Proof. It is tempting to “prove” this by substituting \(x=A\) into \(X_A(x)=\det (A-xI)\) to get \(\det (A-A)=0\). That argument is wrong: \(X_A(x)\) is a scalar polynomial, and replacing the scalar \(x\) by a matrix inside a determinant is not a legitimate operation. A genuine proof uses the adjoint.

The entries of \(\text {adj}(A-xI)\) are determinants of \((n-1)\) by \((n-1)\) submatrices of \(A-xI\), hence polynomials in \(x\) of degree at most \(n-1\). Collecting powers of \(x\), we may therefore write \[\text {adj}(A-xI)=B_0+B_1x+B_2x^2+\cdots +B_{n-1}x^{n-1}\] for some constant matrices \(B_0,\dots ,B_{n-1}\).

The identity established in the proof of Theorem 4.2.10 gives \[(A-xI)\,\text {adj}(A-xI)=\det (A-xI)\,I=X_A(x)\,I.\] Write \(X_A(x)=c_0+c_1x+\cdots +c_nx^n\) and expand the left-hand side: \[(A-xI)\sum _{k=0}^{n-1}B_kx^k =\sum _{k=0}^{n-1}AB_kx^k-\sum _{k=0}^{n-1}B_kx^{k+1}.\] Two polynomials in \(x\) with matrix coefficients are equal only if corresponding coefficients agree, so \[AB_0=c_0I,\qquad AB_k-B_{k-1}=c_kI \ \ (1\leq k\leq n-1),\qquad -B_{n-1}=c_nI.\]

Now multiply the \(k\)-th of these on the left by \(A^k\) and add them all: \[AB_0+\sum _{k=1}^{n-1}\left (A^{k+1}B_k-A^kB_{k-1}\right )-A^nB_{n-1} =c_0I+\sum _{k=1}^{n-1}c_kA^k+c_nA^n.\] The left-hand side telescopes — \(AB_0\) cancels against \(-A^1B_0\), then \(A^2B_1\) against \(-A^2B_1\), and so on — leaving \(0\). The right-hand side is exactly \(X_A(A)\). Hence \(X_A(A)=0\). □

Solution. \begin {align*} X_A(x) &= \begin {vmatrix} 1-x&3\\4&5-x\\ \end {vmatrix}\\\\ &=(1-x)(5-x)-12\\ &=x^2-6x-7 \end {align*}

Now \begin {align*} X_A(A) &=A^2-6A-7I\\ &= \begin {pmatrix} 1&3\\4&5\\ \end {pmatrix} \begin {pmatrix} 1&3\\4&5\\ \end {pmatrix} -6 \begin {pmatrix} 1&3\\4&5\\ \end {pmatrix} -7 \begin {pmatrix} 1&0\\0&1\\ \end {pmatrix}\\\\ &= \begin {pmatrix} 0&0\\0&0\\ \end {pmatrix}\\ \end {align*}

Proof. Follows directly from Theorem 8.3.1 and Lemma 8.2.7. □

Proof. Both directions amount to showing that the roots of \(M(x)\) are exactly the eigenvalues of \(A\), since by Theorem 8.1.5 the roots of \(X_A(x)\) are the eigenvalues.

Every eigenvalue is a root of \(M\)

Let \(\lambda \) be an eigenvalue with eigenvector \(v\neq 0\). Then \(Av=\lambda v\), and applying \(A\) repeatedly gives \(A^kv=\lambda ^kv\), so for any polynomial \(f\), \[f(A)v=f(\lambda )v.\] Taking \(f=M\) and using \(M(A)=0\) gives \(M(\lambda )v=0\). As \(v\neq 0\), this forces \(M(\lambda )=0\).

Every root of \(M\) is an eigenvalue

Let \(M(\lambda )=0\) and factor \(M(x)=(x-\lambda )g(x)\). Since \(\deg g<\deg M\) and \(M\) has least degree among monic annihilating polynomials, \(g(A)\neq 0\). So there is a vector \(v\) with \(w=g(A)v\neq 0\). Then \[(A-\lambda I)w=(A-\lambda I)g(A)v=M(A)v=0,\] so \(w\) is a non-zero vector with \(Aw=\lambda w\), that is, \(\lambda \) is an eigenvalue.

The two halves together show the roots of \(M\) and of \(X_A\) form the same set, so the distinct linear factors coincide. □

Proof. Suppose first that \(A\) has \(n\) distinct eigenvalues. By Theorem 8.3.4 the minimum polynomial has all \(n\) of them as roots, so it is divisible by \((x-\lambda _1)\cdots (x-\lambda _n)\) and therefore has degree at least \(n\). On the other hand \(M\) divides \(X_A\), by Theorem 8.3.1 together with Lemma 8.2.7, so \(\deg M\leq n\). Hence \(\deg M=n\); being monic, of the same degree as \(X_A\), and a divisor of it, \(M=X_A\).

For the general statement, suppose \(A\) is diagonalisable, so it is similar to a diagonal matrix \(D\) carrying the \(\lambda _i\) on its diagonal. By Lemma 8.2.8 similar matrices share a minimum polynomial, so it suffices to compute that of \(D\). For diagonal \(D\), \(f(D)\) is diagonal with entries \(f(\lambda _i)\), so \(f(D)=0\) exactly when \(f\) vanishes at every \(\lambda _i\). The monic polynomial of least degree doing so is \((x-\lambda _1)\cdots (x-\lambda _k)\), which is therefore the minimum polynomial.

Conversely, suppose \(M(x)=(x-\lambda _1)\cdots (x-\lambda _k)\) with the \(\lambda _i\) distinct. For each \(i\) put \(g_i(x)=\prod _{j\neq i}(x-\lambda _j)\). These \(k\) polynomials have no common root, so their greatest common divisor is \(1\) and there are polynomials \(h_i\) with \(\sum _i h_ig_i=1\). Substituting \(A\), \[\sum _{i=1}^{k}h_i(A)g_i(A)=I,\] so every \(v\) decomposes as \(v=\sum _i v_i\) with \(v_i=h_i(A)g_i(A)v\). Each such \(v_i\) satisfies \((A-\lambda _iI)v_i=h_i(A)M(A)v=0\), so it is either \(0\) or an eigenvector for \(\lambda _i\). Hence the eigenvectors span the whole space, and by Theorem 8.2.3(i) \(A\) is diagonalisable. □

Solution. \begin {align*} X_A(x) &= \begin {vmatrix} 1-x&1&0&0\\0&1-x&0&0\\0&0&2-x&0\\0&0&0&2-x\\ \end {vmatrix} =(2-x) \begin {vmatrix} 1-x&1&0\\0&1-x&0\\0&0&2-x\\ \end {vmatrix}\\\\ &=(2-x)^x \begin {vmatrix} 1-x&1\\0&1-x\\ \end {vmatrix}\\\\ &=(2-x)^2(1-x)^2\\ \end {align*}

By Theorem 8.3.4, the minimum polynomial is one of

i.

\((x-2)(x-1)\)

ii.

\((x-2)^2(x-1)\)

iii.

\((x-2)(x-1)^2\)

iv.

\((x-2)^2(x-1)^2\)

So we start with i \begin {align*} f(A) &=(A-I)(A-2I)\\ &= \begin {pmatrix} 0&1&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\\ \end {pmatrix} \begin {pmatrix} -1&1&0\\0&-1&0\\0&0&0\\0&0&0\\ \end {pmatrix}\\\\ &= \begin {pmatrix} 0&-1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\\ \end {pmatrix}\\\\ &\neq 0\\ \end {align*}

ii it fails

Now, we check for iii \begin {align*} f(A) &=(A-I)^2(A-2I)\\ &= \begin {pmatrix} 0&1&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\\ \end {pmatrix} \begin {pmatrix} 0&-1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\\ \end {pmatrix}\\\\ &= \begin {pmatrix} 0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\\ \end {pmatrix}\\\\ &=0 \end {align*}

So \(M(x)=(x-1)^2(x-2)\). \(\implies \) it is not diagonazable since the minimum and characteristic polynomial do not coincide \(\implies \) \(A\) is not diagonazable.