At the risk of getting voted down for duplicating A matrix is similar to its transpose or Proving an $n\times n$ matrix is similar to its transpose, or such, I didn't feel comfortable "answering" their questions with my own question as I believe that's a faux pas worse than duplication, perhaps?
Anyway, the first question on our department's algebra qualifying exam in January 2015 was this, vis:
Prove or give a counterexample: every $n \times n$ complex matrix $A$ is similar to its transpose, $A^t$.
My question is this: what is wrong with my "simple" proof (below)? I'm skeptical of its viability because of the much more sophisticated solutions which accompany this question elsewhere on this forum. In particular, they all appear to appeal to Jordan Normal Form, which I didn't know about at the time I "solved" this. TIA for comments.
Claim: $A \sim A^t$.
Proof: By induction on $n$.
For the base case, $A = (z)$, a single complex number; $A^t = A = (z)$; hence the claim is true if we take $P$ to be the $1 \times 1$ identity matrix, $(1)$.
The base case is established, therefore assume, for $k < n$ we have that all $k \times k$ complex matrices are similar to their transpose, and consider $k = n$. We can break up $A_{n \times n}$ into $m$ blocks of size $i_m \times i_m$ where $i_m < n$. Let $B_i$ denote the $i$-th block of $A$. We have that $A = (B_i)_{1 \leq i \leq m}$.
By the induction hypothesis, each block is similar to its transpose. Hence, there are $P_i$ for $1 \leq i \leq m$ such that $B_i^t = P_i^{-1} B_i P_i$ for $1 \leq i \leq m$.
It should be clear that $A^t = (B_i^t)$. In other words, we can transpose $A$ block-wise and obtain the same result. [Should I be proving this? I'd use another induction, I guess, on $m$, with fixed $n$.]
Consider the matrix $\mathcal{P} = (P_i)_{1 \leq i \leq m}$. We have that: \begin{align*} \mathcal{P}^{-1} A \mathcal{P} &= ( P_i^{-1} B_i P_i) \\ &= ( B_i^t ) \text{ by induction } \\ &= A^t. \end{align*}
Perhaps you can apply your method to $A = \pmatrix{1 & 1 \\ 0 & 1}$. Since in this case $m$ is necessarily $1$, each $1 \times 1$ matrix is similar to its transpose, with each similarity matrix being $\pmatrix{1}$.
How do you propose to construct a $2 \times 2$ similarity matrix for $A$ from these? More generally, for any $n \times n$ matrix, we can break it into $1 \times 1$ blocks and use the same argument. So you should be able to show how to find the similarity matrix $P$, for any matrix, from a large collection of $1 \times 1$ identity matrices.
I think you see by now what's missing from your proof: when you find the smaller matrices, you also have to show how to assemble them into a larger one...and as the $m = 1$ example shows, that's not going to work out.