Suggest edit — NUS-MATH530 Similar to Diagonal

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title: "NUS-MATH530 Similar to Diagonal"
source: https://www.jemoka.com/posts/kbhnus_math530_similar_to_diagonal/
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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 \(\dim n\) &mdash; 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. \(\blacksquare\)