Linear Algebra

388 Chapter Five. Similaritywith respect to the standard basis by(√ )3/2 −1/2T =1/2 √ 3/2and verifying that 0T 4 + 0T 3 + 1T 2 − 2T − 1I equals the zero matrix is easy.1.2 Definition For any polynomial f(x) = c n x n + · · · + c 1 x + c 0 , where t is alinear transformation then f(t) is the transformation c n t n + · · · + c 1 t + c 0 (id)on the same space, and where T is a square matrix then f(T) is the matrixc n T n + · · · + c 1 T + c 0 I.The polynomial of the matrix represents the polynomial of the map: if T =Rep B,B (t) then f(T) = Rep B,B (f(t)). This is because T j = Rep B,B (t j ), andcT = Rep B,B (ct), and T 1 + T 2 = Rep B,B (t 1 + t 2 ).1.3 Remark Most authors write the matrix polynomial slightly differently than themap polynomial. For instance, if f(x) = x − 3 then most authors explicitly writethe identity matrix f(T) = T − 3I but don’t write the identity map f(t) = t − 3.We shall follow this convention.Consider again Example 1.1. Although T ∈ M 2×2 , which is a dimension fourspace, we exhibited a polynomial of T that gave the zero matrix and was ofdegree less than four. So for any particular map or matrix, degree n 2 will sufficebut there may be a smaller degree polynomial that works.1.4 Definition The minimal polynomial m(x) of a transformation t or a squarematrix T is the polynomial of least degree and with leading coefficient one suchthat m(t) is the zero map or m(T) is the zero matrix.The fact that leading coefficient must be one keeps a minimal polynomial frombeing the zero polynomial. That is, a minimal polynomial must have degree atleast one. Thus, the zero matrix has minimal polynomial p(x) = x while theidentity matrix has minimal polynomial ˆp(x) = x − 1.1.5 Lemma Any transformation or square matrix has a unique minimal polynomial.Proof We first show existence. The earlier observation that degree n 2 sufficesshows that there is at least one nonzero polynomial p(x) = c k x k + · · · + c 0 thattakes the map or matrix to zero (p is not the zero polynomial because the earlierobservation includes that at least one of the coefficients is nonzero). From amongall nonzero polynomials taking the map or matrix to zero, there must be at leastone of minimal degree. Divide this p by c k to get a leading one. Thus for anymap or matrix a minimal polynomial exists.We now show uniqueness. Suppose that m(x) and ˆm(x) both take the mapor matrix to zero, are both of minimal degree and are thus of equal degree, and