Linear Algebra

Section IV. Jordan Form 389both have a leading one. In their difference d(x) = m(x) − ˆm(x) the leadingterms cancel. So d is of smaller degree than m and ˆm. If d were to have aleading coefficent that is nonzero then we could divide by it to get a polynomialthat takes the map or matrix to zero and has leading coefficient one. Thiswould contradict the choice of m and ˆm as of minimal degree. Thus the leadingcoefficient of d is zero, so m(x) − ˆm(x) is the zero polynomial, and so the twoare equal.QED1.6 Example We can see that m(x) = x 2 − 2x − 1 is minimal for the matrix ofExample 1.1 by computing the powers of T up to the power n 2 = 4.(T 2 1/2 − √ )3/2= √3/2 1/2( )T 3 0 −1=1 0T 4 =Put c 4 T 4 + c 3 T 3 + c 2 T 2 + c 1 T + c 0 I equal to the zero matrixand use Gauss’ Method.−(1/2)c 4 + (1/2)c 2 + ( √ 3/2)c 1 + c 0 = 0−( √ 3/2)c 4 − c 3 − ( √ 3/2)c 2 − (1/2)c 1 = 0( √ 3/2)c 4 + c 3 + ( √ 3/2)c 2 + (1/2)c 1 = 0−(1/2)c 4 + (1/2)c 2 + ( √ 3/2)c 1 + c 0 = 0c 4 − c 2 − √ 3c 1 − 2c 0 = 0c 3 + √ 3c 2 + 2c 1 + √ 3c 0 = 0(−1/2 − √ )3/2√3/2 −1/2Setting c 4 , c 3 , and c 2 to zero forces c 1 and c 0 to also come out as zero. To geta leading one, the most we can do is to set c 4 and c 3 to zero. Thus the minimalpolynomial is quadratic.Using the method of that example to find the minimal polynomial of a 3×3matrix would be tedious because it would mean doing Gaussian reduction on asystem with nine equations in ten unknowns. We shall develop an alternative.1.7 Lemma Suppose that the polynomial f(x) = c n x n + · · · + c 1 x + c 0 factorsas k(x − λ 1 ) q1 · · · (x − λ z ) q z. If t is a linear transformation then these two areequal maps.c n t n + · · · + c 1 t + c 0 = k · (t − λ 1 ) q 1◦ · · · ◦ (t − λ z ) q zConsequently, if T is a square matrix then f(T) and k · (T − λ 1 I) q1 · · · (T − λ z I) q zare equal matrices.Proof We use induction on the degree of the polynomial. The cases wherethe polynomial is of degree zero and degree one are clear. The full inductionargument is Exercise 1.7 but we will give its sense with the degree two case.A quadratic polynomial factors into two linear terms f(x) = k(x − λ 1 ) · (x −λ 2 ) = k(x 2 + (λ 1 + λ 2 )x + λ 1 λ 2 ) (the roots λ 1 and λ 2 could be equal). We can