Utente:Ming mm/Constraints in the 2 × 2 case

Constraints in the 2 × 2 case

Link da OLD version =Logarithm_of_a_matrix&diff=prev&oldid=1003417720

If a 2 × 2 real matrix has a negative determinant, it has no real logarithm. Note first that any 2 × 2 real matrix can be considered one of the three types of the complex number z = x + y ε, where ε² ∈ { −1, 0, +1 }. This z is a point on a complex subplane of the ring of matrices.

The case where the determinant is negative only arises in a plane with ε² =+1, that is a split-complex number plane. Only one quarter of this plane is the image of the exponential map, so the logarithm is only defined on that quarter (quadrant). The other three quadrants are images of this one under the Klein four-group generated by ε and −1.

For example, let a = log 2 ; then cosh a = 5/4 and sinh a = 3/4.

For matrices, this means that

$A=\exp {\begin{pmatrix}0&a\\a&0\end{pmatrix}}={\begin{pmatrix}\cosh a&\sinh a\\\sinh a&\cosh a\end{pmatrix}}={\begin{pmatrix}1.25&.75\\.75&1.25\end{pmatrix}}$.

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$log(A)=B_{n}~$⇔$~~e^{B_{n}}=A$

$e^{B_{n}}=\sum _{k=0}^{\infty }{1 \over k!}B_{n}^{k}~$ where

$(B_{n})^{0}=1~I_{2},$

$(B_{n})^{1}=(\alpha ){\begin{pmatrix}0&+1\\+1&0\\\end{pmatrix}},$

$(B_{n})^{2}=(\alpha )^{2}~I_{2}$
…

$\sum _{k=0}^{\infty }{1 \over k!}B_{n}^{k}={\begin{pmatrix}\cosh(\alpha )&\sinh(\alpha )\\\sinh(\alpha )&\cosh(\alpha )\\\end{pmatrix}}=A~.$
qed.

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So this last matrix has logarithm

$\log A={\begin{pmatrix}0&\log 2\\\log 2&0\end{pmatrix}}$.

These matrices, however, do not have a logarithm:

${\begin{pmatrix}3/4&5/4\\5/4&3/4\end{pmatrix}},\ {\begin{pmatrix}-3/4&-5/4\\-5/4&-3/4\end{pmatrix}},\ {\begin{pmatrix}-5/4&-3/4\\-3/4&-5/4\end{pmatrix}}$.

They represent the three other conjugates by the four-group of the matrix above that does have a logarithm.

A non-singular 2 x 2 matrix does not necessarily have a logarithm, but it is conjugate by the four-group to a matrix that does have a logarithm.

It also follows, that, e.g., a square root of this matrix A is obtainable directly from exponentiating (logA)/2,

${\sqrt {A}}={\begin{pmatrix}\cosh((\log 2)/2)&\sinh((\log 2)/2)\\\sinh((\log 2)/2)&\cosh((\log 2)/2)\end{pmatrix}}={\begin{pmatrix}1.06&.35\\.35&1.06\end{pmatrix}}~.$

For a richer example, start with a pythagorean triple (p,q,r)

and let a = log(p + r) − log q. Then

$e^{a}={\frac {p+r}{q}}=\cosh a+\sinh a$.

Now

$\exp {\begin{pmatrix}0&a\\a&0\end{pmatrix}}={\begin{pmatrix}r/q&p/q\\p/q&r/q\end{pmatrix}}$.

Thus

${\tfrac {1}{q}}{\begin{pmatrix}r&p\\p&r\end{pmatrix}}$

has the logarithm matrix

${\begin{pmatrix}0&a\\a&0\end{pmatrix}}$ ,

where a = log(p + r) − log q.