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- 1. Prove that every symmetric matrix is diagonalisable by an orthogonal matrix. Solution (i) Suppose ( Sv=lambda v ) Then ( <Sx,v>=(Sx)^Tv=x^TS^Tv = x^TSv = <x,Sv> = lambda <x,v> ) So if x is in W, we have <x,v>=0 and <Sx,v>=0 => Sx is in W. So the linear transformation x -> Sx takes vectors in W to W (ii) Let's define ( T : W -> W ) such that ( T(w) = Sw ) First we know that W has dimension (n-1) and let ( B = {b1,...,b_{n-1}} ) be an orthonormal basis for W. ( Sb_j in W : exists (a_k)_{1leq k leq n-1} s.t. Sb_j = sum_{k=1}^{n-1} a_k b_k ) Then since the basis B is orthonormal we have : ( <b_i,Sb_j> = a_i ) since we have ( <b_i,b_j> = delta_{ij} ) ( T(b_i)=Sb_i=sum_{k=1}^{n-1} <b_k,Sb_j> b_i ) Which proves the first statement as the B-matrix for T is composed of the coefficients of the ( T(b_i) ) that are exactly ( <b_i,Sb_j> ) at the ij-the entry. To conclude : ( <bi,Sbj> = <Sbj,bi>=(Sbj)^Tbi=bj^TS^Tbi=bj^TSbi=<bj,Sbi> )
- 2. (iii) Let's do induction on the size of S. Suppose n=1. Then ( {b_1=v/<v,v>} ) is an orthonormal basis of eigenvectors for S. Now suppose the statement is true for (n-1). We will prove it for n. Suppose S is of size n. We take ( b_1 = v/<v,v> ) one eigenvector of S. Then W as defined in (ii) has size (n-1) and we can apply our induction hypothesis to the symmetric matrix T. So we can find an orthonormal matrix of eigenvectors of T ( {b_2, dots,b_{n}} ) . Note that the eigenvectors of T are eigenvectors of S, since ( T(w)=Sw ) Then we have found an orthonormal basis of eigenvectors of S : ( { b_1, dots, b_n} ).