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Consider the vector space of real $2 x 2$ matrices and take as base $\{{E_{11},E_{12},E_{21},E_{22}}\}$. Where $E_{ij}$ represents the matrix with a $1$ in the $i$-th row and $j$-th column and the remaining space filled with zeros.

Consider in this space the linear transformation of real $2 x 2$ matrices $T : \Bbb R^{2x2} \to \Bbb R^{2x2}$ that $A \to A^T$.

Determine the matrix of this linear transformation against given base.

I'm used to doing transformations with vectors in the base but not with matrices in the base. How do I approach this problem?

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  • $\begingroup$ Consider using \to rather than ->. $\endgroup$
    – Ben Grossmann
    Aug 25, 2015 at 18:19

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Just remember that, according to our choice of base, the coordinate vector corresponding to $a E_{11} + bE_{12} + cE_{21} + dE_{22}$ is $$ \pmatrix{a\\b\\c\\d} $$ Now, we calculate $T(v)$ for each $v$ in our base. The $i$th column is the $i$th coordinate vector of $T(v)$.

For example: the second basis element is $E_{12}$, and $T(E_{12}) = E_{21}$, which is represented by the coordinate vector $$ \pmatrix{0\\0\\1\\0} $$ So, this vector is the second column of our transformation matrix.

All together, we end up with $$ M = \pmatrix{ 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 } $$

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  • $\begingroup$ I don't understand how you get the coordinate vector of $E_{21}$ $\endgroup$
    – Stanko
    Aug 25, 2015 at 18:16
  • $\begingroup$ Note that $E_{21} = 0E_{11} + 0E_{12} + 1E_{21} + 0E_{22}$. Then, apply the first formula. $\endgroup$
    – Ben Grossmann
    Aug 25, 2015 at 18:16
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    $\begingroup$ Ok, I understand it, thank you. $\endgroup$
    – Stanko
    Aug 25, 2015 at 18:19

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