My Math Forum vector as product of matrices, exp function of vector

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 June 9th, 2015, 03:43 AM #1 Newbie   Joined: Oct 2014 From: China Posts: 12 Thanks: 2 vector as product of matrices, exp function of vector A vector (a11x, a22y, a33z) can be expended as: \begin{pmatrix} a_{11}x \\ a_{22}y\\ a_{33}z \end{pmatrix}= \begin{pmatrix} a11 & a12 & a13\\ a21 & a22 & a23\\ a31 & a32 & a33 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 0& 1 &0\\ 0&0 & 1 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} * & * \\ * & * \\ * & * \end{pmatrix} \begin{pmatrix} * & * & *\\ * & * & * \end{pmatrix} \begin{pmatrix} 1 & 0 & 0\\ 0& 1 &0\\ 0&0 & 1 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} but how can I write the following vector as product of matrices: \begin{pmatrix} exp(a_{11})x \\ exp(a_{22})y\\ exp(a_{33})z \end{pmatrix} the annoying exp() prevents me doing any transformation! Maybe the question is: is there a exp() function whose parameter is a vector instead of a single number?
 June 9th, 2015, 07:11 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,820 Thanks: 750 "$exp(a_{11})x$ means $e^{a_{11}}$. Is that really what you meant or did you mean $e^{a_{11}x}$? If it is the former, then you have just the matrix multiplication $\begin{bmatrix}e^{a_{11}} & 0 & 0 \\ 0 & e^{a_{22}} & 0 \\ 0 & 0 & e^{a_{33}}\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}$. If it is the latter, it is a little more complicated. For a diagonal matrix, $A= \begin{bmatrix}a &0 \\ 0=&b=&0 \\ 0=&c\end{bmatrix}=$ then $e^{Ax}= \begin{bmatrix}e^{ax} &0 \\ 0=&e^{bx}=&0 \\ 0=&e^{cx}\end{bmatrix}=$. If A is not diagonal, then you will have to try to "diagonalize" it. That is, find a matrix, B, such that $BAB^{-1}= D$ where D is a diagonal matrix. In that case $A= B^{-1}DB$, of course, and $e^{Ax}= B^{-1}e^{Dx}B$. (Only matrices that have a "full set of eigenvalues", that is, a basis for the vector space that are eigenvectors for the matrix, can be "diagonalized". For others, you have to use the "Jordan Normal Form" and that gets quite a bit more complicated.) Thanks from whitegreen

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