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Orthogonal Diagonalization of symetric matrix

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Martin, Burda

Lecture 10 notes by Y. Burda 1 Towards orthogonal diagonalization of a sym- metric matrix By now we know that if a quadratic form q has coecient matrix A relative to the standard basis, then relative to a dierent basis B it has coecient matrix B = P AP, where P = [I] . E,B If the new basis is orthonormal, then P is orthogonal and B = P AP = T 1 P AP is similar to A. Our hope is that if for some reason we could nd an orthonormal basis of eigenvectors of A, then relative to this basis the coecient matrix of q will be diagonal, i.e. the form itself would be equal to a sum of squares with some coecients. Why should we be able to nd an orthonormal basis of eigenvectors for A? It turns out that the fact that A is symmetric is just enough for that. The next theorem is the rst indication that it might be possible: Theorem. Suppose A is a real symmetric matrix. Then eigenvectors with dierent eigenvalues are orthogonal to each other. Proof. Suppose Av = 1v and 1 1= v , w2ere 2 2 and v ,v1are 2 1 2 non-zero. Then v1 Av 2 = v 1 ( v2 2= 2v v 1 On 2he other hand v Av = v Av = v A v = (Av ) v = (Av ) v = v v . Thus 2 1 2 1 2 1 2 1 2 1 1 2 1 1 v 2 v 2 1 . 2ince and1 are d2stinct, v v = 1. 2 Now we have the following algorithm that works if A is diagonalizable: Choose bases for eigenspaces: v ,...,v 1 for E , 1 n1 1 v1,...,v n for E 2 2 ..., v ,...,v k for E 1 nk k By applying Gram-Schmidt process to each of these bases, we can nd orthonormal bases for each of these spaces: 1 1 w 1...,w n1 for E 1 w ,...,w 2 for E , 1 n2 2 ..., k k w 1...,w nk for E k 1
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