MHB JustCurious's question at Yahoo Answers (Diagonalization)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Diagonalization
AI Thread Summary
The discussion focuses on the diagonalization of matrices with mutually orthogonal eigenvectors, specifically addressing the matrix A represented as A = UDU^-1, where D is diagonal and U consists of the eigenvectors. The key suggestion is to replace U^-1 with U^T, which is valid for orthogonal matrices, and then take the transpose of both sides. This leads to the conclusion that A is equal to its transpose, demonstrating that A is symmetric. The properties of transposition and orthogonality are crucial in this proof. The discussion emphasizes the relationship between diagonalization and symmetry in matrices.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

his problem is about diagonalization of matrices A which have n mutually orthogonal eigenvectors, each of which has length one. It is customary to write U for the matrix with columns constructed from the eigenvectors. The problem is straightforward, but requires you to follow a given suggestion. Here is the problem, followed by the suggestion.
Suppose A = UDU^-1;
where D is diagonal and U is given as above. The entries of D; U are real numbers. Show that A is equal to its transpose matrix.
Suggestion: In the diagonalization formula for A, replace U^-1 by U^T (this is valid for such matrices) and then take the transpose of both sides. Compare.
In the literature, A is called symmetric, and U is called orthogonal.

Here is a link to the question:

Diagnalization with matrices? - Yahoo! Answers


I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello JustCurious,

We have $U^{-1}=U^T$, hence $A=UDU^T$. Then, using well kown properties of transposition $$A^T=(UDU^T)^T=(U^T)^TD^TU^T=UDU^T=A$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top