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The book "Theory of Lie Groups" by Chevalley says

A linear endomorphism $\alpha$ of $C^n$ is determined when the elements $\alpha e_i=\sum\limits_{j=1}^n a_{ji}e_j$ are given. There corresponds to this endomorphism a matrix $(a_{ij})$ of degree $n$.

I don't understand this statement. The elements $e_1,e_2,\dots,e_n$ form a base of $\Bbb{C}^n$ over $\Bbb{C}$. Let $i=1$. Then $\alpha e_1=a_{11}e_1+a_{21}e_2+a_{31}e_3+\dots+a_{n1}e_n$.

So $a_{11}e_{11}+a_{12}e_{12}+\dots+a_{1n}e_{1n}=a_{11}e_{11}+a_{21}e_{31}+\dots+a_{1n}e_{n1}$, where $a_{31}$ is the 1-axis component of $e_3$ and $e_{13}$ is the $3$-axis component of $e_1$? What makes $e_{ij}=e_{ji}$? I know this is true when $e_i=(0,0,\dots,i,\dots 0)$, but why should this be true for all base vectors of $\Bbb{C}^n$?

Thanks in advance!

  • Your base is ${e_1,\ldots ,e_n}$. What is $e_{ij}$ ? And no, the matrix is not symmetric in general. – Dietrich Burde Nov 28 '13 at 12:03
  • $e_{ij}$ is the element on the $i^{th}$ row and $j^{th}$ column of the basis matrix, in which each basis vector $e_j$ takes up the whole of the $j^{th}$ column. –  Nov 28 '13 at 12:20

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