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Let $A=(a_{ij})_{i,j=1}^{4}\in M_4(\mathbb R)$. Apply inverse of the isomorphis $\chi:\mathbb H\to M_4(\mathbb R)$ where $\chi(a)=A$ and $$\chi (a_{0}+a_{1}i+a_{2}j+a_{3}k)=\begin{bmatrix}‎ ‎a_{0} &‎ -‎a_{1} & a_{3} &‎ -‎a_{2} \\‎ ‎a_{1} & a_{0} &‎ -‎a_{2} &‎ -‎a_{3} \\‎ -‎a_{3} & a_{2} & a_{0} &‎ -‎a_{1} \\‎ ‎a_{2} & a_{3} & a_{1} & a_{0}‎ ‎\end{bmatrix}.$$ Since every real matrix with size $4\times 4$ is represented by a quaternion, I want to find $a$ as $a=a_0+a_1 i+a_2 j+a_3 k$.

I don't know how i use inverse of $\chi$.

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    How do you define $\chi$ in the first place? You're not giving us any context at all here. – 2017-02-21
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    O.k. I edit now – 2017-02-21
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    Your edit does not answer my question. I'm looking for something like $$ \chi(a_0 + a_1 i + a_2j + a_3k) = \cdots ? $$ – 2017-02-21
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    Note that there is no isomorphism (in any usual category, other than groups) from $\mathbb{H}$ onto $M_4(\mathbb{R})$ since $\mathbb{H}$ is a real vector space of dimension $4$ while $M_4(\mathbb{R})$ is a real vector space of dimension $16$. – 2017-02-21
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    I edited this..... Yes the $\chi$ is an isomorphism. – 2017-02-21
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    @444: As I said, it's not an isomorphism *onto* $M_4(\mathbb{R})$ for the reasons stated. For example, the diagonal matrix with entries $1,1,1,0$ is not in the image of $\chi$. So $\chi^{-1}$ is not defined on all of $M_4(\mathbb{R})$ in any kind of natural way. – 2017-02-21
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    $\Bbb H$ and $M_4(\Bbb R)$ are **not** isomorphic. There are left and right regular representations $\Bbb H\to M_4(\Bbb R)$, and there is an $\Bbb R$-algebra isomorphism $\Bbb H\otimes_{\Bbb R}\Bbb H\to M_4(\Bbb R)$. – 2017-02-22

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Hint:

$\chi$ is not an isomorphism from $ \mathbb H$ and $M_4(\mathbb R)$, but from $ \mathbb H$ and the subspace $V$ of $M_4(\mathbb R)$ spanned by the matrices: $$ E= \begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{bmatrix} \qquad I= \begin{bmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&0&0&-1\\ 0&0&1&0 \end{bmatrix} $$ $$ J= \begin{bmatrix} 0&0&0&-1\\ 0&0&-1&0\\ 0&1&0&0\\ 1&0&0&0 \end{bmatrix} \qquad K= \begin{bmatrix} 0&0&1&0\\ 0&0&0&-1\\ -1&0&0&0\\ 0&1&0&0 \end{bmatrix} $$ With the correspondence $$ 1 \to E\quad i \to I \quad j \to J \quad k \to K $$ Note that $V$ is a space of dimension $4$ over $\mathbb R$ and $\chi$ is invertible from $V$ to $\mathbb H$ in a an obvious way.