by user_hello1
Last Updated August 13, 2019 21:20 PM

Let $\mathsf{V}$ and $\mathsf{W}$ be vector spaces such that $\dim{\mathsf{V}} = \dim{\mathsf{W}}$, and let $\mathsf{T: V \to W}$ be linear. Show that there exist ordered bases $\beta, \gamma$ such that $[\mathsf{T}]_\beta^\gamma$ is a diagonal matrix.

If $[\mathsf{T}]_\beta^\gamma$ is diagonal, then $\mathsf{T}(\beta) = \{\mathsf{T}(\beta_1), \dots, \mathsf{T}(\beta_n)\}$ is linearly independent. Suppose otherwise, $[\mathsf{T}]_\beta^\gamma$ is diagonal but there is some linearly dependent $\mathsf{T}(\beta_m)$. (Informal) Then $\mathsf{T}(\beta_m)$ can be written in terms of the remaining elements, such that the linear matrix will be undiagonal.

$\mathsf{T}(\beta)$ is linearly independent if and only if $\mathsf{T}(\beta)$ generates $\mathsf{W}$ (and therefore only if $\mathsf{T}$ is onto). If $\mathsf{T}(\beta)$ is linearly independent, then it is a linearly independent set of size $\dim{\mathsf{W}}$. Hence it is a basis of $\mathsf{W}$. If $\mathsf{T}(\beta)$ generates $\mathsf{W}$, then again, given its size, it is a basis for $\mathsf{W}$.

By a Theorem stated in my textbook, having the same conditions as the question, $\mathsf{T}$ is onto if and only if $\mathsf{T}$ is one-one.

**So it seems that if $[\mathsf{T}]_\beta^\gamma$ is diagonal, $\mathsf{T}$ must be one-one? That does not seem right.**

Edit: I am aware that there is an **exact duplicate** Show that there exist ordered bases $\beta$ and $\gamma$ for V and W, such that T is a diagonal matrix) of the question, however I do not wish for an answer as is there provided, but only for *limited guidance*.

Updated January 22, 2018 09:20 AM

Updated May 17, 2018 12:20 PM

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