If I have a finite dimensional inner product space [itex]V = M_{n \times n}(\mathbb{R})[/itex], then one basis of V is the set of n² (n x n)-matrices, [itex]\beta = \{E_1, \dots , E_{n^2}\}[/itex] where [itex]E_i[/itex] has a 1 in the [itex]i^{th}[/itex] position, and zeroes elsewhere (and by [itex]i^{th}[/itex] position, I mean that the first position is the top-left, the second position is just to the right of the top-left, and the last position is the bottom-right). Since these matrices are linearly independent and span V, they certainly form a basis, and since there are n² of them, dim(V) = n². Therefore, if I have some linear operator T on V, then [itex]A = [T]_{\beta}[/itex] is an (n² x n²)-matrix, right? However, if v is some element of V, then T(v) = Av, but Av is not even possible, since it involves multiplying two square matrices of different dimension. Now, if I had made a mistake earlier, then maybe A is supposed to be an (n x n)-matrix. But that doesn't seem right.(adsbygoogle = window.adsbygoogle || []).push({});

My textbook proves:

If V is an N-dimensional vector space with an ordered basis [itex]\beta[/itex], then [itex][I_V]_{\beta} = I_N[/itex], where [itex]I_V[/itex] is the identity operator on V. Now, in our case, N = n², but if I was wrong before, and in the previous example, A should have been an (n x n)-matrix, then the equality above essentially states that an (n x n)-matrix is equal to an (n² x n²)-matrix. Where have I (or my book) made a mistake?

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# Homework Help: Vector Space of Matrices

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