Proving #2

[franz32]

Hello , it's me again.

It's about proving linear aalgebra concepts.

1. Show that if A = A^-1, then det (A) = 1 or -1.

2. Show that if A^T = A^-1, then det (A) = 1 or -1.

[master_coda]

\begin{align}
\det(AB)&=\det(A)\det(B) \\
\det(A^t)&=\det(A)
\end{align}

These are all you need.

[franz32]

Hello again,

So the two are enough to prove the two theorems.

But, I got only +1, how about -1?

[Hurkyl]

How did you do it?

[franz32]

For "If A = A^-1, then det A = +1 or -1."

1. det A = det (A^-1) \\ determinating both sides
2. det A = 1 / (det A) \\ proeprty: det (A^-1) = 1 / (det A)
3. (det A)(det A) = 1 \\ Multiplying det A both sides
4. (det A^2) = 1 \\ simplifying
5. det A = + or 1 (square root of 1 ) \\ extracting sq.root
6. Thus, det A = +1 or -1.

For "If A^t = A^-1, then det A = 1 or -1."

1. (A^T) A = (A^-1) A \\ If the inverse of A is said to be existing, then A itself must exist and the product of the two must be an identity matrix. I multiplied A both sides of equation.

2. (A^T) A = I \\ Identity matrix.
3. det [ (A^T) A ] = det I \\ determinating both sides.
4. det(A^T) det A = det I \\ Prop: det AB = det A det B
5. det A det A = det I \\ Prop: det (A^T) = det A.
6. det A^2 = 1 \\ determinant of identity matrix is 1.
\\ simplifying
7. det A = + or - square root of 1 \\ extracting a square root.
8. Thus det A = 1 or -1.

[Muzza]

The first proof can be done using only property (1) that master_coda mentioned.

1 = det(I) = det(A * A^-1) = det(A)det(A^-1) = det(A)det(A) = det(A)^2

det(A)^2 = 1
det(A) = +/- 1

[franz32]

[:D] Hi!

Hmmm, that's correct. =) Thank you. =)

Well, was my proving to both of the two in the previous "replies" correct? (I want to know if I am doing fine in my proving).

[franz32]

Hello Muzza,

I have read your answer. It seems correct but I doubt on one point.

The det A^-1 is equal to [1 / (det A)], how come it turns out to be det A?[:))]

[Muzza]

Your proofs seem correct to me.

det(A) = 1/det(A) does hold, for det(A) = 1, or det(A) = -1 (1 = 1/1 and -1 = 1/(-1) are both true statements), which is well... what you showed that the possible values for det(A) could be :P