# Determinant

#### George Saliaris

##### New member
If A (n x n) matrix
B (n x b) matrix
AA^T = ΒΒ^Τ = I
Prove that:
det[(A^T *B) - (B^T * A)] = det[A^2 - B^2]
I proved that AB = BA but apart from that I am doing a bunch of useless algebra and thus I am stuck.. Could somebody give me an insight? Ty
P.S1 : It's not a homework exercise/extra credit etc. I am doing in order to practice in determinants
PS2 : I found this online and I thought It could be solved with determinant properties, inverse matrix and properties and the fact that (AB) ^ T =B^T * A^T and not something else.. Correct me If I am wrong..

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#### Jomo

##### Elite Member
Look up orthogonal matrices.

#### blamocur

##### Full Member
If A (n x n) matrix
B (n x b) matrix
AA^T = ΒΒ^Τ = I
Prove that:
det[(A^T *B) - (B^T * A)] = det[A^2 - B^2]
I proved that AB = BA but apart from that I am doing a bunch of useless algebra and thus I am stuck.. Could somebody give me an insight? Ty
P.S1 : It's not a homework exercise/extra credit etc. I am doing in order to practice in determinants
PS2 : I found this online and I thought It could be solved with determinant properties, inverse matrix and properties and the fact that (AB) ^ T =B^T * A^T and not something else.. Correct me If I am wrong..
AB = BA ? Even for [imath]b \neq n[/imath] ?

#### blamocur

##### Full Member
If A (n x n) matrix
B (n x b) matrix
AA^T = ΒΒ^Τ = I
Prove that:
det[(A^T *B) - (B^T * A)] = det[A^2 - B^2]
I proved that AB = BA but apart from that I am doing a bunch of useless algebra and thus I am stuck.. Could somebody give me an insight? Ty
P.S1 : It's not a homework exercise/extra credit etc. I am doing in order to practice in determinants
PS2 : I found this online and I thought It could be solved with determinant properties, inverse matrix and properties and the fact that (AB) ^ T =B^T * A^T and not something else.. Correct me If I am wrong..
It seems that you theorem makes sense only when [imath]b = n[/imath], otherwise how do you define [imath]B^2[/imath] ?

#### George Saliaris

##### New member
Yes, I made a mistake b = n

#### Subhotosh Khan

##### Super Moderator
Staff member
If A (n x n) matrix
B (n x b) matrix
AA^T = ΒΒ^Τ = I
Prove that:
det[(A^T *B) - (B^T * A)] = det[A^2 - B^2]
I proved that AB = BA but apart from that I am doing a bunch of useless algebra and thus I am stuck.. Could somebody give me an insight? Ty
P.S1 : It's not a homework exercise/extra credit etc. I am doing in order to practice in determinants
PS2 : I found this online and I thought It could be solved with determinant properties, inverse matrix and properties and the fact that (AB) ^ T =B^T * A^T and not something else.. Correct me If I am wrong..
What does A^T or B^T mean?

#### George Saliaris

##### New member
What does A^T or B^T mean?
A^Τ means that A is the transpose matrix of A (If I am saying that correct)..

#### Jomo

##### Elite Member
No, A^T is the transpose of A. If as you said, A is the transpose matrix of A, then A = A^T. Is there any reason to assume that A=A^T?
Now since A*A^T = I, what else can you say about A^T? What if A*C = I, what can you say about C?

#### blamocur

##### Full Member
AB = BA ? Even for [imath]b \neq n[/imath] ?
If A (n x n) matrix
B (n x b) matrix
AA^T = ΒΒ^Τ = I
Prove that:
det[(A^T *B) - (B^T * A)] = det[A^2 - B^2]
I proved that AB = BA but apart from that I am doing a bunch of useless algebra and thus I am stuck.. Could somebody give me an insight? Ty
P.S1 : It's not a homework exercise/extra credit etc. I am doing in order to practice in determinants
PS2 : I found this online and I thought It could be solved with determinant properties, inverse matrix and properties and the fact that (AB) ^ T =B^T * A^T and not something else.. Correct me If I am wrong..
I don't know how you proved that [imath]AB = BA[/imath] since there is a simple counterexample with rotation matrices by 90 degrees around the axes, e.g. around X and Z:
$A=\left(\begin{array}{rrr}1 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{array}\right) \;\;\;B=\left(\begin{array}{rrr}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 1\end{array}\right)$

#### George Saliaris

##### New member
No, A^T is the transpose of A. If as you said, A is the transpose matrix of A, then A = A^T. Is there any reason to assume that A=A^T?
Now since A*A^T = I, what else can you say about A^T? What if A*C = I, what can you say about C?
Ughhh yes,my mistake..C is the inverse matrix of A, in this case A^T = A^(-1)
I don't know how you proved that [imath]AB = BA[/imath] since there is a simple counterexample with rotation matrices by 90 degrees around the axes, e.g. around X and Z:
$A=\left(\begin{array}{rrr}1 & 0 & 0\\ 0 & 0 & 1\\ 0 & -1 & 0\end{array}\right) \;\;\;B=\left(\begin{array}{rrr}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 1\end{array}\right)$
From my tries I honestly can't see what thing led me to that..(I am still searching)

#### blamocur

##### Full Member
Hint : if [imath]X[/imath] is an arbitrary square matrix and [imath]U[/imath] is an arbitrary orthogonal matrix of the same dimension as [imath]X[/imath] then [imath]\det(AX)=\det(XA) = \det(X)[/imath]

#### George Saliaris

##### New member
Hint : if [imath]X[/imath] is an arbitrary square matrix and [imath]U[/imath] is an arbitrary orthogonal matrix of the same dimension as [imath]X[/imath] then [imath]\det(AX)=\det(XA) = \det(X)[/imath]
Um where is U ?

#### blamocur

##### Full Member
In this case [imath]U[/imath] and [imath]X[/imath] do not refer to anything in your post. I wanted to point out a general matrix property which I applied to solve your problem.

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#### blamocur

##### Full Member
Turns out you cannot prove the statement in the original post because it is wrong -- and don't ask me why it took me so long
Here is a counterexample:
$A=\left(\begin{array}{cc} 0&1\\1&0\end{array}\right) \;\;B= \left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)$
I believe what you can prove is that $\left|\det\left(A^TB-AB^T\right)\right| = \left|\det\left(B^2-A^2\right)\right|$The absolute values in the last identity can be removed iff [imath]\det(A)\det(B) > 0[/imath]
Does this make sense?

#### George Saliaris

##### New member
Well so far I have only proved that A^TB = A^T B^T B^2
B^TA = (B^T) ^2 B A^T A^2
But i doubt that's useful...

#### blamocur

##### Full Member
Just a hint: matrices don't have to be identical to have the same determinant.
Well so far I have only proved that A^TB = A^T B^T B^2
B^TA = (B^T) ^2 B A^T A^2
But i doubt that's useful...

#### blamocur

##### Full Member
Turns out you cannot prove the statement in the original post because it is wrong -- and don't ask me why it took me so long
Here is a counterexample:
$A=\left(\begin{array}{cc} 0&1\\1&0\end{array}\right) \;\;B= \left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)$
I believe what you can prove is that $\left|\det\left(A^TB-AB^T\right)\right| = \left|\det\left(B^2-A^2\right)\right|$The absolute values in the last identity can be removed iff [imath]\det(A)\det(B) > 0[/imath]
Does this make sense?
Also, I am curious: where does this incorrectly stated problem come from. Is this a mistake in a textbook, or a mistake in a derivation from another problem?

#### George Saliaris

##### New member
Also, I am curious: where does this incorrectly stated problem come from. Is this a mistake in a textbook, or a mistake in a derivation from another problem?
Last exercise from a 2014 university examination...
I also tried squaring both sides of what I am supposed to prove (the matrices inside the the determinants) but that did not work..

#### blamocur

##### Full Member
Last exercise from a 2014 university examination...
I also tried squaring both sides of what I am supposed to prove (the matrices inside the the determinants) but that did not work..
Take a look at [imath]A\left(A^TB-AB^T\right)B[/imath]