# Alternative definition of diagonalisable transformation

Supposedly a transformation $$T: V \to V$$ is diagonalisable iff there exists a basis of $$V$$ consisting only of eigenvectors of $$T$$. Can someone show me why this is true? I don't really know where to even start other than "Let $$[v_1, ..., v_n]$$ be a basis of $$V$$ with $$v_i$$ eigenvectors".

• If you want to show that two definitions are equivalent, it's useful to include the other definition you are using - what is that? – StackTD Mar 11 at 8:26

You have the right idea. If $$[v_1, ..., v_n]$$ is a basis of $$V$$ with $$v_i$$ eigenvectors, let us try to understand how $$T$$ would look in this basis (I'm assuming you already know how to find a representing matrix of a linear transofrmation, given some basis). Since $$T$$ acts on the basis element $$v_1$$ by scaling it by a factor of $$\lambda_1$$ (since it is an eigenvector), the first column of $$T$$ would be - $$\begin{pmatrix} \lambda_1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$$. Similarly, the second column of $$T$$ would be - $$\begin{pmatrix} 0 \\ \lambda_2 \\ \vdots \\ 0 \end{pmatrix}$$, and so on. Overall, you will get a diagonal matrix.
For the other way around - if $$T$$ is diagonal in some basis, you know that it acts on the first basis vector by scaling it by the factor on the corresponding part of the diagonal (again, I'm assuming you know the general method of finding a representing matrix, given some basis). Similarly, each basis vector is scaled by the corresponding part of the diagonal. Since $$T$$ acts on each basis vector by scaling it, we conclude that the basis consists only of eigenvectors of $$T$$.