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1. Eigen Values and Eigen Vectors V Is a Vector Space Over R, the Real Numbers (Or C , the Complext Numers). Let T : V → V Be

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1. Eigen Values and Eigen Vectors V Is a Vector Space Over R, the Real Numbers (Or C , the Complext Numers). Let T : V → V Be 1. Eigen Values and Eigen Vectors V is a vector space over R, the real numbers (or C , the complext numers). Let T : V ! V be a linear transformation. A scalar λ is called an eigenValue of T if there is a non zero vector v in V such that T (v) = λv. This nonzero vector v is called an eigenvector of T with the eigen value λ. If A is a square matrix of size n over R, the real numbers (or C , the complext numers), then the eigen values and eigen vectors of A are the eigen values and the eigen vectors of the linear transformation on Rn (or Cn defined by multiplication by A. Let V = Rn (or Cn A scalar λ is called an eigenValue of A if there is a non zero vector v in V such that Av = λv. This nonzero vector v is called an eigenvector of A with the eigen value λ. The characteristic polynomial PA(λ) of A is the polynomial n n n−1 determinant (A − λI) = (−1) λ + an−1λ + ··· + ao For a matrix A, and a scalar λ, Eλ(A) = fvjA(v = λvg is the subspace of all eigen vectors with eigen value λ. This is called the eigen space of λ. Thus, dimension of Eλ(A) is not zero if and only if λ is an eigen value of A. Theorem 1. : x is an eigen value of A if and only if x is the root of the characteristic polynomial of A, that is if and only if PA(x) = 0. Theorem 2. Similar matrices have the same characteristic polynomial and hence the same eigen values Theorem 3. A and At, the transpose of A, have the same characteristic polynomial and the same eigen values. As a result of Theorem 2, the eigen values and eigen vectors of a linear transformation T as defined above can be computed by finding those of the matrix of T with respect to some basis B of the vector space V . Theorem 2 says that the eigen values and eigen vectors do not depend on the choice of the basis. To Compute the eigen values and the eigen vectors of a matrix A. 1. Compute A-xI 2. Compute determinant (A-xI) = PA(x) 3. Solve PA(x) = 0 to find the roots x = λ1; ··· λt λ1; ··· λt are the eigen values of A. 4. Proceed to find the eigen vectors . Eλi (A) = Null space of (A − λiI). Use Gaussian elimination to find a basis for the null space of A − λiI for each i. Theorem 4. If v1; v2; ··· vn are eigen vectors of A with distinct eigen values λ1; ··· λn, then v1; v2; ··· vn are linearly independent. Proof: Suppose not. Let n be the smallest positive integer such that there exist n linearly dependent eigenvectors v1; v2; ··· vn with distinct eigenvalues λ1; ··· λn. Then we have c1v1 + c2v2 + ··· + cnvn = 0 1 , with none of the ci = 0. This is because if any of the ci = 0, we can drop that vi and then we will have only n − 1 vectors which are linearly dependent, contradicting the minimality of n. Now, applying A to both sides, we get A(c1v1 + c2v2 + ··· + cnvn) = A0 = 0 , That is, λ1c1vi + λ2c2v2 + ··· + λncnvn) = 0 . Eliminating the vn from the two equations by multiplying the first by λn and sub- tracting from the second, we get (λ1 − λn)c1v1 + (λ2 − λn)c2v2 + ··· +)(λn−1 − λn)cn−1vn−1) = 0 . Since λn is different from λi for all i < n, this contradicts the minimality of n. This proves the theorem. An eigen value λ of a matrix A is said to be of multiplicity k if (λ − x)k, divides the k+1 characteristic polynomial PA(x) and (λ − x) does not divide PA(x). That is, λ is an eigen value with multiplicity k if λ is a root of PA(x) = 0 with multiplicity k. Thoeorem 5: If λ is an eigen value of A with multiplicity k then dim Eλ(A) ≤ k To compute the multiplicity of the eigen value λ and dim Eλ(A). Factor the characterisitic polynomial PA(x). Then the multiplicity of λ equals the power of (x − λ) in the factorizartion of PA(x). Compute a row echelon form of matrix A − λI. Dimension of the eigne space of λ is the number of columns without a leading 1. 2. Diagonalization A matrix A is diagonalizable if there is an invertible matrix X such that X−1AX is a diagonal matrix. Theorem 6. An n x n matrix A is diagonalizable over R (or C ) if and only if there is a basis for Rn (respectively Cn) consisting of eigen vectors of A. To determine if a given matrix A = (aij)n×n is diagonalizable: 1. Compute the eigen values and their multiplicties as in the previous section. 2. Let λ1; ··· λt be the distinct eigen values of A with multiplicities k1; k2; ··· kn respectively. Compute the basis for the eigen spaces Eλi (A) as before. If dim (Eλi (A) < ki for any i, then A is not diagonalizable. If dim Eλi (A) = ki for all i, then A is diagonalizable. 2 Collect all the vectors in the bases of Eλi (A) for each i, to get a v1; v2; ··· vn consisting −1 of eigen vectors. Let X = v1; v2; ··· vn. Then X AX is diagonal. 3. Orthogonal diagonalization A matrix A is said to be orthogonally diagonalizable if there is an orthogonal matrix X such that XAX−1 is diagonal. A is orthogonally diagonalizable if there is an orothonormal basis consisting of eigen vectors. Theorem 7. Suppose that A is a symmetric matrix. If v1 and v2 are eigen vectors with distinct eigen values t1 and t2 respectively, then v1 and v2 are orthogonal to each other. Theorem 8. If A is orthogonally diagonalizable then A must be symmetric. Suppose that A is symmetric. To orthogonally diagonalize A: 1. Proceed as in section 2 to find the eigen values and bases for eignen spaces. 2. Use Gram-Schmidt process to find orthonormal basis for each eigen space. Putting these together gives us an orthonormal basis of eigen vectors. 4. Jordan Normal Form Let A be an n×n matrix over the complex number. Let the characterisitic polynomial Qt α Qt of A be PA(x) = i=1(x − λi)i and the minimal polynomial of A be mA(x) = i=1(x − β λi)i . Now, 1 ≤ βiαi; 1 ≤ i ≤ t. Let ri = dimEλi (A). It is indeed possible for two matrices to have the same characterisitic and minimal polynomials and the dimension of the eigenspaces and yet be not similar. Infact, r = dimEλ(A)= number of blocks with λ as eigen value in the Jordan form of A. Suppose the number of blocks of size d with eigenvalue λ is sd. Then we have, β j X j X n − rank(A − λ) = d = 1 dsd + jsd d=j+1 j j+1 Pβ So, rank(A − λ) − rank(A − λ) = d=j+1 sd d−1 d d+1 So, sd = rank(A − λ) − 2rank(A − λ) + rank(A − λ) and β = j where, rank(A − λ)j = rank(A − λ)j+1. Thus, if we know the ranks of (A−λ)j, for all j, and all eigenvalues λ, we can determine the Jordan normal form of A. 7 3 Example: If the PA(x) = (x − 2) ; mA(x) = (x − 2) ; and dim E2(A) = 3, we cannot determine A upto similarity. But if we knew that rank (A − 2I)2 = 1, we can say that the Jordan blocks of A must be of sizes 2, 2, and 3. If however, rank (A − 2I)2 = 2, then the Jordan blocks must be of sizes 1, 3, 3. For, s2 = 4 − 2(1) − 0 = 2 in the first case, and s2 = 4 − 2(2) = 0 in the second case. 3 5 1 2 7 Example: 1. PA(x) = (x − 3) (x − 1) 0 Ranks of (A − I); (A − I) ; ··· (A − I) are respectively, 11, 9, 7, 5, 5, 5, 5. Determine A upto similarity. 5. Rational Canonical Form: Over the rational numbers, a matrix A is similar to a matrix in rational canonical form: d d−1 If q(x) = x + qd−1x + ··· q0 is a polynomial of degree d, then we call the d × d matrix whose last column is formed by the coefficients −qi; 0 ≤ i ≤ d − 1 and the (j + 1; j) entry is 1, for 1 ≤ j ≤ d − 1, is called the companion matrix to the polynomial q(x). In fact, the characteristic polynomial of this matrix is precisely q(x). Now, associated to any n×n matrix A, there are unique polynomials of degree at least 1, a1; a2; ··· am such that ai divides ai+1 and the product of the ai's is the characteristic polynomial of A and that am is the minimal polynomial of A. The concatanation of the companion matrices of ai's is the rational canonical form of A. That is if the companion matrix of ai is Ci, the rational canonical form of A is the n × n matrix with diagonal blocks C1;C2 :::Cm. Example: The characteristic polynomial of A is (x + 3)3(x − 2)2 and the minimal polynomial of A is (x + 3)(x − 2)2 = x3 − x2 − 8x + 12. 2 Then the only possibility for ai is (x + 3); (x + 3); (x + 3)(x − 2) . Thus the rational canonical form is 2 −3 0 0 0 0 3 6 0 −3 0 0 0 7 6 7 6 0 0 0 0 −12 7 4 0 0 1 0 8 5 0 0 0 1 1 4.
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