What is the meaning of the symmetrical matrix


Example 3 from the Examples sheet has shown that matrices with real elements can have complex eigenvalues. Complex eigenvalues ​​always occur in pairs, namely conjugate complex, ie if $ \ lambda = a + b \ cdot i $ is an eigenvalue of $ {\ rm \ bf A} $ then $ \ overline {\ lambda} = a - b \ cdot i $ has an eigenvalue of $ {\ rm \ bf A} $. For the following we recall the definition in the sheet Special matrices: A square matrix is ​​called symmetric if \ begin {equation *} {\ rm \ bf A ^ T = A} \ end {equation *} or \ begin {equation *} a_ {ij} = a_ {ji}. \ end {equation *}
\ begin {equation *} \ mathbf {A} = \ begin {pmatrix} 8 & 1 & 2 \ cr 1 & 5 & 4 \ cr 2 & 4 & 3 \ end {pmatrix} \ quad \ hbox {is symmetric.} \ end {equation *}
Remember the relationships: \ begin {eqnarray *} (\ rm \ bf A \ cdot B) ^ T & = & \ rm \ bf B ^ T \ cdot A ^ T \ qquad \ hbox {or} \ cr \ cr (\ rm \ bf A \ cdot B \ cdot C) ^ T & = & \ rm \ bf C ^ T \ cdot B ^ T \ cdot A ^ T \ hbox {etc.} \ end {eqnarray *}
The eigenvalues ​​of a real symmetric matrix $ {\ rm \ bf A} $ are real.
Proof: (cf. beckmannII: 1973, p.92) The characteristic polynomial of $ {\ rm \ bf A} $ $ \ vert {\ rm \ bf A} - \ lambda {\ rm \ bf I} \ vert $ has conjugate complex roots: $ \ lambda $ and $ \ overline {\ lambda} $ \ begin {eqnarray *} \ lambda & = & a + bi \ \ overline {\ lambda} & = & a - bi \ end {eqnarray *} assumption : $ \ lambda $ and $ \ overline {\ lambda} $ are complex (i.e. $ b \ ne 0 $). Then the following applies: \ begin {equation *} {\ rm \ bf A} \ cdot \ vec {x} = \ lambda \ cdot \ vec {x} \ qquad \ hbox {and} \ qquad {\ rm \ bf A} \ cdot \ overline {\ vec {x}} = \ overline {\ lambda} \ cdot \ overline {\ vec {x}} \ end {equation *} A suitable multiplication gives: \ begin {equation *} \ overline {\ vec {x}} ^ T \ cdot {\ rm \ bf A} \ cdot \ vec {x} = \ lambda \ cdot \ overline {\ vec {x}} ^ T \ cdot \ vec {x} \ quad (* ) \ qquad \ vec {x} ^ T \ cdot {\ rm \ bf A} \ cdot \ overline {\ vec {x}} = \ overline {\ lambda} \ cdot \ vec {x} ^ T \ cdot \ overline {\ vec {x}} \ quad (**) \ end {equation *} With $ \ rm \ bf A = A ^ T $ it follows: \ begin {eqnarray *} \ overline {\ vec {x}} ^ T \ cdot {\ rm \ bf A} \ cdot \ vec {x} & = & \ left (\ vec {x} ^ T \ cdot {\ rm \ bf A} \ cdot \ overline {\ vec {x}} \ right) ^ T \ \ vec {x} ^ T \ cdot {\ rm \ bf A} \ cdot \ overline {\ vec {x}} & = & \ left (\ overline {\ vec {x}} ^ T \ cdot {\ rm \ bf A} \ cdot \ vec {x} \ right) ^ T \ end {eqnarray *} So by subtracting (*) and (**): \ begin {equation *} 0 = \ left (\ lambda - \ overline {\ lambda} \ right) \ cdot \ left (\ vec {x} ^ T \ overline \ cdo t {\ vec {x}} \ right) \ end {equation *} So $ \ lambda = \ overline {\ lambda} $, since the scalar product of $ \ vec {x} ^ T \ cdot \ overline {\ vec {x}} $ is real. So $ \ lambda $ is real.
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Let $ {\ rm \ bf A_ {n, n}} $ be a real symmetric matrix with the eigenvalues ​​$ \ lambda_1, \ dots, \ lambda_n $. Then for $ \ lambda_i \ ne \ lambda_j $ the eigenvectors $ \ vec {x} ^ {\ lambda_i} $ and $ \ vec {x} ^ {\ lambda_j} $ are orthogonal.
For a proof see beckmannII: 1973, p. 92. If these vectors are normalized, they are orthonormal (i.e. they are orthogonal and have the length 1).
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Let $ \ rm \ bf A_ {n, n} $ be a regular symmetric real matrix, then the eigenvectors form an orthonormal basis of $ \ mathbb {R} _n $.