Prove the Cayley-Hamilton theorem.
Prove that the minimal polynomial divides the characteristic polynomial.
Prove that the cokernel of \(A\in \Mat(n\times n, \ZZ)\) is finite \(\iff \det A \neq 0\), and show that in this case \(\abs{\coker(A)} = \abs{\det(A)}\).
Show that a nilpotent operator is diagonalizable.
Show that if \(A,B\) are diagonalizable and \([A, B] = 0\) then \(A,B\) are simultaneously diagonalizable.
Does diagonalizable imply invertible? The converse?
Does diagonalizable imply distinct eigenvalues?
Show that if a matrix is diagonalizable, its minimal polynomial is squarefree.
Show that a matrix representing a linear map \(T:V\to V\) is diagonalizable iff \(V\) is a direct sum of eigenspaces \(V = \bigoplus_i \ker(T -\lambda_i I)\).
Show that if \(\theset{\vector v_i}\) is a basis for \(V\) where \(\dim(V) = n\) and \(T(\vector v_i) = \vector v_{i+1 \mod n}\) then \(T\) is diagonalizable with minimal polynomial \(x^n-1\).
Show that if the minimal polynomial of a linear map \(T\) is irreducible, then every \(T\dash\)invariant subspace has a \(T\dash\)invariant complement.
The fourth part states that a nilpotent operator is diagonalizable; by step <1>4 this holds only for the zero operator. The tenth part holds over \(\CC\), or over any field in which \(x^n-1\) has \(n\) distinct roots.
Solution
Throughout, \(A\in M_n(F)\) for a field \(F\), with characteristic polynomial \(p(t)=\det(tI-A)=\sum_{j=0}^nc_jt^j\).
1. \(p(A)=0\) (Cayley–Hamilton).
Write \(\operatorname{adj}(tI-A)=\sum_{j=0}^{n-1}B_jt^j\) with \(B_j\in M_n(F)\). The adjugate identity \((tI-A)\operatorname{adj}(tI-A)=p(t)I\) gives, on comparing coefficients of \(t^j\), \[B_{n-1}=c_nI,\qquad B_{j-1}-AB_j=c_jI\ (1\le j\le n-1),\qquad -AB_0=c_0I .\] Multiplying the equation for \(t^j\) on the left by \(A^j\) and summing over \(j\), the right sides sum to \(p(A)\) and the left sides telescope to \(0\).
2. The minimal polynomial \(m_A\) divides \(p\).
\(m_A\) generates the ideal \(\{f\in F[t]:f(A)=0\}\) of the PID \(F[t]\), and \(p\) lies in it by step 1.
3. For \(A\in M_n(\ZZ)\), \(\operatorname{coker}A\) is finite if and only if \(\det A\neq0\), and then \(|\operatorname{coker}A|=|\det A|\).
By Smith normal form there are \(P,Q\in\operatorname{GL}_n(\ZZ)\) with \(PAQ=D=\operatorname{diag}(d_1,\dots,d_n)\), \(d_i\ge0\). Since \(P,Q\) are automorphisms of \(\ZZ^n\), \(\operatorname{coker}A\cong\operatorname{coker}D\cong\bigoplus_i\ZZ/d_i\ZZ\), which is finite if and only if every \(d_i\neq0\), and then has order \(\prod_id_i\). Since \(\det P,\det Q=\pm1\), \(|\det A|=|\det D|=\prod_id_i\).
4. A nilpotent operator \(N\) is diagonalizable only if \(N=0\).
If \(N^k=0\) and \(Nv=\lambda v\) with \(v\neq0\), then \(\lambda^kv=0\), so \(\lambda=0\). A diagonalizable \(N\) is therefore similar to the zero matrix, hence \(N=0\).
5. If \(A\) and \(B\) are diagonalizable and \(AB=BA\), then they are simultaneously diagonalizable.
For \(v\in E_\lambda(A)=\ker(A-\lambda I)\), \(A(Bv)=B(Av)=\lambda Bv\), so each eigenspace \(E_\lambda(A)\) is \(B\)-invariant. The minimal polynomial of \(B|_{E_\lambda(A)}\) divides that of \(B\), which has distinct linear factors, so \(B|_{E_\lambda(A)}\) is diagonalizable. The union over \(\lambda\) of eigenbases of the restrictions is a basis of \(V=\bigoplus_\lambda E_\lambda(A)\) of common eigenvectors.
6. Diagonalizability neither implies nor is implied by invertibility, and does not imply distinct eigenvalues.
\(\begin{pmatrix}1&0\\0&0\end{pmatrix}\) is diagonal and singular. \(\begin{pmatrix}1&1\\0&1\end{pmatrix}\) is invertible with minimal polynomial \((x-1)^2\), so it is not diagonalizable. \(I_n\) (\(n\ge2\)) is diagonal with the single eigenvalue \(1\).
7. \(T\colon V\to V\) is diagonalizable if and only if \(V=\bigoplus_i\ker(T-\lambda_iI)\), and then \(m_T=\prod_i(x-\lambda_i)\) is squarefree.
Eigenvectors for distinct eigenvalues are linearly independent, so the sum \(\sum_i\ker(T-\lambda_iI)\) is direct. \(T\) is diagonalizable exactly when \(V\) has a basis of eigenvectors, that is, when this sum is \(V\). In that case \(\prod_i(T-\lambda_iI)\) kills every basis vector, so \(m_T\) divides the squarefree polynomial \(\prod_i(x-\lambda_i)\).
8. If \(T(v_i)=v_{i+1\bmod n}\) for a basis \(v_0,\dots,v_{n-1}\), then \(m_T=x^n-1\), and \(T\) is diagonalizable over \(\CC\).
\(T^n=I\), so \(m_T\mid x^n-1\); and \(v_0,Tv_0,\dots,T^{n-1}v_0\) is a basis, so no nonzero polynomial of degree less than \(n\) kills \(T\). Hence \(m_T=x^n-1\). Over \(\CC\) it has the \(n\) distinct roots \(e^{2\pi ik/n}\), so by step 7 \(T\) is diagonalizable. Over a field where \(x^n-1\) has a repeated root or does not split, \(T\) is not diagonalizable.
9. If \(m_T\) is irreducible, every \(T\)-invariant subspace \(W\) has a \(T\)-invariant complement.
\(E=F[x]/(m_T)\) is a field, and \(V\) is an \(E\)-vector space with \(x\) acting as \(T\). \(T\)-invariant subspaces are exactly the \(E\)-subspaces, and an \(E\)-linear complement of \(W\) is \(T\)-invariant.