General linear groups of dimension at least 2 are nonabelian

Show that GL_n(F) is nonabelian for all n \geq 2 and all fields F.

Recall that every field contains 0 and 1, and that 0 \neq 1. Suppose now that A, B are matrices in GL_n(F) such that the first row of A is [1,0,\ldots,0,1], the first column of A is [1,0,\ldots,0,0], the first row of B is [1,0,\ldots,0,0], and the first column of B is [1,0,\ldots,0,1]. Such matrices always exist in GL_n(F); for instance, take the identity matrix and change the (1,n) or (n,1) entry from 0 to 1. The resulting matrix is either upper or lower triangular, so that the determinant is the product of the diagonal entries. This product is 1, so that A and B are invertible.

With A and B having this form, the (1,1)-entry of AB is 2 and the (1,1)-entry of BA is 1. If 1 = 2 in F then we have 0 = 1, a contradiction. Since matrices are equal precisely when their corresponding entries are equal, we have AB \neq BA. Thus GL_n(F) is nonabelian for n \geq 2 and for all fields F where 0 \neq 1. \blacksquare

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  • dayana cepeda  On May 15, 2010 at 3:27 pm

    hola espero que esten bien
    solo queria decirles que esta pagina es muy interesante los felicito.
    por otro lado quisiera que me respondieran una pregunta referente al ejercicio 1.4 # 8 queria saber como asegurar que esas entradas que me dan para la matriz me forme una matriz invertible o no. gracias

    • nbloomf  On May 15, 2010 at 10:21 pm

      Gracias por leer!

      Mi español es muy malo, así que utilicé Google Translate. Si he entendido bien, no es inmediatamente obvio que las matrices invertibles A y B existen.

      Sí, esto podría ser más claro. Las matrices A y B existe siempre, consideras la matriz identidad con dimensiones (n,n) que el (1, n) o (n, 1) entrada 1 en lugar de 0. Estas matrices son triangulares superior e inferior, respectivamente, por lo que su factor determinante es el producto de las entradas diagonales. Dado que el determinante es 1, A y B son invertibles.

      He editado la solución para hacer de este explícito.

      Thank you for reading!

      My spanish is very bad, so I used Google Translate. If I understand correctly, it is not immediately obvious that invertible matrices A and B exist.

      Yes, this could be made more clear. Matrices A and B always exist; take the n by n identity matrix and make the (1,n) or (n,1) entry 1 instead of 0. These matrices are upper and lower triangular, respectively, so their determinant is the product of the diagonal entries. Since the determinant is 1, A and B are invertible.

      I edited the solution to make this explicit.

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