12

Which $2\times 2$ matrices satisfy the equation $$\det e^A=e^{\det A}?$$

I know that $\det e^A=e^{\operatorname{trace}A}$ so assuming $A$ is real we get $$\operatorname{trace}A=\det A.$$ Then, $$\det(A-\lambda I)=\lambda^2-2\operatorname{trace}(A)\lambda+\det(A)=(\lambda-\det(A))^2$$ so the only eigenvalue is $$\lambda=\det(A)=\operatorname{trace}A.$$ Hence, $$\operatorname{trace}A=2\lambda=2\operatorname{trace}A\implies\operatorname{trace}A=\det A=0.$$ Write $$A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}$$ so that $\operatorname{trace}A=0$ is already taken into account. Then, $$\det A=-a^2-bc=0$$ so $a^2=-bc$. Thus, $$A=\begin{pmatrix}\sqrt{-bc}&b\\c&-\sqrt{-bc}\end{pmatrix}$$ Is that the end of the solution?

asdasdasdasd
  • 121
  • 4
    Did you accidentally replace $\det(A)$ with $\det(A)^2$ when factoring the characteristic polynomial? There's plenty of non-zero solutions to $\lambda_1 \lambda_2 = \lambda_1 + \lambda_2$. – Erick Wong Aug 17 '15 at 21:55
  • 2
    And $\det(A-\lambda I)$ is equal to $\lambda^2 - \operatorname{trace}(A)\lambda + \det(A)$, but you have mistakenly doubled the trace in your question. – user1551 Aug 17 '15 at 22:10

3 Answers3

4

The condition $\lambda_1 + \lambda_2 = \lambda_1 \lambda_2$ can be written as $(\lambda_1 - 1)(\lambda_2 - 1) = 1$. Thus the pair of eigenvalues $(\lambda_1, \lambda_2)$, if real, are on a hyperbola in the $\lambda_1,\lambda_2$ plane. You also have solutions where $(\lambda_1, \lambda_2)$ are a pair of complex conjugates $\alpha \pm i \beta$, where $(\alpha - 1)^2 + \beta^2 = 1$ (this describes a circle in the $\alpha, \beta$ plane). An example of a real matrix with the real eigenvalues $\lambda_1, \lambda_2$ is $$ \pmatrix{\lambda_1 & 0\cr 0 & \lambda_2\cr}$$ An example of a real matrix with the complex eigenvalues $\alpha \pm i \beta$ is $$ \pmatrix{0 & -\alpha^2 - \beta^2\cr 1 & 2\alpha}$$ Of course you may take anything similar to one of these, so multiply by $S$ on the left and $S^{-1}$ on the right, where $S$ is any invertible $2 \times 2$ matrix.

Robert Israel
  • 448,999
0

By the equation $\det(A)=\mathrm{trace}(A)$ so if $\lambda_1,\lambda_2$ are the eigenvalues of $A$ so $$ \lambda_1\lambda_2=\lambda_1+\lambda_2 $$ Then $$ \lambda_1=\frac{\lambda_2}{\lambda_2-1} $$ So $\sigma(A)=\{t,\frac{t}{t-1} \}$ if $t\in\mathbb{R}-\{0,2\}$ $A$ will be diagonalisable and $A=PDP^{-1}$ where $P$ is any invertible matrix and $D=\mathrm{Diag}(t,\frac{t}{t-1})$

If $t=0$ or $t=2$ we can use the jordan normal form to write : $$ A=PJ_0P^{-1} \qquad \qquad \textrm{ Or } \qquad A=PJ_2P^{-1} $$

Fabrosi
  • 673
Hamza
  • 3,781
  • I'm not familiar with the notation $J_0$ and $J_2$: is it specifically referring to a $2\times 2$ Jordan block or does it allow $J_0$ to be diagonal? Also, you should probably exclude $t=1$ as well. – Erick Wong Aug 18 '15 at 00:32
  • yeah @ErickWong $J_x=\left(\begin{array}{c} x&1\ 0& x \end{array}\right)$ the case $t=1$ can't be a solution by the equality $\det(A)=trace(A)$ – Hamza Aug 18 '15 at 00:56
  • That means your solution does not include the cases $A = O$ or $A = 2I$: these should be included in the diagonalisable case ($t \in {0,2}$). – Erick Wong Aug 18 '15 at 00:57
  • Huh? Not one claim in your last comment is true. A non-identity projection matrix is never invertible (and the $P$ you give is not a projection). No similarity transformation will ever turn a Jordan block into a diagonal matrix. No non-zero matrix can possibly be similar to the zero matrix. $O$ and $2I$ are definitively missing. – Erick Wong Aug 18 '15 at 07:24
  • you have right a delete my last comment. – Hamza Aug 18 '15 at 12:40
0

Since any $n\times n$ matrix $A\in\mathbb{C}$ is unitarily similar to upper triangular matrix, i.e. $P^{-1}AP=D, \:\overline{P}^T=P^{-1}=P$, there is $$ \det e^A=\det e^{PDP^{-1}}=\det{P^{}}\det e^{D}\det{P^{-1}}=e^{\sum_{i=1}^n\lambda_i}=e^{\operatorname{trace}A} $$ Since $\det A=\prod_{i=1}^n\lambda_i$ and $e^{\lambda_i}$ are eigenvalues of $e^A$ $$ \det e^A=\prod_{i=1}^ne^{\lambda_i}=e^{\sum_{i=1}^n\lambda_i}=e^{\det A}\iff \prod_{i=1}^n\lambda_i=\sum_{i=1}^n\lambda_i $$

Eugene Zhang
  • 16,805