Lecture 14. Diagonalization

Transcription

1 International College of Economics and Finance (State University Higher School of Economics) Lectures on Linear Algebra by Vladimir Chernya Lecture. Diagonalization To be read to the music of Flowers on the wall by Statler Brothers We present here a technique for using eigenvalues and eigenvectors to find a diagonal form for a given square matrix. EXAMPLE. Consider again the 3 3 matrix A = from Examples and from lecture 3. In Example, we solved for the eigenvalues λ = and λ = 0 of A. We also found eigenvectors x = (, 3, 0 ), and y = (, 0, ) corresponding to λ = and an eigenvector z = (,,) for λ = 0. We will use these three vectors as columns for a 3 3 matrix P = 3 0 Now, P = (verify!), and so P is nonsingular. A quic calculation yields P = We now use A, P, and P to compute a new matrix D = P AP, which is a diagonal matrix: D P = AP = = = Notice that d ii, the i-th entry on the main diagonal of D, is an eigenvalue with a corresponding eigenvector in the i-th column of P. The diagonalization process presented in Example wors for many matrices. In fact, we have the following theorem:

2 Theorem. Let A be n n matrix, and let P be an n n matrix such that each column of P is an eigenvector for A. If P is non-singular, then D = P AP is a diagonal matrix, with its i-th main diagonal entry dii equal to the eigenvalue for the eigenvector which is i-th column of P. The proof of Theorem is not difficult, and we leave it for you to do. Note also that since D = P AP, then A = PDP. (Why?) Theorem requires that you first create a non-singular matrix P whose columns are eigenvectors for A. We did this in Example, but it is not always possible in general. Thus, we have the following definition: Definition. Let A be an n n matrix. A is diagonalizable if and only if there exists a non-singular n n matrix P such that D = P AP is a diagonal matrix. Although not explicitly stated in the definition, it can be shown that the columns of P are eigenvectors of A. Method for diagonalizing an Step. Calculate. n n matrix A (if possible). Step. Find all real roots of (that is, all real solutions to = 0 eigenvalues λ, λ, λ 3,, λ, for A. Step 3. For each eigenvalue λ m in turn perform the following: (a) Row reduce the augmented matrix for the homogeneous system I A) x = 0 ( λ m n. These are the (b) Solve for one particular solution for this system for each independent variable in the system. The j-th such solution, x j, is found by setting the j-th independent variable equal to, and setting all other independent variables equal to 0. (c) Eliminate fractions, if desired, from the vectors in part (b) by replacing them with nonzero scalars multiples of themselves. Step. If after repeating Step 3 for each eigenvalue, you have a total of less than n eigenvectors for A, then A cannot be put into diagonal form. Stop. Step 5. Otherwise, form a matrix P whose columns are these n eigenvectors. Step 6. The matrix P from Step 5 is non-singular, and D = P AP is a diagonal matrix whose dii entry is the eigenvalue corresponding to the eigenvector which is the i-th column of P. Also note that A = PDP. The set of eigenvectors found in Step 3 is not the entire eigenspace E λ. Instead, it is a special set of particular solutions to the homogeneous system having properties that mae the method wor The

3 assertions in Step that A cannot be diagonalized, and in Step 6 that P is non-singular, will not be proved here, but will follow from results in one of the future lectures. Application: Large Powers of a Matrix If D is a diagonal matrix, any positive integer power of D can be obtained by merely raising the diagonal entries of D to that power (Why?) For example, Example = = 0 ( ) Now, suppose that A is an n n matrix for which there exists a non-singular matrix P such that P AP = D, a diagonal matrix. We now A = PDP. But then, A = AA = ( PDP )( PDP ) = PD( P P) DP = PDInDP = PD P More generally, a straightforward proof by induction can be used to show that, for all positive integers, A = PD P Hence, if P and D are nown, calculating positive powers of A is much easier. The technique illustrated in Example can also be adapted to calculate square roots and cube roots of matrices. Nondiagonalizable Matrices. Not every square matrix can be diagonalized, as we see in the following examples. EXAMPLE 3. Consider the matrix 7 B = 3 8 from Example 3 (lecture 3). In that example, we found p B = ( x + ) ( x ), thus giving us the eigenvalues λ = and λ =. Using Step 3 of the Diagonalization Method produces the eigenvectors (, 7, ) for λ = and (,, ) for λ =. Since the method yielded only two eigenvectors for this 3 3 matrix, B cannot be diagonalized. Algebraic Multiplicity of an Eigenvalue Definition. Let A be n n matrix and let λ be an eigenvalue for A. Suppose that pa = ( x λ) g( x), where g ( λ) 0. (That is, ( x λ) is the highest power of ( x λ) that divides. Then is called algebraic multiplicity of λ. EXAMPLE. Let matrix A be 6 A =

4 whose characteristic polynomial is = x 3x x = x( x )( x + ). The algebraic multiplicity of λ = is (because the factor ( x +) appears to the second power in, while the algebraic multiplicities λ = and λ = 0 3 are both. The Diagonalization Method produced exactly two eigenvectors corresponding to λ = and one eigenvector each for λ = and λ = 0. 3 Definition. The maximum number of linearly independent eigenvectors corresponding to the eigenvalue λ is the geometric multiplicity of λ. We will prove further that, for any eigenvalue, the number of eigenvectors produced by Step 3 of the Diagonalization Method cannot be larger than the algebraic multiplicity of the eigenvalue. EXAMPLE 5 The matrices A and B from the lecture 3 has = ( x 6)( x + 5) and p B = ( x + ) ( x ). Both eigenvalues λ = 6 and λ 5 = for A have algebraic multiplicity, as does the eigenvalue λ = for B. However, the eigenvalue λ = for B has algebraic multiplicity because the factor ( x + ) appears to the second power in p B. However, in Example 6, we found that Step 3 of the Diagonalization Method applied to this matrix B produced only one eigenvector for a λ =. EXAMPLE 6. Consider the 3 3 matrix 3 A = pa = xi3 A = x 6x + 5x = x( x 6x + 5). Since x 6x + 5 has no real solutions (try the quadratic formula!), A has only eigenvalue, λ = 0, which has algebraic multiplicity. Thus, Step 3 of the Diagonalization Method can produce only one eigenvector for λ. and hence a total of only eigenvector overall. Hence, according to Step, A cannot be diagonalized. Example illustrates that if the sum of the algebraic multiplicities of all the eigenvalues for an n n matrix A is less than n, then A cannot be diagonalized. Examples 6 and 9 show, however, that even when the sum equals n, there is no guarantee that the matrix is diagonalizable. Let us finalise our investigation of the diagonalization problem by two theorems. Theorem. The geometric multiplicity of an eigenvalue is no more than its algebraic multiplicity. Theorem 3. If the sum of all algebraic multiplicities is n, and for all eigenvalues the geometric multiplicity of an eigenvalue is equal to its algebraic multiplicity then the matrix is diagonalizable.

5 The Cayley-Hamilton Theorem (optional) We conclude this section with an interesting relationship between a matrix and its characteristic polynomial. Theorem. (Cayley-Hamilton Theorem) Let A be an n n matrix, and let be its characteristic polynomial. Then p ( A) 0. A = 3 EXAMPLE 7 Let A =. Then = x x. The Cayley-Hamilton Theorem states that p A ( A) = 0. To verify this, we calculate ( A) = A A I = = Bibliography Carl P. Simon, Lawrence Blume. Mathematics for Economists. W.W.Norton&Company. New-Yor, London. 99. Chapter 3 5