Determinant Properties

[hkus10]

1) If det(AB) = 0, is det(A) or det(B) = 0? Give reasons for your answer.

 

Q1) First, cannot both det(A) or det(B) be 0? If it can, is this statement false. In any case, how can I prove that this is true for all statement since I only know how to find an example to show this is true, which cannot represent all the possibility.

 

2) Show that if A is singular and Ax = b, b is not equal to 0, has one solution, then it has infinitely many.

 

Q2) How to approach this question?

 

3) Let A^2 = A. Prove that either A is singular or det(A) = 1.

 

Q3) How can I approach this question?

[timo]

1) If det(AB) = 0, is det(A) or det(B) = 0? Give reasons for your answer.

 

Q1) First, cannot both det(A) or det(B) be 0?

You probably meant "both, det(A) and det(B), be zero"? Sure. Take A=B=0 (0 meaning a matrix with all entries being zero), for example.

If it can, is this statement false.

A mathematician visits a car dealer. He looks around a bit, and then he's approached by the salesman. "Good day, sir. Already found something you like?" The mathematician answers: "Yes indeed. I'll either take the blue one here or the red one over there - but not both".

I think I got this one from this forum's jokes thread. The point is: in mathematics, an "or" usually means an inclusive or, so "det A=0 or det B=0" does include the case that "det A=0 and det B=0".

 

In any case, how can I prove that this is true for all statement since I only know how to find an example to show this is true, which cannot represent all the possibility.

The determinant of a matrix M being zero means that there are non-zero vectors v for which Mv=0, because the determinant is the product of the eigenvalues.

 

2) Show that if A is singular and Ax = b, b is not equal to 0, has one solution, then it has infinitely many.

Q2) How to approach this question?

Incidently, my statement about the zero eigenvalue (and the corresponding eigenvectors) should help a lot on this one.

[ajb]

1) If det(AB) = 0, is det(A) or det(B) = 0? Give reasons for your answer.

 

Q1) First, cannot both det(A) or det(B) be 0? If it can, is this statement false. In any case, how can I prove that this is true for all statement since I only know how to find an example to show this is true, which cannot represent all the possibility.

 

So does [math]\det(AB) = \det(A)\det(B)[/math]?

 

 

2) Show that if A is singular and Ax = b, b is not equal to 0, has one solution, then it has infinitely many.

 

Q2) How to approach this question?

 

There is a more slightly more general theorem you should prove.

 

Theorem A system of linear equations as a) no solutions, b) exactly one solution or c) infinite number of solutions.

 

You can prove this quite directly. See what you can do, ask for more hints if you get stuck.

 

You then have to show that for a singular matrix there is more than one solution. It will be worth proving that if your matrix were invertible then it has one solution.

 

3) Let A^2 = A. Prove that either A is singular or det(A) = 1.

 

Q3) How can I approach this question?

 

See if my hints for Q1) help.

Edited February 23, 2011 by ajb