Proof suppose a and b are similar matrices by

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Nature of Mathematics
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Chapter 16 / Exercise 1
Nature of Mathematics
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Proof. Suppose A and B are similar matrices. By definition, the nullity of A is the dimension of its null space. But N ( A ) = E 0 ( A ), the 0-eigenspace of A . By the last sentence in the statement of Proposition 4.2, E 0 ( A ) and E 0 ( B ) have the same dimension. Hence A and B have the same nullity. Since rank +nullity = n , A and B also have the same rank.
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Nature of Mathematics
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Chapter 16 / Exercise 1
Nature of Mathematics
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152 16. SIMILAR MATRICES 4.3. Intrinsic and extrinsic properties of matrices. It is not un- reasonable to ask whether a matrix that is similar to a symmetric ma- trix is symmetric. Suppose A is symmetric and S is invertible. Then ( SAS - 1 ) T = ( S - 1 ) T A T S T = ( S T ) - 1 AS T and there is no obvious reason which this should be the same as SAS - 1 . The explicit example 1 2 0 1 2 3 3 4 1 2 0 1 - 1 = 8 11 3 4 1 - 2 0 1 = 8 - 5 3 - 2 shows that a matrix similar to a symmetric matrix need not be symmetric. PAUL - More to say. 5. Diagonalizable matrices 5.1. Diagonal matrices. A square matrix D = ( d ij ) is diagonal if d ij = 0 for all i 6 = j . In other words, D = λ 1 0 0 · · · 0 0 0 λ 2 0 · · · 0 0 0 0 λ 3 · · · 0 0 . . . . . . . . . 0 0 0 · · · λ n - 1 0 0 0 0 · · · 0 λ n for some λ 1 , . . . , λ n R . We sometimes use the abbreviation diag( λ 1 , . . . , λ n ) = λ 1 0 0 · · · 0 0 0 λ 2 0 · · · 0 0 0 0 λ 3 · · · 0 0 . . . . . . . . . 0 0 0 · · · λ n - 1 0 0 0 0 · · · 0 λ n . The identity matrix and the zero matrix are diagonal matrices. Let D = diag( λ 1 , . . . , λ n ). The following facts are obvious: (1) the λ i s are the eigenvalues of D ; (2) det( D ) = λ 1 λ 2 · · · λ n ; (3) the characteristic polynomial of D is ( λ 1 - t )( λ 2 - t ) · · · ( λ n - t ); (4) De j = λ j e j , i.e., the standard basis vectors e j are eigenvectors for D . Warning: Although the axis R e j is contained in λ j -eigenspace for D its λ j -eigenspace might be bigger. For example, if λ i = λ j , then the E λ i contains the plane R e i + R e j .
5. DIAGONALIZABLE MATRICES 153 5.2. Definition. An n × n matrix A is diagonalizable if it is similar to a diagonal matrix, i.e., if S - 1 AS is diagonal for some S . A diagonal matrix is diagonalizable: if D is diagonal, then I - 1 DI is diagonal! Since similar matrices have the same characteristic polynomials, the characteristic polynomial of a diagonalizable matrix is a product of n linear terms. See “obvious fact” (3) above. 5.3. Example. If A = 5 - 6 3 - 4 and S = 2 1 1 1 , then S - 1 AS = 1 - 1 - 1 2 5 - 6 3 - 4 2 1 1 1 = 2 0 0 - 1 so A is diagonalizable. 5.4. The obvious questions are how do we determine whether A is di- agonalizable or not and if A is diagonalizable how do we find an S such that S - 1 AS is diagonal. The next theorem and its corollary answer these questions. Theorem 5.1 . An n × n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Proof. We will use the following fact. If B is a p × q matrix and C a q × r matrix, then the columns of BC are obtained by multiplying each column of C by B ; explicitly, BC = [ BC 1 , . . . , BC r ] where C j is the j th column of C .

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