty

Answer 22

I assume the \(e_i\) are vectors in the standard basis for \(\mathbb R^n\). Your question is basically why \(\mathbf P^{-1}\) multiplied by the \(i\)th column of \(\mathbf P\) reduces to the corresponding \(e_i\).

This is perfectly natural since
\[\begin{align*}
\begin{bmatrix}\mathbf P^{-1}\mathrm {col}_1(\mathbf P)&\mathbf P^{-1}\mathrm {col}_2(\mathbf P)&\cdots&\mathbf P^{-1}\mathrm {col}_{k-1}(\mathbf P)&\mathbf P^{-1}\mathrm {col}_k(\mathbf P)\end{bmatrix}&=\mathbf P^{-1}\mathbf P\\[1ex]
&=\mathbf I_k
\end{align*}\]and the \(i\)th column of \(\mathbf I_k\) is \(e_i\) with \(1\le i\le k\).

Answer 23

Thanks... :)