Question

hey I have MATHS q. Show that if A^2 =0 then I-A in invertible and (I-A)^-1=I+A

Answer 1

it is pretty easy
\[ \large (I-A)(I+A)=I^2-A^2=I-0=I \]
\[ \large (I+A)(I-A)=I^2-A^2=I-0=I \]
this shows that \(I-A\) is invertible and that \((I-A)^{-1}=I+A\)

Answer 2

got it??

Answer 3

orh yes thanks!

Answer 4

but how does this show (I−A)−1=I+A??

Answer 5

well. by definition, a square matrix A is invertible if there exists a matrix (also square) B such that AB=BA=I. when such B exists, it's called the inverse of A and denoted by A^-1

Answer 6

okay so the inverse of (I-A) will be equal to I+A since I-A is invertible??

Answer 7

no. since I-A multiplied on the left and the right by I+A equals I (the identity matrix) then I-A is invertible and hence I+A is the inverse of I-A

Answer 8

orh I got it! thanks!!!

Answer 9

then what if is Show that if A^3 =0 then I-A in invertible and (I-A)^-1=I+A+A^2??

Answer 10

it is exactly the same. compute
\[ \large (I-A)(I+A+A^2)= \]
and
\[ \large (I+A+A^2)(I-A)= \]

Answer 11

Ya i got it!can i ask you 2 more q??

Answer 12

Let D be a diagonal matrix.i.e a square matrix of order n whose non-diagonal entries are all zero. Let d1,d2,...dn be the diagonal entries of D. So for any non-negative integer k, show that D^k is a diagonal matrix whose diagonal entries are (d1)^k, (d2^k),...(dn^k).

Answer 13

first you should prove that the product of two diagonal matrices is a diagonal matrix. and then use induction to prove this.

Answer 14

i mean if
\[ \large D_1=\hbox{diag}(d_1,d_2,\dots,d_n) \]
and
\[ \large D_2=\hbox{diag}(c_1,c_2,\dots,c_n) \]
then
\[ \large D_1D_2=\hbox{diag}(d_1c_1,d_2c_2,\dots,d_nc_n) \]

Answer 15

So I should do D^k=(D^k-1)*D?

Answer 16

yes this is the easy part.

Answer 17

but first prove that the product of diagonal matrices is diagonal

Answer 18

Can i just let n=2,3,4...n to see the general pattern that the diagonal matrix of D^k is (d1)^k, (d2^k),...(dn^k) instead?

Answer 19

yeah. that would count as if u did "induction".

i guess it depends on how strict is your professor.

Answer 20

Then if A^n=0 for n ≥4, is I-A invertible?Is it the same method as previous one?

Answer 21

yes. do u remember this from school
\[ \large a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}) \]

Answer 22

well if A^n=0 then I-A is invertible.

Answer 23

yup thanks for your help:)

Answer 24

u r welcome

Answer 25

when do I know I have to use induction to prove the formula??

Answer 26

usually when then proposition depends on a variable that is a natural number. in your case u were supposed to prove that if
\[ \large D=\hbox{diag}(d_1,\dots,d_n) \]
then
\[ \large D^k=\hbox{diag}(d_1^k,\dots,d_n^k) \]
where k is natural. then u should have used induction over k

Answer 27

okay thanks:)

Answer 28

u r welcome

Answer 29

Hi, please help me solve this question:)
let a>0.Prove that limx-->a(x^1/4)=a^1/4