Question

Can someone prove that "if there exists C such that CA = I, then AC=I"?

Answer 1

That would imply that C is invertible.

Answer 2

AC= A*I*C=A*(A^(-1)*C^(-1))*C=I*I=I

Answer 3

xyy: why I = A^(-1)*C(-1), you cannot assume the existence of the matrix inverses in the first place.

Answer 4

And to the original question: I checked it up on my linear algebra book, and its too long for me to write here, but the proof exists, and is not just a simpe artithmetic.

Answer 5

Tumas, you are right. If A is not square matrix, I donot think this is a right conclusion. For simple counter example just try Q which is orthonormal matrix, and Q transpose as C.

Answer 6

xyy: Yeah its applicable to square matrices of the same size. (AB=I => BA=I)

Answer 7

Suppose CA = I ------ (1)
Suppose Ax = b ------ (2)

From (2)
Ax = b
=> (AC)Ax = (AC)b
=> A(CA)x = (AC)b
=> Ax = (AC)b ------- (3)

(2) = (3)
Ax = b = (AC)b
Thus, I = AC

Got the idea from Zareon at http://www.scienceforums.net/topic/21570-inverse-of-a-matrix/
Not sure if it's correct though

Answer 8

philip, I think the proof is flawed.
b = (AC)b => I=AC
for one, if b = 0 the above statement is false.
The proof introduced the variable b in the Ax=b statement.
I don't think one can infer anything about b here.
(one can't be sure if it is not 0 vector)

Answer 9

Since CA = I, we know that A is an invertible matrix.
Since A is invertible, Ax = b is solvable for all b's.
So, we can just assume that b is non-zero.

Answer 10

philip, the point of original statement is. (We are assuming square matrix)
if we know a left inverse exist, can we say that the left inverse also is actually a right inverse?

"A is an invertible matrix" = "There is a left inverse and right inverse and they are the same."
"CA = I" doesn't say there exists a right inverse. Let alone their equivalence.

Anyway thanks for reply. good to see someone is taking the same course roughly at the same time :p

Answer 11

actually equivalence part comes automatically.

if AC = I and BA = I

BAC = B and BAC = C
Therefore B=C

Answer 12

Did you just prove it? Looks correct.

Anyway, let me try again.
If CA = I, then the combination of the rows of A spans the whole vector space. So,
rref(A) = I
=> the columns of A are independent (still haven't really wrapped my head around how row independence implies column independence. neat fact)
=> the columns of A span the whole vector space
=> Ax = b is solvable for any b (since the columns of A span the whole space)

So let b be a non-zero vector. And proceed with the proof I showed previously. Does this work?

Answer 13

If CA = I, then the combination of the rows of A spans the whole vector space. So,
rref(A) = I

Why is that?
I guess you are assuming C is a product of elementary row operation.
Can an arbitrary matrix be considered as a product of elementary row operation?
What if a diagonal entry is zero?

Answer 14

To pcompassion:
If CA=I, then I is the combination of the rows of A, then the row space of I is a subspace of the row space of A. Since the row space of I is the whole vector space, the row space of A must be the whole vector space too.
Suppose C is m by n, A is n by m, I is m by m. Then Rm is a subspace of Rm. So A’s rank is m and it has m independent columns and rows. If n=m, then A is square, rref (A) is I (m by m). if m<n, rref(A) is I (m by m) with zero in other rows. M cannot be larger than n.

Answer 15

Thanks yadi.
Since we are assuming square matrix, combining your answer with philips completes the proof.

Answer 16

Look to this:
If AC=I, obviously A=C^(-1)
so AC = CA = I
ELL ( Easy like life )

Answer 17

Closed

Look C = A^(-1) and A=C^(-1)
then AC=CA

Answer 18

philip, where in the book or lecture says linear independence of rows implies independence of columns? I saw them too, but can't find it.

Augusto.ACM look previous discussions,
the question asks if there is a left inverse, would it be also an right inverse?
you can't assume existence of A(-1)