Need HELP w/ Linear Algebra Proof!

[ChineseGuy]

So I got this from my Math professor yesterday, and I am suppose to present this in class tomorrow. I'm suppose to make a proof for the statement.

"Let A be an nxn matrix such that Ax = x for every n-vector x. Then A=I "(I being the identity matrix)

To me, this just seems so obvious that it doesn't need a proof. It's like saying A*B=B what does A have to be? umm.. 1? I don't see how I can write a proof for this. Can someone help me? Where do I even start?

 

[esun]

Do it by contradiction, then. Say A were not the identity matrix. Then Ax = [a_11x_1 a_22x_2 a_33x_3 ... a_nnx_n], where a_ii != 1 for some 1 <= i <= n. Then Ax != x (since x = [x_1 x_2 ... x_n]), so our original assumption was false, so A must be the identity matrix.

 

[ChineseGuy]

I followed most of the proof, except for
"where a_ii != 1 for some 1 <= i <= n."
Computer keyboard is not the easiest thing to use when you want to type math.

So you are saying for a_ii, where i and n are greater than or equal to 1?

 

[CycloWizard]

1<=i<=n means 'i is greater than or equal to one and less than or equal to n.'

An alternative proof would be to show that x*x^-1 (that is, x times its own inverse) is equal to the identity matrix, though this is much more involved than esun's simple contradiction.

 

[ChineseGuy]

Hmm... CycloWizard's way of doing this proof is interesting. Might do it just for the heck of it Thanks for the clarification CycloWizard.

 

[esun]

You could also say something like this:

Given: Ax = x

By the definition of I, we know Ix = x, therefore

Ax = Ix
Ax - Ix = 0 (zero matrix)
(A - I)x = 0

This can only be true for A = I if it is satisfied for all x.

 

[CSMR]

Originally posted by: ChineseGuy
So I got this from my Math professor yesterday, and I am suppose to present this in class tomorrow. I'm suppose to make a proof for the statement.

"Let A be an nxn matrix such that Ax = x for every n-vector x. Then A=I "(I being the identity matrix)

To me, this just seems so obvious that it doesn't need a proof. It's like saying A*B=B what does A have to be? umm.. 1? I don't see how I can write a proof for this. Can someone help me? Where do I even start?

It's important to understand what steps are required even for the smallest deductions; that you know when one step is required or when no steps are required. A lot of important mathematical arguments rely on slights of hand which you can easily miss the importance of.

Now you don't quite understand vector spaces. A.B=B does not imply A=I.

You need to know what the definitions are.

I would assume that by "identity matrix" your professor means the matrix with 1's on the diagonal and 0's elsewhere.
(So you have to prove that the identity transformation on R^n is represented by this matrix.)
And you know the definition of matrix multiplication.
So you know what you have to do?

 

[ChineseGuy]

Cyclowizard's idea wouldn't run. Because you cannot take an inverse of a vector... I ended up proving using contradiction like esun said. Professor got the idea from reading up how google does its page ranking. Quite interesting actually. Thanks for all your help guys.