{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# u7 - Unit VII Symmetric Matrices 1 Spectral Theorem A...

This preview shows pages 1–3. Sign up to view the full content.

Unit VII: Symmetric Matrices 1. Spectral Theorem A linear transformation on a subspace V of R n is a function T from V to V that satisfies T ( a x + b y ) = aT ( x ) + bT ( y ) , x , y V, a, b scalars . The matrix of T with respect to the basis { v 1 , . . . , v m } of V is the m × m matrix A satisfying T ( v k ) = m X j =1 A jk v j , 1 k m. These definitions are virtually the same as those given earlier for a linear transformation on R n . Again note that the k th column of A contains the coordinates of T ( v k ) with respect to the basis { v 1 , . . . , v m } . Also, note that if c 1 , . . . , c m are the components of u V with respect to the basis { v 1 , . . . , v m } , and d 1 , . . . , d m are the components of T ( u ), then d j = k A jk c k . Lemma If { u 1 , . . . , u m } is an orthonormal basis for V , then the matrix A for T with respect to the basis is given by A jk = T ( u k ) · u j , 1 k m. Proof. This follows from T ( u k ) · u j = r A rk u r · u j = A jk . Definition: A symmetric linear transformation on a subspace V of R n is a linear trans- formation T on V that satisfies T ( x ) · y = x · T ( y ) , x , y V. Lemma. Let T be a linear transformation on a subspace V of R n , and let { u 1 , . . . , u m } be an orthonormal basis for V . Then T is a symmetric linear transformation if and only if the matrix of T with respect to the basis { u 1 , . . . , u m } is a symmetric matrix. Proof. This follows from A jk = T ( u k ) · u j = u k · T ( u j ) = T ( u j ) · u k = A kj . Theorem 1. Let A be a (real) symmetric n × n matrix. Then the roots of the character- istic polynomial p A ( λ ) = det ( λI - A ) of A are real. Thus A has n real eigenvalues (counting multiplicity). Proofsketch. Let λ be a root of the characteristic polynomial of A . Then there is a nonzero vector z = ( z 1 , . . . , z n ) C n that satisfies ( λI - A ) z = 0 . We express each z j = x j + iy j as the sum of its real and imaginary parts. Then z = x + i y , where x = ( x 1 , . . . , x n ) and y = ( y 1 , . . . , y n ) are vectors in R n . We denote x - i y = z . Then from z j z j = | z j | 2 , we obtain z · z = | z j | 2 6 = 0. Since A z = λ z , we have λ X | z j | 2 = X j λz j z j = X j ( Az ) j z j = X j,k A kj z k z j . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Since the matrix A is symmetric and has real entries, this is equal to X j,k z k A jk z j = X k z k ( Az ) k = X k z k λz k = λ X k | z k | 2 . Dividing by | z j | 2 , we conclude that λ = λ , and λ is real. Theorem 2. If T is a symmetric linear transformation, then eigenvectors of T corre- sponding to different eigenvalues are orthogonal. Proof. If T ( u ) = λ u and T ( v ) = μ v , then λ u · v = T ( u ) · v = u · T ( v ) = μ u · v . Thus either λ = μ or u · v = 0. Theorem 3. Let T be a symmetric linear transformation on R n . If V is a subspace of R n such that T ( V ) V , then T ( V ) V . Proof. Let w V . If v V , then T ( v ) V , so 0 = T ( v ) · w = v · T ( w ). Since this holds for all v V , we obtain w V . Theorem 4 (Spectral Theorem). Let T be a symmetric linear transformation on R n . Then there is an orthonormal basis for R n consisting of eigenvectors of T .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

u7 - Unit VII Symmetric Matrices 1 Spectral Theorem A...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online