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. \(\blacksquare\)