Proof: determinants of orthogonal matrices

RM5152

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Hello ?. I wonder if anyone could look at my answer and workings to this question and tell me if I have properly established the proof .. if not could you tell me where I went wrong please. Thanks very much. Ps 3(a) answer is in a different attachment to 3(b) and 3(c).
 

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I just read your proof and I am going to much harsher than blamocur and say that your proof is not good as all. You assumed something that has not been proven, namely that (I−P)=det(I)−det(P). For the record, if this has been proven in class, then you should mention that in your proof.
I would also like to see this proof.
 
Looks good to me except for the part where you say [imath]\det(I_n - P) = \det(I_n) - \det(P)[/imath] -- don't you think it has to be proven
Yes I suppose I just assumed that because det (In) is just 1 .... I have tried to prove this but am really struggling. I’ve attached my workings.. could you help me out on where to go with this proof please. Thank you
 

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Yes I suppose I just assumed that because det (In) is just 1 .... I have tried to prove this but am really struggling. I’ve attached my workings.. could you help me out on where to go with this proof please. Thank you
[imath]\det(I_n - P) = \det (P^T (I_n - P)) = ...[/imath]
 
Any further hints ? I just can’t get it - this is my workings
This was to prove that [imath]\det (I_n-P) = 0[/imath] :
[math]\det(P^T(I_n-P)) = \det (P^T - P^TP) = \det ((P - P^TP)^T))=...[/math]
 
Hint: If A is an odd square matrix and det(A) = det(-A) then det(A) = ? (and why?)
 
Hint: If A is an odd square matrix and det(A) = det(-A) then det(A) = ? (and why?
This is what I have gotten .. I think this proves that det(P) = - det(P) because I know P is an odd square matrix… but can I use this to prove that this applies to the matrix In-P even though I have not been told that In-P is an odd square matrix? Thanks for the help
 

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This is what I have gotten .. I think this proves that det(P) = - det(P) because I know P is an odd square matrix… but can I use this to prove that this applies to the matrix In-P even though I have not been told that In-P is an odd square matrix? Thanks for the help
I was hoping that you'd realize that my hint was wrong. Please correct the hint for odd square matrices.
 
Can you please give me another hint on how to prove that det(In-P) = det(In) - det(P) ... I am really struggling and I don't know where to go from here. If I find det (P^T(ln-P) = P^T-PP^T ... then this just equals P^T - In? I am so stuck
 
Can you please give me another hint on how to prove that det(In-P) = det(In) - det(P) ... I am really struggling and I don't know where to go from here. If I find det (P^T(ln-P) = P^T-PP^T ... then this just equals P^T - In? I am so stuck
[imath]\det (I-P) = \det (P^T(I-P)) = \det(P^T-I) = \det((P-I)^T) = \det (P-I) = -\det (I-P)[/imath]. Note that only for odd dimensions we can claim that [imath]\det(-M) = -\det(M)[/imath].
 
[imath]\det (I-P) = \det (P^T(I-P)) = \det(P^T-I) = \det((P-I)^T) = \det (P-I) = -\det (I-P)[/imath]. Note that only for odd dimensions we can claim that [imath]\det(-M) = -\det(M)[/imath].
I'm still unsure how to split the - det (In-P) into det(In) - det(P) .. because how can I find det (In-P) without splitting them up ?? for my proof and to prove that det (In-P) = 0 I think I must have to prove that det (In-P)= det (In) - det(P) but I just can't figure out how to prove this... I'm sorry for so many questions.
 
You cannot split them in general case, not even when [imath]\det(P) = 1[/imath]. Your matrix ``splits'' only because it is orthonal [imath]P[/imath] and the result is 0, as has been proven. Here is an example of a 3x3 matrix which "does not split":
[math]P = \left( \begin{array}{rrr} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array}\right)[/math]for which
[math]\det(P) = 1[/math][math]\det(I_3-P) = -1[/math]
 
You are missing the whole point which we have been trying to tell you.
det(I−P)=det(P-I) which blamocur showed you the proof of.
Now note that det(P-I) = det(-(I-P)) = -det(I-P) since P-I is an odd square matrix.

Now you need to put these two facts together. I will even tell you the exact two facts.
det(I−P)=det(P-I) and det(P-I) =-det(I-P).
Just using the two equations in bold you should be able to make some conclusion. (Hint: the conclusion is what you want to show!)
Further hint: If a=b and b=c, then a=c, where a,b and c are real numbers.
 
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