1. ## maths questions

I have three questions for which I need help:

1/ Show that if a matrix has an inverse then it must be square.

2/ Let and B be square matrices such that AB = 0. Show that A cannot be invertible unless B = 0.

3/ Let B be an invertible matrix. If X and B are matrices such that AX = B, show that X = A-1B. Hence or otherwise, solve the system of equations:

2x - 3y = 1, 2x - 3y = 5
x + y = -4, x + y = 3

In question 3, where it says "X = A-1B", the '-1' is supposed to be inverted.

2. 1/ Show that if a matrix has an inverse then it must be square.
The inverse matrix was only defined for squares

2/ Let A and B be square matrices such that AB = 0. Show that A cannot be invertible unless B = 0.
I do not know if you discussed that but the ring of matricies over the reals has no divisors of zero.
Meaning that if,
AB=Z where Z is the zero matrix then,
A or B must be zero matrices.
A zero matrix never has an inverse (because anything multiplied by itself is zero matrix, just like dividing by zero).
Thus, if we want A to have an inverse it cannot be zero this causes B to be a zero matrix (disjuctive inference).

3. Originally Posted by ThePerfectHacker
The inverse matrix was only defined for squares

I do not know if you discussed that but the ring of matricies over the reals has no divisors of zero.
Meaning that if,
AB=Z where Z is the zero matrix then,
A or B must be zero matrices.
A zero matrix never has an inverse (because anything multiplied by itself is zero matrix, just like dividing by zero).
Thus, if we want A to have an inverse it cannot be zero this causes B to be a zero matrix (disjuctive inference).

Thankyou

5. Originally Posted by kgpretty
-=USER WARNED=-
---
3)
You have AX=B
You are told it is invertible.
Let A^-1 be that matrix.
Multiply left sides by it:
A^1(AX)=A^-1B
Matrices over a field form a ring.
Thus, the following move is associate.
(A^-1A)X=A^-1B
That produces identity matrix.
IX=A^-1B
By definition of Identity matrix.
X=A^-1B

The reason why you multiply the left is because matrix multiplcation IS NOT commutative.

6. I have another question for which I need help:

4. Show that if A is invertible and AB = AC, then B = C

Code:
Here's what I did:
Let A = |a b|, B = |e f|
|c d|    |g  h|

AB = |ae+bg af+bh|
|ce+dg cf+dh|

AC = |ae+bg af+bh|
|ce+dg cf+dh|

C = |e f|
|g h|

B = C

7. A small correction... matrices in general do have plenty of zero divisors. AB = 0 in general will happen whenever the image of B (as a linear transformation, eating vectors from the right) falls into the kernel of A. For example,

A = (1, 0 | 0, 0) B = (0, 0 | 1, 0)

and the product AB = 0. (sorry, I don't know how to use tex in this forum, A and B are 2 x 2 matrices, with the first row of A being (1,0), the second row (0,0), likewise for B).

Recall that an n x n matrix A is invertible precisely when the dimension of its image is n, or alternatively, when the dimension of the kernel is zero. If A is invertible and AB = 0, this then forces the image of B to be contained in the kernel of A, so the image of B is {0}. So B kills every vector, and the only matrix that does this is the zero matrix.

Question 1 is somewhat obvious, but requires some care... an n x m matrix A would be invertible if there were an m x n matrix B such that

(i) ABv = v for any (column) vector v in R^n
and
(ii) BAw = w for any w in R^m

Suppose that say, m < n. I claim then that condition (i) can never hold. Since B has m columns, its image has at most dimension m. The image of AB thus has at most dimension m (since the image of a linear transformation can never have dimension bigger than the original space). And thus the image of AB cannot be all of R^n, which it would be if B were the inverse of A.

For problem 3), I think you are being asked to show...

"if AX = B and B is invertible, then A is invertible and X = A^(-1)B"

The second part is I think obvious assuming A is invertible. Again, this follows from the observation that, if either A or X were non-invertible, it would have a non-trivial kernel, and so would any product involving them (namely B), and would thus be non-invertible.

8. Originally Posted by kgpretty
I have three questions for which I need help:

1/ Show that if a matrix has an inverse then it must be square.

.
To flesh Perfect Hacker's answer out

Suppose you have an n by m matrix A which is not a square matrix (n not equal to m)
Suppose further that it has an inverse B

Then AB = BA = I

I is a square matrix, so it is either a n by n matrix or a m by m matrix by the rules of matrix multiplication

AB would have to become a m by m matrix by the rules of matrix multiplication
BA would have to become a n by n matrix by the rules of matrix multiplication

Contradiction as they both give different sized matrices and therefore cannot equal each other

Therefore A must have been a square matrix

9. Originally Posted by kgpretty
I have another question for which I need help:

4. Show that if A is invertible and AB = AC, then B = C

Code:
Here's what I did:
Let A = |a b|, B = |e f|
|c d|    |g  h|

AB = |ae+bg af+bh|
|ce+dg cf+dh|

AC = |ae+bg af+bh|
|ce+dg cf+dh|

C = |e f|
|g h|

B = C
No

AB = AC

A is invertible so has an inverse A^(-1)

A^(-1)AB = A^(-1)AC

IB = IC

B = C

10. Originally Posted by Glaysher
To flesh Perfect Hacker's answer out

Suppose you have an n by m matrix A which is not a square matrix (n not equal to m)
Suppose further that it has an inverse B

Then AB = BA = I

I is a square matrix,
The problem is I is not the identity matrix for the non-square matrices. You can only speak of inverses when they create a group. Non-square matrices do not.

11. Originally Posted by ThePerfectHacker
The problem is I is not the identity matrix for the non-square matrices. You can only speak of inverses when they create a group. Non-square matrices do not.
Not necessarily as you can have inverses when the other axioms of groups do not hold.* Though you have pointed out the part I wasn't really happy with. Really needs a proof that no identity exists for non square matrices.

But the point that AB cannot equal BA still stands. I should have argued from that point of view as the definition of inverse means that you would have to get the same from either calculation. I vaguely remember that a proof exists that shows that you can't have two different identities

* Multiplication with 1, 2, 1/2,

Has inverses 1 inverse of 1, 2 inverse of 1/2, 1/2 inverse of 2
Has identity 1
It is associative
But it is not closed as 2* 2 = 4 which is not an element of the group

12. Originally Posted by Glaysher
I vaguely remember that a proof exists that shows that you can't have two different identities
Simple.
Let (G,*) we a group.

If ex=xe=x for all x in G
If Ex=xE=x for all x in G
Then,
ex=Ex
Since (G,*) is a group the right cancellation law stands.
Thus,
e=E
This shows uniqueness.
~~~~~~~~~~
For the other topic. It seems you and I have different definitions of an identity matrix. I take it to mean that it plays the role of the identity element in a group which does. You simply take it to mean that is preserves multiplication from at least one side. Are you sure that in linear algebra your definition is used.

13. Originally Posted by ThePerfectHacker
Simple.
Let (G,*) we a group.

If ex=xe=x for all x in G
If Ex=xE=x for all x in G
Then,
ex=Ex
Since (G,*) is a group the right cancellation law stands.
Thus,
e=E
This shows uniqueness.
~~~~~~~~~~
For the other topic. It seems you and I have different definitions of an identity matrix. I take it to mean that it plays the role of the identity element in a group which does. You simply take it to mean that is preserves multiplication from at least one side. Are you sure that in linear algebra your definition is used.
I recall it has to be from both sides and I have done work in the past when that was the definition of an identity but the field I was working in at the time long escaped my mind. However I do remember that identities exist for structures that aren't groups. Fields and rings are things that come to mind but their definitions currently escapes me. I think fields have identities. I can't remember about rings.

14. Originally Posted by Glaysher
However I do remember that identities exist for structures that aren't groups.
Magma's I believe they are called. Thus, you are treating the matrices are magmas and thus are saying they are having an identity, right?

I think fields have identities.
Definition: A field is a commutative division ring. So yes it must have an inverse.

Not necessarily when it does it is called ring with unity. (The more important rings have unity to them).

15. Originally Posted by ThePerfectHacker
Magma's I believe they are called. Thus, you are treating the matrices are magmas and thus are saying they are having an identity, right?
Never come across the term before

Originally Posted by ThePerfectHacker
Definition: A field is a commutative division ring. So yes it must have an inverse.

Not necessarily when it does it is called ring with unity. (The more important rings have unity to them).
Rings a bell. Perhaps when I retire I'll actually have time to relearn all I used to know and do a pHd