View Full Version : Proofs about Matrix
athrun200
Sep18-11, 02:14 AM
1. The problem statement, all variables and given/known data
See question 5
2. Relevant equations
3. The attempt at a solution
For part a, it is very easy.
Multiply the inverse of A 2 times on both side, we can see the B=inverse of A.
i.e. The required B is inverse of A, then the proof is finished.
But how about part b?
It seems it is the same part a.
Is part b also correct?
HallsofIvy
Sep18-11, 05:27 AM
Part b is quite different from part a- and the difference is important to learn. Mathematics must be very very precise in its wording- unlike science we don't have observations and experiments to fall back on. In other words, we can't just look at the real world- words are everything!
In part a it ask if, given a non-singular matrix A, there exist a matrix B such that AB^2= A. You are right- just multiply, on the left, on both sides by A^{-1}, which exists because A is non-singular, and the equation becomes AB= I. Yes, B exists and is the inverse of A.
In part B, it asks if there exists a matrix B such that, for any non-singular matrix, A, A^2B= A. "Any" is the crucial word there. Is there a single matrix B that is the inverse of all invertible matrices?
athrun200
Sep18-11, 05:35 AM
Part b is quite different from part a- and the difference is important to learn. Mathematics must be very very precise in its wording- unlike science we don't have observations and experiments to fall back on. In other words, we can't just look at the real world- words are everything!
In part a it ask if, given a non-singular matrix A, there exist a matrix B such that AB^2= A. You are right- just multiply, on the left, on both sides by A^{-1}, which exists because A is non-singular, and the equation becomes AB= I. Yes, B exists and is the inverse of A.
In part B, it asks if there exists a matrix B such that, for any non-singular matrix, A, A^2B= A. "Any" is the crucial word there. Is there a single matrix B that is the inverse of all invertible matrices?
Well, after listening to your explaination, I know part b is obvious wrong.
However, I wonder how to write it out.
HallsofIvy
Sep18-11, 05:37 AM
"No, there does not exist a single matrix, B, such that A^2B= A for all non-singular matrices, A."
athrun200
Sep18-11, 05:43 AM
"No, there does not exist a single matrix, B, such that A^2B= A for all non-singular matrices, A."
Oh, this is the prove?
athrun200
Sep18-11, 05:47 AM
Let me try for part d.
Since A is non-singular, A^{-1} exists.
So \vec{x}=A^{-1}\vec{y} exists.
So, there exists \vec{x} s.t. A\vec{x}=\vec{y}
Again, how to disprove part c?
By simply saying NO, there doesn't exist?
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