DETERMINANTS AND EIGENVALUES 1. Introduction Gauss ...

102 II. DETERMINANTS AND EIGENVALUES(The reader should check explicitly in this case that[1 −11 1] −1 [2 11 2][ ]1 −1=1 1[ ] )3 0.0 1By means of these steps, the matrix A has been expressed in terms of a diagonalmatrix with its eigenvalues on the diagonal. This process is called diagonalizationWe shall return in Chapter III to a more extensive discussion of diagonalization.Example 2. Not every n×n matrix A is diagonalizable. That is, it is not alwayspossible to find a basis for R n consisting of eigenvectors for A. For example, letThe characteristic equation isA =[ ]3 1.0 3[ ]3 − λ 1det=(3−λ) 2 =0.0 3−λThere is only one root λ = 3 which is a double root of the equation. To find thecorresponding eigenvectors, we solve the homogeneous system (A − 3I)v = 0. Thecoefficient matrix [ ]3 − 3 1=0 3−3[ ]0 10 0is already reduced, and the corresponding system has the general solutionThe general solution vector isv =v 2 =0, v 1 free.[ ] [ ]v1 1= v0 1 = v0 1 e 1 .Hence, the eigenspace for λ = 3 is one dimensional with basis {e 1 }. There are noother eigenvectors except for multiples of e 1 . Thus, we can’t possibly find a basisfor R 2 consisting of eigenvectors for A.Note how Example 2 differs from the examples which preceded it; its characteristicequation has a repeated root. In fact, we have the following general principle.If the roots of the characteristic equation of a matrix are all distinct, then thereis necessarily a basis for R n consisting of eigenvectors, and the matrix is diagonalizable.In general, if the characteristic equation has repeated roots, then the matrixneed not be diagonalizable. However, we might be lucky, and such a matrix maystill be diagonalizable.