Question

show that if A is a linear transformation such that A^2 +2A +3 = 0, then A is invertible
Please, help

Answer 1

What are the properties of a linear transformation?

Answer 2

you mean definition?

Answer 3

It is just have product 's property like left/right distribute apply on addition and multiplication, nothing special

Answer 4

And there is a zero property.

Answer 5

yes

Answer 6

And for invertibility one must have in inverse transformation f(g(v)) = v AND g(f(u)) = u.

Answer 7

I am sorry for being dummy. I know that thing but don't see the link between that and the problem

Answer 8

are all square matrices linear transformation matrices?

Answer 9

yes, invertible or not is another topic

Answer 10

?? Why? Are we just ignoring the problem statement? "...then A is invertible".

Answer 11

this equation can be factored into (A-r1 I)(A-r2 I)=0, yielding 2 trivial invertible solution. i'm not quite sure how to prove these are the only possible solutions though

Answer 12

is it not that we have to simplify the left hand side to have the form of A* something = 1 to get that something is A inverse. then conclude that A is invertible

Answer 13

Do we agree that A MUST be square?

Answer 14

I have a simpler problem like A^2 -A +1 =0 --> A -A^2 =1 --> A(1-A) =1 and conclude that 1-A is inverse of A

Answer 15

that makes sense.
A^2+2A+4I=I
factor & solve

Answer 16

and one more back ward to get A is inverse of (1-A) ... so that A is invertible

Answer 17

@eashy yyyyyyyyyyyyyyyyyyes!! I am so tired with it to not know how to square it

Answer 18

solve the quadratic equation x^2+2x+4=0

Answer 19

oh my god.

Answer 20

Thanks a lot.

Answer 21

:)

Answer 22

ha!! no no , it's not that , just A^2 +2A +1 =-2 or (A+1)^2 = -2

Answer 23

\(A^2+2A=-3\Rightarrow -\frac{1}{3}A^2-\frac{2}{3}A=I\Rightarrow A\left(-\frac{1}{3}A-\frac{2}{3}I\right)=I\Rightarrow -\frac{1}{3}A-\frac{2}{3}I=A^{-1}\)

Answer 24

\(-\frac{1}{3}A-\frac{2}{3}I=A^{-1}\)
BTW, \(A\in \{n\times n \text{ matrices with complex elements, }n \ge 1\}\) otherwise \(A^2\) is undefined.