A student has asked me go in more detail of the model answers for parts 2009 (c)(iii) and (d)(ii)

First let me recall some generalities about the Jordan Chevalley decomposition. In the (complex) Jordan form, the diagonal part of the matrix is semi-simple - as it obviously has a diagonal complex Jordan form - and the remaining off-diagonal part is nilpotent (as it is upper or lower triangular with zero diagonal; one easily verifies that taking powers of such matrices eventually always results in the 0 matrix). One also easily verifies in Jordan form that the diagonal part and the upper or lower triangular part of the matrix commute with each other. A very convenient fact is that the properties "semi-simple", "nilpotent" and "commutation" are intrinsic and do not depend on the choice of coordinates:

If \(A^k=0\) then \((TAT^{-1})^k=0\) for any invertible matrix \(T\).

If \(A\) is complex diagonalizable, then so is \(TAT^{-1}\) for any invertible matrix \(T\).

If \(A\) and \(B\) commute, i.e. \(AB=BA\), then also \((TAT^{-1})(TBT^{-1})=(TBT^{-1})(TAT^{-1})\) for any invertible matrix \(T\).

So we observe that we can prove the Jordan-Chevalley decomposition (and its uniqueness) directly from the Jordan normal form.

We proved in an exercise that \(\exp(A+B)=\exp(A)\exp(B)\) if \(AB=BA\). So in particular, if \(A=A_s+A_n\) is the Jordan-Chevalley decomposition, then \(\exp(At)=\exp(A_st)\exp(A_nt)\). This is very useful since the first part \(\exp(A_st)\) depends only on the eigenvalues of \(A\), and thus contains terms depending only on \(e^{\lambda_i t}\) where \(\lambda_i\) denote eigenvalues of \(A\) (if eigenvalues are complex this leads to terms with dependencies \(e^{\Re(\lambda_i)t}\cos(\Im(\lambda_i) t)\) and \(e^{\Re(\lambda_i)t}\sin(\Im(\lambda_i) t))\), were \(\Re\) and \(\Im\) denote the real and imaginary parts, respectively).

We know that sometimes we also have polynomial terms appearing in the expression \(\exp(At)\). These polynomials come from the second part \(\exp(A_nt)\) since \(\exp(A_n t)=\sum_{m=0}^{k-1}\frac{A_n^m}{m!} t^m\) (this follows from the fact that \(A_n^k=0\)).

The question (c)(iii) is about the Jordan-Chevalley decomposition of \(\exp(At)\). The only thing to check is that we can write this as sum of a nilpotent and a semi-simple matrix which commute with each other. (The Jordan-Chevalley decomposition theorem than asserts that this decomposition is in fact unique.)

The question contains the hint that \(\exp(A_s t)\) is semi-simple. We can see this by verifying that if \(TA_sT^{-1}\) is (complex) diagonal, then so is \(T\exp(A_st)T^{-1}=\exp(TA_sT^{-1}t)\).

Let us check whether indeed the semi-simple part of \(\exp(At)\) is equal to \(\exp(A_st)\) (in the sense of the Jordan-Chevalley decomposition). We write \(\exp(At)=\exp(A_st)+N\) where \(N:=[\exp(At)-\exp(A_st)]\). Now we recall that \(\exp(At)=\exp(A_st)\exp(A_nt)\) so \(N=\exp(A_st)[\exp(A_nt)-1]\) and as these two factors commute we have \(N^k=\exp(A_s t k)[\exp(A_nt)-1] ^k\) and if \(A_n^k=0\) we also have \([\exp(A_n t)-1]^k=0\) since \([\exp(A_n t)-1]=p(A_n)\) is a polynomial in \(A_n\) with \(p(0)=0\). Thus \(N\) is nilpotent, and it is readily checked that \(N\) also commutes with \(\exp(A_s t)\). So \(\exp(A t)=\exp(A_s t)+N\) is the Jordan-Chevalley decomposition of \(\exp(A t)\) where \(\exp(A_s t)\) is the semi-simple part and \(N=\exp(A_s t)[\exp(A_n t)-1]\) is the nilpotent part.

In part d(ii), \(B\) is the projection to the eigenspace for eigenvalue \(-1\) with as kernel the generalised eigenspace for eigenvalue +1, and \(D=A_n\). As there is a Jordan block for eigenvalue +1 (and not for eigenvalue -1), the range of A_n is the eigenspace of A for eigenvalue +1 (check this by writing a 2-by-2 matrix with a Jordan block); the kernel of \(A_n\) is spanned by the eigenspaces of \(A\). Since the range of \(D\) lies inside the kernel of \(B\), and the range of \(B\) in the kernel of \(D\), it follows that \(DB=BD=0\).

First let me recall some generalities about the Jordan Chevalley decomposition. In the (complex) Jordan form, the diagonal part of the matrix is semi-simple - as it obviously has a diagonal complex Jordan form - and the remaining off-diagonal part is nilpotent (as it is upper or lower triangular with zero diagonal; one easily verifies that taking powers of such matrices eventually always results in the 0 matrix). One also easily verifies in Jordan form that the diagonal part and the upper or lower triangular part of the matrix commute with each other. A very convenient fact is that the properties "semi-simple", "nilpotent" and "commutation" are intrinsic and do not depend on the choice of coordinates:

If \(A^k=0\) then \((TAT^{-1})^k=0\) for any invertible matrix \(T\).

If \(A\) is complex diagonalizable, then so is \(TAT^{-1}\) for any invertible matrix \(T\).

If \(A\) and \(B\) commute, i.e. \(AB=BA\), then also \((TAT^{-1})(TBT^{-1})=(TBT^{-1})(TAT^{-1})\) for any invertible matrix \(T\).

So we observe that we can prove the Jordan-Chevalley decomposition (and its uniqueness) directly from the Jordan normal form.

We proved in an exercise that \(\exp(A+B)=\exp(A)\exp(B)\) if \(AB=BA\). So in particular, if \(A=A_s+A_n\) is the Jordan-Chevalley decomposition, then \(\exp(At)=\exp(A_st)\exp(A_nt)\). This is very useful since the first part \(\exp(A_st)\) depends only on the eigenvalues of \(A\), and thus contains terms depending only on \(e^{\lambda_i t}\) where \(\lambda_i\) denote eigenvalues of \(A\) (if eigenvalues are complex this leads to terms with dependencies \(e^{\Re(\lambda_i)t}\cos(\Im(\lambda_i) t)\) and \(e^{\Re(\lambda_i)t}\sin(\Im(\lambda_i) t))\), were \(\Re\) and \(\Im\) denote the real and imaginary parts, respectively).

We know that sometimes we also have polynomial terms appearing in the expression \(\exp(At)\). These polynomials come from the second part \(\exp(A_nt)\) since \(\exp(A_n t)=\sum_{m=0}^{k-1}\frac{A_n^m}{m!} t^m\) (this follows from the fact that \(A_n^k=0\)).

The question (c)(iii) is about the Jordan-Chevalley decomposition of \(\exp(At)\). The only thing to check is that we can write this as sum of a nilpotent and a semi-simple matrix which commute with each other. (The Jordan-Chevalley decomposition theorem than asserts that this decomposition is in fact unique.)

The question contains the hint that \(\exp(A_s t)\) is semi-simple. We can see this by verifying that if \(TA_sT^{-1}\) is (complex) diagonal, then so is \(T\exp(A_st)T^{-1}=\exp(TA_sT^{-1}t)\).

Let us check whether indeed the semi-simple part of \(\exp(At)\) is equal to \(\exp(A_st)\) (in the sense of the Jordan-Chevalley decomposition). We write \(\exp(At)=\exp(A_st)+N\) where \(N:=[\exp(At)-\exp(A_st)]\). Now we recall that \(\exp(At)=\exp(A_st)\exp(A_nt)\) so \(N=\exp(A_st)[\exp(A_nt)-1]\) and as these two factors commute we have \(N^k=\exp(A_s t k)[\exp(A_nt)-1] ^k\) and if \(A_n^k=0\) we also have \([\exp(A_n t)-1]^k=0\) since \([\exp(A_n t)-1]=p(A_n)\) is a polynomial in \(A_n\) with \(p(0)=0\). Thus \(N\) is nilpotent, and it is readily checked that \(N\) also commutes with \(\exp(A_s t)\). So \(\exp(A t)=\exp(A_s t)+N\) is the Jordan-Chevalley decomposition of \(\exp(A t)\) where \(\exp(A_s t)\) is the semi-simple part and \(N=\exp(A_s t)[\exp(A_n t)-1]\) is the nilpotent part.

In part d(ii), \(B\) is the projection to the eigenspace for eigenvalue \(-1\) with as kernel the generalised eigenspace for eigenvalue +1, and \(D=A_n\). As there is a Jordan block for eigenvalue +1 (and not for eigenvalue -1), the range of A_n is the eigenspace of A for eigenvalue +1 (check this by writing a 2-by-2 matrix with a Jordan block); the kernel of \(A_n\) is spanned by the eigenspaces of \(A\). Since the range of \(D\) lies inside the kernel of \(B\), and the range of \(B\) in the kernel of \(D\), it follows that \(DB=BD=0\).

Thanks for sharing these questions and their answers. It would be of great help for the students who are prepairing for their final examination on their own.

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