Let I: V->V be the identity operator on an n-dimensional vector space V defined by I(v)=v for every v in V. Show that the matrix of I with respect to a basis S for V is $\displaystyle I_{n}$.
Well it is the identity. So in particular, it takes the first basis element to the first basis element ($\displaystyle 1e_1 + 0e_2 + ... + 0e_n$), the second basis element to the second basis element ($\displaystyle 0e_1 + 1e_2 + ... + 0e_n$), etc.
Thus the first column is [1,0,0...0]
The second Column is [0,1,0,...0]
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The nth column is [0,0,...,1]
So $\displaystyle a_{ii}=1$ and $\displaystyle a_{ij}=0 $ for $\displaystyle i \not = j$
$\displaystyle a_ij=\delta_{ij}=In$