Linear Algebra

Section IV. Jordan Form 393First we find its characteristic polynomial c(x) = (x − 1)(x − 2) 3 with theusual determinant. Now the Cayley-Hamilton Theorem says that T’s minimalpolynomial is either (x − 1)(x − 2) or (x − 1)(x − 2) 2 or (x − 1)(x − 2) 3 . We candecide among the choices just by computing⎛⎞ ⎛⎞ ⎛ ⎞1 0 0 1 0 0 0 1 0 0 0 01 1 0 21 0 0 2(T − 1I)(T − 2I) = ⎜⎟ ⎜⎟⎝0 0 1 −1⎠⎝0 0 0 −1⎠ = 1 0 0 1⎜ ⎟⎝0 0 0 0⎠0 0 0 0 0 0 0 −1 0 0 0 0and⎛ ⎞ ⎛⎞ ⎛⎞0 0 0 0 0 0 0 1 0 0 0 0(T − 1I)(T − 2I) 2 1 0 0 11 0 0 2= ⎜ ⎟ ⎜⎟⎝0 0 0 0⎠⎝0 0 0 −1⎠ = 0 0 0 0⎜⎟⎝0 0 0 0⎠0 0 0 0 0 0 0 −1 0 0 0 0and so m(x) = (x − 1)(x − 2) 2 .Exerciseš 1.13 What are the possible minimal polynomials if a matrix has the given characteristicpolynomial?(a) 8 · (x − 3) 4 (b) (1/3) · (x + 1) 3 (x − 4) (c) −1 · (x − 2) 2 (x − 5) 2(d) 5 · (x + 3) 2 (x − 1)(x − 2) 2What is the degree of each possibility?̌ 1.14 Find the minimal polynomial of each matrix.⎛ ⎞ ⎛ ⎞ ⎛ ⎞3 0 03 0 03 0 0(a) ⎝1 3 0⎠(b) ⎝1 3 0⎠(c) ⎝1 3 0⎠0 0 40 0 30 1 3⎛⎞−1 4 0 0 0⎛ ⎞2 2 10 3 0 0 0(e) ⎝0 6 2⎠(f)0 −4 −1 0 0⎜⎟0 0 2 ⎝ 3 −9 −4 2 −1⎠1 5 4 1 41.15 Find the minimal polynomial of this matrix.⎛0 1⎞0⎝0 0 1⎠1 0 0(d)⎛2 0⎞1⎝0 6 2⎠0 0 2̌ 1.16 What is the minimal polynomial of the differentiation operator d/dx on P n ?̌ 1.17 Find the minimal polynomial of matrices of this form⎛⎞λ 0 0 . . . 01 λ 0 00 1 λ. .. ⎜⎟⎝λ 0⎠0 0 . . . 1 λwhere the scalar λ is fixed (i.e., is not a variable).