Timeline for $I+B$ is invertible if $B^{k} = 0$
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2015 at 21:25 | history | edited | Kim Jong Un | CC BY-SA 3.0 | edited body |
| Nov 23, 2015 at 21:25 | comment | added | Brian Tung | @KimJongUn: I think it may be better to use $I$ in the numerator of your first fraction, perhaps? | |
| Nov 23, 2015 at 21:24 | comment | added | Brian Tung | @Lisa: $1+x^4 = (1-\sqrt{2}x+x^2)(1+\sqrt{2}+x^2)$. But what is being used here is the idea that if $B^k = 0$, then obviously $I-B^k = I$. But $I-B^k$ can be factored as $(I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1})$. Therefore $(I+B)(I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}) = I$, and thus $(I+B)^{-1}$ exists and is equal to $I-B+B^2+\cdots+(-1)^{k-1}B^{k-1}$. | |
| Nov 23, 2015 at 19:59 | comment | added | Lisa | So what does $(1+x^4)$ equal to? | |
| Nov 23, 2015 at 18:22 | comment | added | Andrew Dudzik | @Lisa This answer doesn't claim that it is. But $(1+x)(1-x+x^2-x^3) = 1-x^4$. | |
| Nov 23, 2015 at 18:05 | comment | added | Lisa | This is not true for $(1+x^{4})$ which is not equal to $(1+x)(1-x+x^2-x^3)$ | |
| Nov 23, 2015 at 17:10 | history | answered | Kim Jong Un | CC BY-SA 3.0 |