Talk:Adjugate matrix
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[edit] adjoint is not used?
The adjugate has sometimes been called the "adjoint", but that terminology is ambiguous and is not used in Wikipedia. Today, "adjoint" normally refers to the conjugate transpose.
A more generic example is this: given matrix A its adjoint is ...
[edit] transpose required
after the co-factoring break up is done as shown in the picture, you need to take the transpose of that entire matrix to get the real adjoint, this is not mentioned on the picture or the text
I'm a mere student using this site to help with coursework, so take this with a pinch of salt. But when you say
"Today, "adjoint" normally refers to the conjugate transpose."
Might I suggest that you mean the conjugate transpose of the cofactor matrix?
Please ignore if I'm wrong or if you feel this is implied.
ta,
Th
--I don't think they mean conjugate transpose of the cofactor matrix. The cofactor matrix is only involved in the *classical adjoint* or adjugate, whereas the *adjoint* is precisely the conjugate (i.e. take the complex conjugate of each element in the matrix) transpose of the matrix. 74.104.1.193 06:08, 4 February 2007 (UTC) Jordan
[edit] correction
the last adjugate example, the 3 by 3 with A subscripts, is incorrect. the result needs to be transpose.
Fixed. TooMuchMath 19:14, 28 January 2006 (UTC)
[edit] q(A)
I read If p(t) = det(A - tI) is the characteristic polynomial of A and we define the polynomial q(t) = (p(0) - p(t))/t, then adj(A) = q(A)., but q(A) = (p(0) - p(A))/A and p(A) = det(A - AI) = 0.. Maybe you mean qA(t) --151.28.36.120 07:43, 27 September 2006 (UTC)
There was actually no problem here. I clarified this in the text by noting the standard way to understand q(A) with q a polynomial is as the sum q_0 + q_1 A + ...+ q_n A^n where q_n are the coefficients of q(t). You are correct that p(A)=0, however this doesn't imply q(A) = 0, rather q(A) = q(0)/ A = (deta)/A = adj(A)! (Incidentally, the proof that p(A)=0 since det(A-A I) =0 is incorrect, as there is a priori no reason that p(A) = det (A-AI)!. To explain: for an arbitrary matrix B one defines p(B) = p_0 + p_1 B + ... + p_n B^n with p_j the coefficients of p(t) = det (A -t I). And indeed, it is not necessarily true that p(B) = det (A -B) I! A simple example is
A = ( 0 1 // 0 0 ) and B = (0 1 // 1 0 ) // = new row .
Then p(A)= det (A -t I) = t^2 so p(B) = B^2 = I. However, det (A - B) = 0!) --Jhschenker 16:30, 18 October 2006 (UTC)
