Determinant

George Saliaris

New member
Joined
Dec 15, 2019
Messages
43
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 :D
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..
 
Last edited:

blamocur

Full Member
Joined
Oct 30, 2021
Messages
562
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 :D
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
Joined
Oct 30, 2021
Messages
562
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 :D
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] ?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
25,852
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 :D
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?
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
11,462
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
Joined
Oct 30, 2021
Messages
562
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 :D
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:
[math]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)[/math]
 

George Saliaris

New member
Joined
Dec 15, 2019
Messages
43
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:
[math]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)[/math]
From my tries I honestly can't see what thing led me to that..(I am still searching)
 

blamocur

Full Member
Joined
Oct 30, 2021
Messages
562
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]
 

blamocur

Full Member
Joined
Oct 30, 2021
Messages
562
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.
 
Last edited:

blamocur

Full Member
Joined
Oct 30, 2021
Messages
562
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:
[math]A=\left(\begin{array}{cc} 0&1\\1&0\end{array}\right) \;\;B= \left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)[/math]
I believe what you can prove is that [math]\left|\det\left(A^TB-AB^T\right)\right| = \left|\det\left(B^2-A^2\right)\right|[/math]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
Joined
Dec 15, 2019
Messages
43
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
Joined
Oct 30, 2021
Messages
562
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:
[math]A=\left(\begin{array}{cc} 0&1\\1&0\end{array}\right) \;\;B= \left(\begin{array}{cc} 0&-1\\1&0\end{array}\right)[/math]
I believe what you can prove is that [math]\left|\det\left(A^TB-AB^T\right)\right| = \left|\det\left(B^2-A^2\right)\right|[/math]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
Joined
Dec 15, 2019
Messages
43
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..
 
Top