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# Class Note for MATH 215 with Professor Dostert at UA

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Date Created: 02/06/15
THE UNIVERSITY Q OF ARIZONA Math 215 Introduction to Linear Algebra Section 43 Eigenvalues and Eigenvectors of n x n Matrices Paul Dostert September 30 2008 At Eigenvalues The eigenvalues of a square matrix A are the solutions A of the equation det A AI 0 When det A AI is computed the result is a polynomial in A which we call the characteristic polynomial of A The equation det A AI 0 is the characteristic equation of A Ex Find the characteristic polynomial for A a Z How many d1 0 0 eigenvalues will there be What about A 0 d2 0 0 0 d3 For an n gtlt n matrix the characteristic polynomial will be of degree n How many distinct roots can this have How many distinct eigenvalues A Finding Eigenvalues and Eigenspaces We use the following procedure to find the eigenvalues and eigenvectors eigenspaces of an n gtlt n matrix A 1 Compute the characteristic polynomial det A AI of A 2 Find the eigenvalues of A by solving the characteristic equation det A AI 0 for A 3 For each eigenvalue A find the eigenspace EA this is the null space of A AI by solving Ax Ax 0 4 Find a basis for each eigenspace 2 0 0 Ex Find the eigenvalues and eigenspaces for A 1 2 1 Do the 1 2 0 same for A l lllN l lllN mwo A Multiplicity 1 2 1 Ex Find the eigenvalues and eigenspaces for A 2 0 2 Do the 1 2 3 5 2 2 same for A 2 5 2 2 2 5 Note that in the above examples we only have two distinct eigenvalues If we count with multiplicity then each matrix has 3 eigenvalues We define the algebraic multiplicity of an eigenvalue to be its multiplicity as a root of the characteristic equation For each of the previous examples what is the algebraic multiplicity of the eigenvalues Note that for the first example the eigenspace corresponding to A 2 consists of only 1 vector For the second example the eigenspace corresponding to A 3 has 2 vectors in the eigenspace We define the geometric multiplicity of an eigenvalue to be the the dimension of the eigenspace Ah Special matrices Ex Find the eigenvalues and eigenspaces for 1 0 0 0 1 0 0 0 1 OOll 0000 00 and 0 0 and 3 0 1 1 Thm The eigenvalues of a triangular matrix are the entries on its main diagonal A The Fund Thm of Invertible Matrices V3 Thm A square matrix A is invertible iff 0 is NOT an eigenvalue of A Pf A invertible iff det A 3A 0 FT of Invert Matrices V3 Let A be an n gtlt n matrix TFSAE a A is invertible b Ax b as a unique solution for every b e Rquot c Ax 0 has only the trivial solution d The rref of A is In e A is a product of elementary matrices f rank A n g nullity A 0 h The column vectors of A are linearly independent i The column vectors of A span Rquot j The column vectors of A form a basis for Rquot k The row vectors of A are linearly independent I The row vectors of A span Rquot m The row vectors of A form a basis for R n det A 3A 0 o 0 is not an eigenvalue of A Pf We have already shown a ltgt n and a ltgt o is given by the previous theorem A Formulas for Evals of Powers and Inverses Thm Let A be a square matrix with eigenvalue A and corresponding eigenvector x a For any positive integer n A is an eigenvalue of Aquot with corresponding eigenvector x b If A is invertible then 1 is an eigenvalue of A 1 with corresponding eigenvector x c For any integer n A is an eigenvalue of A with corresponding eigenvector x Pfhints For a use induction b Is pretty straight fonvard I would use a and b to try to prove c Ex ForA 01 21 and b 11 compute A20b by writing b in terms of the eigenvectors of A u Linear Combos and Linear Independence Thm Suppose the n gtlt n matrix A has eigenvectors v17 7vm with corresponding eigenvalues A17 7 Am If x is a vector in R that can be expressed as X01V1 Cme then for any integer k Akx Clklfvl kalfnvm Thm Let A be an n gtlt n matrix and let A17 A with corresponding eigenvectors v17 vm Then v17 independent 7Am be distinct eigenvalues of 7vm are linearly Pfhints Assume the v2 are linearly dependent Write one vj as a linear combination of v2 for 239 lt j Multiply this equation by A Also multiply this equation by Aj Subtract the two equations so you have a RHS of 0 Use linear independence of the v2 with z lt j to conclude that each cz 0 and hence vj 0 This is a contradiction A Matlab Eigenvalues and Eigenspaces Finding the algebraic multiplicity of an eigenvalue is easy but how would we find the geometric multiplicity We can do this by reducing A A to RREF 5 2 2 Consider A 2 5 2 We write 2 2 5 A52 225 2 2 25 v D eigA rrefA eye33D11 rrefA eye33D33 We find that the geometric multiplicity for A 9 is 1 since the rref has one zero row and the geometric multiplicity for A 3 is 2 since the rref has two zero rows

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