A is square, A^2 = A -> det(A) = 0 or 1

[Jamin2112]

A is square, A^2 = A ----> det(A) = 0 or 1

Homework Statement

I need to prove the thread title.

Homework Equations

The only relevant equation I can think of is
det(A) = 0 <---> A * inv(A) = I

The Attempt at a Solution

Been toying around with it but not getting anywhere. Help, brahs?

 

[Dick]

Your 'only relevant equation' makes no sense. Can't you do a little better than that? How is det(AB) related to det(A) and det(B)?

 

[Jamin2112]

Your 'only relevant equation' makes no sense. Can't you do a little better than that?

The determinant of a matrix A is zero if and only if there exists a matrix B such that A*B = I. In that case we call B in the inverse of A, and we denote it A^-1.

 

[Dick]

The determinant of a matrix A is zero if and only if there exists a matrix B such that A*B = I. In that case we call B in the inverse of A, and we denote it A^-1.

If the determinant of a square matrix is zero, then it has no inverse. Check your book again.

 

[Jamin2112]

If the determinant of a square matrix is zero, then it has no inverse. Check your book again.

hahah sorry, brah

I took this class over 2 years ago. My buddy asked me this question and I guess I remembered some facts incorrectly.

So det(A) = 0 iff A has no inverse. In that case I'll need another relevant equation with which I can launch into a rigorous and elegant proof. Got a suggestion, brah?

 

[Dick]

hahah sorry, brah

I took this class over 2 years ago. My buddy asked me this question and I guess I remembered some facts incorrectly.

So det(A) = 0 iff A has no inverse. In that case I'll need another relevant equation with which I can launch into a rigorous and elegant proof. Got a suggestion, brah?

Look at my post #2. Product of matrices is related to product of determinants, isn't it?

 

[Jamin2112]

Look at my post #2. Product of matrices is related to product of determinants, isn't it?

It's not obvious to me that it is. Prove it.

 

[HallsofIvy]

det(AB)= det(A)det(B) is true but I would not use it. If $A^2= A$, then $A^2- A= A(A- I)= 0$. So either A or A- I is not invertible.

 

[Ray Vickson]

Here are some hints.
(1) Go to the library.
(2) Get a book.
(3) Read it.

RGV