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Let $C$ be a complex $n \times n$ matrix with $\det C \neq 0$. Prove that there is a complex matrix $B$ such that $C = e^B$
Hint: use the Jordan form matrices for comlexas

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2 Answers 2

up vote 4 down vote accepted

Suppose $A$ is the Jordan block of size $n$ for some non-zero eigenvalue $\lambda$. That is, $$ A = \pmatrix{ \lambda&1&&\\ &\lambda&1&\\ &&\ddots\\ &&&\lambda } $$ We compute $e^A = e^\lambda K$, where $K$ is the matrix given by $$ K=\pmatrix{ 1&1&\cdots&1\\ &1&\cdots&1\\ &&\ddots&\vdots\\ &&&1 } $$ Note that $K$ is similar to the Jordan block of size $n$ corresponding to $1$.

Let $M$ be the matrix given by $$ M = \pmatrix{ 1&e^{-\lambda}&\\ &1&e^{-\lambda}&\\ &&\ddots&\\ &&&1 } $$ Note that $M$ is also similar to the Jordan block of size $n$ corresponding to $1$. Thus, $M$ and $K$ are similar. Take $S$ such that $K = SMS^{-1}$. We have $$ e^A = e^\lambda K = e^{\lambda}SMS^{-1} = S(e^{\lambda}M)S^{-1}\\ S^{-1}e^A S = e^{S^{-1}AS} = e^\lambda M $$ That is, we have $$ e^{S^{-1}AS} = e^{\lambda}M = \pmatrix{ e^{\lambda}&1&\\ &e^{\lambda}&1&\\ &&\ddots&\\ &&&e^{\lambda} } $$ So, suppose you begin with a matrix of the form $$ C = \pmatrix{ e^{\lambda}&1&\\ &e^{\lambda}&1&\\ &&\ddots&\\ &&&e^{\lambda} } $$ We may take $B = S^{-1}AS$, with $S$ and $A$ as defined above.

Note that any matrix with a non-zero determinant has a Jordan form consisting of blocks of the above form.

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I do not understand why the K is that form –  João Apr 12 at 20:19
    
Compute $e^A$ and it should be clear –  Omnomnomnom Apr 12 at 20:23
    
Or do you mean why $K$ has that Jordan canonical form? Note that $K$ has only $1$ eigenvector. –  Omnomnomnom Apr 12 at 20:24
    
got it, thank you –  João Apr 12 at 20:29
    
I do not understand why you choose C only at the end, the C would not have to be a generic matrix? –  Croos Apr 12 at 20:57

The fact that $\det C \ne 0$ implies that all the eigenvalues of $C$ are nonzero; thus when $C$ is transformed to Jordan form via a similarity transformation by the appropriate nonsingular matrix $P$, $C \to P^{-1}CP = D$, we see that $D = \text{diag}(D_1, D_2, \ldots, D_l)$ is block diagonal consisting of Jordan blocks of the form $D_k = \lambda I_m + N_m$, where $\lambda \ne 0$ is an eigenvalue of $C$ and $N$ may be written

$N_m = [n_{ij}] \; \text{with} \; n_{i, i + 1} = 1; \; n_{ij} = 0, j \ne i + 1, 1 \le i, j \le m, \tag{1}$

and $I_m$ is the $m \times m$ identity matrix. For each such block $D_k$ we may find a matrix $E_k$ such that $e^{E_k} = D_k$ in the following manner: set

$F_k = -\sum_1^{m - 1} \dfrac{(-N_m)^p}{p \lambda^p}; \tag{2}$

$F_k$ is derived by considering the power series for $\ln(1 + s)$ for small $s$ and substituting $N_m / \lambda$ for $s$; since $N_m^p = 0$ for $p \ge m$ the resulting series becomes the mere polynomial in $N_m$ given by (2); thus it is well-defined. We put

$E_k = (\ln \lambda) I_m + F_k, \tag{3}$

and we find, since $I_m$ and $F_k$ commute, that is $I_m F_k = F_k I_m$, that

$e^{E_k} = e^{\ln \lambda} e^{F_k} = \lambda e^{F_k}; \tag{4}$

but

$e^{F_k} = I_m + \dfrac{1}{\lambda}N_m, \tag {5}$

which may be readily seen by noting that $e^{\ln (1 + s)} = 1 + s$, since the algebraic maneuvers required in evaluating the coefficients of the power series of $e^{\ln (1 + s)}$ are the same for $s$ and $N_m / \lambda$, given that $N_m^m = 0$. Thus

$e^{E_k} = \lambda I_m + N_m = D_k, \tag{6}$

which holds for any Jordan block of $D$; taking

$E = \text{diag}(E_1, E_2, \ldots, E_l) \tag{7}$

thus yields a matrix such that

$D = e^E; \tag{8}$

then

$C = PDP^{-1} = Pe^EP^{-1} = e^{PEP^{-1}}; \tag{9}$

QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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