MAS4105Su10d - MAS 4105 Test 4 1(8 pts Let T be a linear...

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

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAS 4105 Test 4 1. (8 pts) Let T be a linear operator on a finite-dimensional vector space V . Indicate whether the following are true or false. i. If V is an inner product space, then every basis of V can be transformed into an orthonormal basis of V . T F ii. If f ( t ) is the characteristic polynomial of T , then f ( T ) = 0. T F iii. The Gram-Schmidt process generates an orthonormal set of vectors. T F iv. If v 1 and v 2 are eigenvectors of T , then v 1 + v 2 is also an eigenvector. T F v. If V is an inner product space, x, y, z ∈ V , and h x, y i = h x, z i for all x ∈ V , then y = z . T F vi. If V is an inner product space and S is an orthogonal subset of V consisting of nonzero vectors, then S is linearly independent. T F vii. If λ is an eigenvalue of T , then the multiplicity of λ is greater than or equal to the dimension of E λ . T F viii. If β and γ are bases of V , then the characteristic polynomials of [ T ] β and [ T ] γ are always equal. T F 2. (7 pts) Explain the underlined equality: Theorem 5.1. A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis β for V consisting of eigenvectors of T . Furthermore, if T is diagonalizable, β = { v 1 , v 2 , . . . , v n } is an ordered basis of eigenvectors of T , and D = [ T ] β , then D is a diagonal matrix and D jj is the eigenvalue corresponding to v j for 1 ≤ j ≤ n . Proof. Let T be a linear operator on a finite-dimensional vector space V . First suppose that T is diagonalizable. By definition, there exists an ordered basis β = { v 1 , v 2 , . . . , v n } such that D = [ T ] β is diagonal. For any v j ∈ β , the image of v j under T is given by T (v j ) = n X i=1 D ij v i = D jj v j = λ j v j where λ j = D jj . Hence by definition, v j is an eigenvector of T and λ j is the corresponding eigen- value; β therefore is an ordered basis consisting of eigenvectors of...
View Full Document

{[ snackBarMessage ]}

Page1 / 11

MAS4105Su10d - MAS 4105 Test 4 1(8 pts Let T be a linear...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online