Talk:Square root of a matrix

message: I think the proof of the unitary freedom of roots is not correct. It assumes that the rows of B, the positive root of T, span Cn. This is equivalent to saying B is invertible, which need not to be the case. B won't be invertible if T isn't, and T will only be invertible if it is positive-definite. If it is positive-semidefinite, having some null eigenvalues, T will be singular. For an example, pick T = 0, then B = 0, but the columns of B don't span anything at all. Daniel Estévez (talk) 12:46, 6 December 2008 (UTC)
replies: text: in the finite dimensional setting, a partial isometry can be extended to a unitary. for the same reason, U can be taken to be unitary in the polar decomposition UP of a matrix. Mct mht (talk) 17:16, 6 December 2008 (UTC)
replies: text: I really don't see how that helps. In the proof it is said that each column of A is linear combination of only the columns of B, because the columns of B span Cn. Moreover, I dont see how B can be an orthogonal basis for Cn. If T diagonalizes as T = UDU*, with eigenvalues real and not negative, because T is positive semidefinite. Then you pick B as B = UD1/2U*. U is an orthonormal basis for Cn, but B isn't. What I'm saying is that if T (and then B) is positive definite, then the proof is OK, but if T is only positive semidefinite, then B has kernel, and so its columns cannot span the whole Cn. Of course I may be completely wrong, as I am not really into linear algebra/functional analysis Daniel Estévez (talk) 18:11, 6 December 2008 (UTC)
replies: text: if A*A = B*B, the proof constructs a partial isometry whose initial and final subspaces are Ran(A) and Ran(B). when A (and therefore B) is not invertible, their ranges are not all of C^n, as you say. this is where one uses "in the finite dimensional setting, a partial isometry can be extended to a unitary." call this unitary U, then AU*UA = B*B. Mct mht (talk) 21:30, 6 December 2008 (UTC)
replies: text: Thanks. Now I understand how the proof works, provided you can construct the partial isometry. But I think this is not stated very clearly in the current proof. I would be glad if you reread the proof, adding this details you have told me. Daniel Estévez (talk) 10:22, 7 December 2008 (UTC)
thread_title:  Proof of unitary freedom 

message: This article states that Cholesky decomposition gives the square root.
This is, I think, a mistake. B.B = A and L.L^T = A are not the same, and B will not equal L, unless L is diagonal. c.f. http://yarchive.net/comp/sqrtm.html
I have edited this article appropriately
--Winterstein 11:03, 26 March 2007 (UTC)
replies: text: well, the Cholesky decomposition gives a square root. the term "square root" are used in different senses in the two sections. Mct mht 11:09, 26 March 2007 (UTC)
replies: text:  At the beginning of section Square root of positive operators the article defines the square root of A as the operator B for which A = B*B and in the next sentence it denotes the operator T½ to mean the matrix for which T = (T½)2. This was very confusing until I realized that T½ is necessarily self-adjoint and that it is therefore also a square-root in the former sense and that only the self-adjoint square root is unique. Is this reasoning correct? --Drizzd (talk) 11:17, 9 February 2008 (UTC)
replies: text: yes, more precisely the positive square root is unique. Mct mht (talk) 17:23, 6 December 2008 (UTC)
replies: text:  surely this should be changed then. Cholesky decomposition does not give a unique root. If L is considered a root because L.L^T = A then L.R where R is orthogonal is also a root since L.R.R^T.L^T = L.L^T = A .  —Preceding unsigned comment added by Parapunter (talk • contribs) 05:01, 3 March 2010 (UTC)
thread_title:  Cholesky vs square root 

message: I think the proof has some flaws. In fact it says that the property of having  2^n  different square root matrices stems from the fact that the matrix is diagonalizable (ie it can be written as  P^{-1}DP); but it is clearly wrong, as the identity matrix is diagonalizable but, as reported above, it has infinite square roots. Maybe it should be make explicit in the proof that the key property is that every eigenvalue is distinct ? (I didn't check if it would work that way, though.. Anyhow it is wrong as currently stated )  — Preceding unsigned comment added by 80.180.2.4 (talk) 21:15, 11 December 2014 (UTC)
replies: text: I moved your comment to the bottom of the list, as is customary, from the top. It must be midnight in Parabiago.  What part of the up-front qualifying statement of having n distinct eigenvalues is not clear? Does any identity matrix have more than one distinct eigenvalue? Cuzkatzimhut (talk) 21:43, 11 December 2014 (UTC)

text: The statement "where  ½ is any square root matrix of , which must be diagonal with diagonal elements equal to square roots of the diagonal elements of  " fails for equal eigenvalues. So, e.g., you may convince yourself that, for 2×2, it holds for diag(1,a), as long as a≠1. Cuzkatzimhut (talk) 01:37, 13 December 2014 (UTC)

text: I reiterated the condition, somewhat repetitively, to demarcate the correct statement provided. No, personally, the leading discussion of non-diagonal square roots for the identity suffices.  If you wish to provide a pithy, compact footnote, or parenthetical statement, preferably with references, justifying the evident to your satisfaction, you may propose it here, on this page. Why don't you get a WP account? Cuzkatzimhut (talk) 17:09, 13 December 2014 (UTC)

text: I am not sure what you want... I disagree that it is wrong. I can't bring myself to imagine diagonalizability of D as confusable with distinct eigenvalues. Maybe you can propose something that is less confusing to you. You want to skip "This is because" and just Provide the correct statement? I cannot second guess how one can get confused. I will be more repetitive, but it is going to get weird, I fear.... Maybe you could propose what you want to see... Cuzkatzimhut (talk) 17:09, 14 December 2014 (UTC)
thread_title:  Proof at the beginning of the Properties section