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These notes closely follow the presentation of the material given in David C. Lay’s textbook Linear Algebra and its Applications (3rd edition). These notes are intended primarily for in-class presentation and should not be regarded as a substitute for thoroughly reading the textbook itself and working through the exercises therein. Eigenvalues and Eigenvectors Recall that if A is an n n (square) matrix, then the mapping x A x is a linear transformation from n into n . For a given n n matrix, A , it may or may not be the case that there are non–zero vectors v n such that A v is a scalar multiple of v . Any non–zero vector v n such that A v is a scalar multiple of v is called an eigenvector of the matrix A . Definition If A is an n n matrix, v is a non–zero vector in n , and is a scalar such that A v v , then v is said to be an eigenvector of the matrix A and is said to be an eigenvalue of the matrix A. More specifically, v is said to be an eigenvector of A associated with the eigenvalue . If is an eigenvalue of A, then the eigenspace of , denoted by eig , is defined to be eig v n | A v v 0 n . Our use of the term eigen space is justified by the following theorem. Theorem If is an eigenvalue of an n n matrix, A, then eig is a subspace of n . Proof Suppose that is an eigenvalue of the n n matrix, A . By definition, eig v n | A v v . We will show that eig is closed under addition and scalar multiplication. Let v 1 and v 2 be vectors in eig . Then A v 1 v 2 A v 1 A v 2 v 1 v 2 v 1 v 2 . This shows that v 1 v 2 eig . Therefore, eig is closed under vector addition. Let v eig and let c be a scalar. Then A c v c A v c v c v . This shows that c v eig . Therefore, eig is closed under scalar multiplication. We have now proved that eig is a subspace of n . Remark If is an eigenvalue of an n n matrix, A, then eig v n | A v v . It is clear from this definition that 0 n eig because A 0 n 0 n . However, note 1

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