By D. A. Suprunenko; translated by K.A. Hirsch
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Example text
Indeed, we see 4 3 that −6 6 3 4 3 4 2 0 = , −12 11 2 3 2 3 0 3 so that P−1 AP is diagonal, where P is the matrix whose columns are the eigenvectors of A. Furthermore, one can find two projection matrices whose column spaces are the eigenspaces, namely P1 = 9 12 −6 , −8 P2 = −8 −12 6 . 4. DIAGONALISABILITY Check directly that P12 = P1 , P22 = P2 , P1 P2 = P2 P1 = 0, P1 +P2 = I, and 2P1 +3P2 = A. This expression for a diagonalisable matrix A in terms of projections is useful in calculating powers of A, or polynomials in A.
Putting x = 0, we find that the constant term is cα (0) = det(−A) = (−1)n det(A). If α is diagonalisable then the eigenvalues are the roots of cα (x): cα (x) = (x − λ1 )(x − λ2 ) · · · (x − λn ). Now the coefficient of xn−1 is minus the sum of the roots, and the constant term is (−1)n times the product of the roots. 54 CHAPTER 4. LINEAR MAPS ON A VECTOR SPACE Chapter 5 Linear and quadratic forms In this chapter we examine “forms”, that is, functions from a vector space V to its field, which are either linear or quadratic.
Now in this case it is easily checked that T and ∑ λi πi agree on every vector in V , so they are equal. So (b) implies (c). Finally, if α = ∑ λi πi , where the πi satisfy the conditions of (c), then V is the direct sum of the spaces Im(πi ), and Im(πi ) is the λi -eigenspace. So (c) implies (b), and we are done. Example Our matrix A = −6 −12 6 is diagonalisable, since the eigenvectors 11 3 2 and are linearly independent, and so form a basis for R. Indeed, we see 4 3 that −6 6 3 4 3 4 2 0 = , −12 11 2 3 2 3 0 3 so that P−1 AP is diagonal, where P is the matrix whose columns are the eigenvectors of A.