Expert: Paul Klarreich - 7/6/2007

Question
QUESTION: Im  trying to prove that the inverse of a matrix is  unique i think i came up with a solution but im not sure  i just wanted you to check it out  for me

let A be a n*n matrix with  B and C as inverses
the  by  definition A*B= 1   and A*C=1 where 1 is the identity matrix
THEN A*B=C*A
B=C ???
or
A*B =1
C(A*B)=C
but C(A*B)= B(AC)=B
Which  gives B=C
let me know

ANSWER: Questioner:   andrew
Category:  Calculus
Private:  No
 
Subject:  linear algebra
Question:  Im  trying to prove that the inverse of a matrix is  unique i think i came up with a solution but im not sure  i just wanted you to check it out  for me

let A be a n*n matrix with  B and C as inverses
the  by  definition A*B= 1   and A*C=1 where 1 is the identity matrix
THEN A*B=C*A
B=C ???
or
A*B =1
C(A*B)=C
but C(A*B)= B(AC)=B

>> this step is not so clear.  


Which  gives B=C
let me know
...........................
Hi, Andrew,

This is really a group-theory exercise.  Matrices are just an example.

I think you want to start with this theorem:

The left- and right- inverses of an element of a group are the same.

Proof: Suppose (big)B is a right-inverse, and (little)b is the left-inverse:

AB = I  and  bA = I
b(AB) = bI
(bA)B = b
IB = b
B = b
[]  << end-of-proof symbol.

NOW you can proceed with your proof.
Proof: Suppose B, C are both inverses of A.
That means  AB = BA = I  and  AC = CA = I
Then  B = BI = B(AC) = (BA)C = IC = C
[]

---------- FOLLOW-UP ----------

QUESTION: hi  thanks ..the soulution is clear ....I just wondered how you would interpet the  Equation  Ax=Yx   where  x is the  eigenvector and Y is the eigenvectr  if the eigenvector if the  value  was  0  as oppoesed to 2 or any other number

Answer
Hi, Andrew,

QUESTION: hi  thanks ..the solution is clear ....I just wondered how you would interpet the  

Equation  Ax=Yx   where  x is the  eigenvector and Y is the eigenvectr

>> Do you mean that  Y is the eigenVALUE?

if the eigenvector if the  value  was  0  as oppoesed to 2 or any other number
..........................
I don't think your Ax = Yx is at all clear.  Suppose that A,Y are 2*2 and x is 2*1, and .

Now the equation  Ax = Yx is equivalent to  (A-Y)x = ZERO.   (ZERO means the zero vector)  

If the vector x is not ZERO, then there is a solution only if the matrix  A-Y is NOT invertible, i.e. does not have an inverse.  (Why? because then  (A-Y)'(A-Y)x = (A-Y)' ZERO, and then  x = ZERO.)

So what you normally do (this is your standard eigenvalue stuff) is to assume  Y = kI, where I is the identity matrix.  Then A - kI is non-invertible if its determinant is zero.  So you write:

det(A - kI) = 0.

which becomes a polynomial equation in k, which you solve using your basic algebra (and a lot of luck).  Then for each solution k, you compute an eigenvector.  Takes about 2 hours.  

Have fun.