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)_{1\\leq 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> \\) (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\\} \\). .

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} ).