Find $B$ if $A=e^B$ and $A=\begin{bmatrix} 2&1&0\\ 0&2&0\\ 0&0&4\\ \end{bmatrix}$.
Besides, I would be very happy if give some general remark(Best approach). I have seen the wiki article on log of a matrix but it was too complicated(for me).
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Can you add any logic you have. – Archis Welankar Dec 16 '15 at 17:52
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Is it $log A=B$ after taking logs – Archis Welankar Dec 16 '15 at 17:54
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@ArchisWelankar: yes, isn't it? – marya Dec 16 '15 at 17:55
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So now its just taking log of a matrix where you can use net to see power series as i dont remember it for now – Archis Welankar Dec 16 '15 at 18:05
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You can make use of the block structure of $A$: $$ A = \begin{pmatrix} C & 0 \\ 0 & 4 \end{pmatrix} = e^B = \sum_{k=0}^\infty \frac{1}{k} B^k \Rightarrow B = \begin{bmatrix} D & 0 \\ 0 & x \end{bmatrix} $$ so we can assume $4 = e^x \Rightarrow x = \ln(4)$. For the block matrices we get $$ C = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = e^D = \sum_{k=0}^\infty \frac{1}{k!} D^k $$ and try an upper triangular matrix $$ D = \begin{pmatrix} y & z \\ 0 & y \end{pmatrix} $$ and get the powers $$ D^2 = \begin{pmatrix} y & z \\ 0 & y \end{pmatrix} \begin{pmatrix} y & z \\ 0 & y \end{pmatrix} = \begin{pmatrix} y^2 & 2 y z \\ 0 & y^2 \end{pmatrix} \\ D^3 = \begin{pmatrix} y^2 & 2 y z \\ 0 & y^2 \end{pmatrix} \begin{pmatrix} y & z \\ 0 & y \end{pmatrix} = \begin{pmatrix} y^3 & 3 y^2 z \\ 0 & y^3 \end{pmatrix} \\ \vdots \\ D^k = \begin{pmatrix} y^k & k y^{k-1} z \\ 0 & y^k \end{pmatrix} \quad (k \ge 1) $$ which suggest $$ C = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} = e^D = \begin{pmatrix} e^y & e^y z \\ 0 & e^y \end{pmatrix} $$ so $y = \ln(2)$ and $z = 1/e^y = 1/2$. This gives $$ B = \begin{pmatrix} \ln(2) & 1/2 & 0 \\ 0 & \ln(2) & 0 \\ 0 & 0 & \ln(4) \end{pmatrix} $$
mvw
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@ mvw , in fact there is a unique real solution. To prove that, you must prove the implication in line 2 and the fact that $D$ is necessarily upper-triangular. – Dec 17 '15 at 14:53
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I used without proof that diagonal matrices are closed under matrix addition, matrix multiplication and multiplication with a scalar. The same for upper triangular matrices. The powers of $D$ might be proofed by induction and the result for $C$ by comparing component-wise. The given matrix is a lucky case where one can calculate the result like above. – mvw Dec 17 '15 at 19:12
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Yes I agree. That I wanted to say is that $B,D$ are necessarily in the above form because $AB=BA$ and $CD=DC$. – Dec 17 '15 at 19:42
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Find a general expression for $(A-2I)^n$ and apply the power series for $\log x$ in powers of $(x-2).$
Justpassingby
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2Yes but only in the lower right corner so you have a log power series in the lower right corner of the result. – Justpassingby Dec 16 '15 at 18:26
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@ Justpassingby , your series do not converge because $||(A-2I)^n||$ tends to $\infty$ with $n$. – Dec 17 '15 at 14:51
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That is because you evaluate the full matrix. You should note that it is block diagonal; the lower right entry of the result is simply $\ln 4$ and you only need to take powers of the upper left $2\times 2$ matrix. – Justpassingby Dec 17 '15 at 14:58
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In your answer to Sucre, you speak about $(A-2I)^n$; anyway you're a big boy; do as you want. – Dec 17 '15 at 18:23