# Duplicate Question: Show that there exist ordered bases β and γ for V and W, such that T is a diagonal matrix.

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.

• I don't know how you came to the conclusion that if $[T]$ is diagonal then $T$ is one-to-one. As a counterexample, the zero matrix is technically diagonal but the transformation it describes is not one-to-one. – Omnomnomnom Aug 13 at 21:12
• This depends on your definition, if you exclude the trivial case, that is the zero matrix, the columns of your diagonal matrix are linearly independent and thus $T$ is bijective by your Theorem 2.5 – Riquelme Aug 13 at 21:39

If there exists basis $$\beta$$ and $$\gamma$$ such that $$[\textsf T] _\beta^\gamma$$ is a diagonal matrix, and each of its diagonal entries is a non-zero entry, then $$\textsf T$$ is one-to-one.
Set $$\beta = \{ v_1,v_2,\dots,v_n \}$$, $$\gamma = \{ u_1,u_2,\dots,u_n \}$$ and $$A=[\textsf T]_\beta^\gamma$$. Suppose $$\textsf T (x) = 0_{ \textsf W}$$ for some $$x \in \textsf V$$. We want to prove that $$x = 0_{ \textsf V }$$ (and hence $$\textsf T$$ is one-to-one).
Since $$x$$ can be written as $$x = \sum_{j=1}^n a_jv_j$$ for some scalars $$a_1,a_2,\dots,a_n$$, then $$0_{ \textsf W} = \textsf T (x) = \sum_{j=1}^n a_j \textsf T (v_j) = \sum_{j=1}^n a_j \left( \sum_{i=1}^n A_{ij}u_i \right) = \sum_{j=1}^n (a_jA_{jj})u_j$$ since $$\gamma$$ is linearly independent, it follows that $$a_j A_{jj} = 0$$ for all $$j=1,2,\dots,n$$. Therefore, $$a_j =0$$ for all $$j$$ and we can conclude that $$x=0_{\textsf V}$$.