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i wanted to know if it was possible to raise a matrix to a matrix and i wanted to confirm if i have it. i tested it out with numbers and notation and i want to know if i'm right. $$ x =\left [ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right ] $$

$$ \left [ \begin{matrix} 2 & 1 \\ 1 & 2 \\ \end{matrix} \right ] ^ x $$

now, i use e^ln(x) = x, and i diagonalize my matrix to take an easy ln.

$$ e^(ln( \left [ \begin{matrix} 2 & 1 \\ 1 & 2 \\ \end{matrix} \right ] ^ x )) $$ $$ e ^ (x*ln(diag(\left [ \begin{matrix} 2 & 1 \\ 1 & 2 \\ \end{matrix} \right ])) $$ my diag of my matrix is as follows: $$ \left[\begin{matrix} 1 & 0\\ 0 & 1\\ \end{matrix}\right] $$ natural log of a diag matrix is as follows $$ \left[\begin{matrix} ln(1)=0 & 0\\ 0 & ln(1)=0\\ \end{matrix}\right] $$ $$ x * \left[\begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix}\right] = \left[\begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix}\right] $$ than i have $$ e^\left[\begin{matrix} 0 & 0\\ 0 & 0\\ \end{matrix}\right] $$ and if that wasn't a zero matrix id apply the Taylor series. did i do that right? would my next step be correct. i'd love to know. thank in advance!!

Mauro ALLEGRANZA
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Pie Pie
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  • How would you define the $\ln$ of a matrix (even a diagonal one) ? Moreover, your diagonalization is not good, the only matrix similar to identity is the identity itself. – TheSilverDoe Apr 14 '19 at 17:25
  • i found online in several place places that the log of a diagonal matrix is the log of the diagonal elements. also i learned that to diagonalize a matrix you take the eigenvalues and in any order place them down the diagonal – Pie Pie Apr 14 '19 at 17:29
  • i thought my math seemed very weird too, thats why im here asking – Pie Pie Apr 14 '19 at 17:29
  • I think you are on the right lines if you properly diagonalise the matrix (and you'll need the diagonalising matrix to undo it, too after you take logs). You ought to then get a matrix which you can diagonalise to work out the exp's in a similar way. – ancient mathematician Apr 14 '19 at 17:33
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    When you are doing a "can I define?" exercise, you should first set out what properties you want to obtain. You can define a matrix raised to another matrix to be anything you like, but it won't be good for anything. – Marc van Leeuwen Apr 14 '19 at 19:09
  • @PiePie how do you define the $\ln$ if the diagonal elements are non positive ? – TheSilverDoe Apr 14 '19 at 19:43
  • What are you going to do if the matrix can’t be diagonalized? – amd Apr 14 '19 at 21:05
  • @TheSilverDoe thats simple ill just simply take the complex number and do everything that same. that is how math logic work, and if it creates a problem than theres a rule that fixes it. math has always been like that. – Pie Pie Apr 14 '19 at 23:06
  • @amd if the matix cant be diagonalized i am aware that there are rule in differential math that can be used to solve it, but the only non-diag matrices are non square ans pre existing diag matrices – Pie Pie Apr 14 '19 at 23:08
  • one last thing, if i wanted to take the sin(diag(matrix)) can i just apply sin function to the diagonal elements? – Pie Pie Apr 14 '19 at 23:12
  • I don’t understand what you‘re trying to get at by “but the only non-diag matrices are...” $\small{\begin{bmatrix}1&1\0&1\end{bmatrix}}$ is square but not diagonalizable. – amd Apr 14 '19 at 23:16
  • To echo @MarcvanLeeuwen’s earlier comment, you can probably come up with a coherent definition of this operations, but what are its applications? – amd Apr 14 '19 at 23:17

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Defining the exponential function of a (complex) matrix $A$ is not too difficult, it is just $\exp A:=\sum_{n=0}^{\infty} \frac{A^n}{n!}$, which can be shown to converge. The same is therefore true for $\cos A$ and $\sin A$.

Working out what $\exp A$ is another question. In the "easy" case, when $A$ is diagonalizable we have $A=P^{-1}DP$, where $D$ is diagonal with entries $d_1,d_2,\dots,d_n$. Then, using $P^{-1}(X+Y)P=P^{-1}X P+P^{-1}Y P$ and $P^{-1}(X \cdot Y)P=P^{-1}X P\cdot P^{-1}Y P$ we can check that $\exp A=P^{-1}\Delta P$, where $\Delta$ is a diagonal matrix with entries $\exp d_1, \exp d_2,\dots,\exp d_n$.

(I think you are forgetting about the matrix $P$, which you must not do.)

Now lets talk about $\log A$. It easier to talk about $\log(I+ A)$, so lets do that.Now the power series for $\log(1+z)$ only converges for small $z$, so we can only proceed if $A$ is in some sense "small". But when it is, we can define $\log(I+A)$ as $\sum_{n=0}^{\infty} \frac{(-A)^{n+1}}{(n+1)}$.

If we are in the happy situation when $A$ is diagonalisable, and its eigenvalues $d_i$ are small, we will get $\log (I+A)=P^{-1}\Lambda P$, where $\Lambda$ is diagonal with entries $\log (1+d_i)$.

Once all this machinery has been set up, you can go on to define - in some restricted circumstances - the matrix $X^Y$. It is going to be $\exp(Y\cdot \log X)$. The restrictions will include that $(X-I)$ and its eigenvalues are "small".

Again, working it out is in general not easy, but if the right things are diagonalisable then the process is diagonalise, apply the usual functions to the diagonal elements, and undo the diagonalisation.

ancient mathematician
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  • thank you, after a lot more research i actually ended up coming to the same conclusion, i was ignoring matrix P, and i was messing up the basic way of calculating eigenvalues, as well as i was not undoing the denationalization. i was messing up in a lot of spots, thank you!!!!! im self taught up to this point and my high school offers no vector classes at all so im in the dark. people like u really help confirm what i learn and think is right. thank you – Pie Pie Apr 15 '19 at 23:25