NUS-MATH530 Similar to Diagonal

Prove but T is diagonalizable if and only if the matrix of T is similar to a diagonal matrix.
Try 2.
Given similarity:
So we have that:

\begin{equation} D = S^{-1} A S \end{equation}

where, D is diagonal. We apply S to both sides to yield:

\begin{equation} SD = AS \end{equation}

Now, note that S is invertible. This means that its column s are linearly independent (as it is an operator, which means it is injective, and hence has a zero null space; that indicates that the dimension of its range is that of the whole space: indicating its columns vectors are spanning; there is dim\ V such columns, so it is a basis and hence linearly independent).
Let S = [v_1 | \dots | v_{n}]; now, SD = [\lambda_{1} v_1 | \dots | \lambda_{n} v_{n}].
By that same definition above course, A[v_1 | \dots | v_{n}] = [\lambda_{1} v_1 | \dots | \lambda_{n} v_{n}].
Finally, then, by definition, v_1 \dots v_{n} are eigenvectors of A. Note again that, per the above, this is n linearly independent eigenvectors in a space of \dim n — this makes them a basis of V. Having made a basis of eigenvectors of A, it is diagonalizable.
Given diagonalizability:
Construct S= [v_1 | \dots | v_{n}], a basis of eigenvectors of A which is diagonalizable. Now, AS would send each of the vectors to their corresponding scales, meaning: AS = [\lambda_{1} v_{1} | \dots | \lambda_{n} v_{n}].
Lastly, applying S^{-1} again would send each vector to each of the standard basis encoded in the original space given homogeneity of the \lambda; leaving the vector of \lambda_{j} scaled by the identity: creating a diagonal D matrix. \blacksquare