# "Simple" proof that an n x n matrix is similar to its transpose

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*}

• You can't transpose block wise. Imagine you have 4 blocks, then all the blocks get transposed, but the 1, 2 and 2, 1 block switch places. Try your inductive step on a $2\times 2$ matrix to see the difficulty in defining $P$. Nov 25, 2023 at 23:52
• "We can break up $A_{n \times n}$ into $m$ blocks of size $i_m \times i_m$ where $i_m < n$." This is not specific, and as you think about what precisely this would mean, I think you'll find that it's actually not true. Nov 26, 2023 at 2:54

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.
• The (1, 2) similarity matrix could be zero, because $(0)^t (0) (0) = (0)$, but it could also be $(1)$, because $(1)^t (0) (1) = (0)$ as well. I chose to make all my similarity matrices be the same because it was easier to write (and to show that the general solution could not be derived from the block-solutions). Nov 26, 2023 at 14:41
• Thank you. I don't take the question to mean that I have to exhibit an actual similarity matrix $P$. I just have to show one exists, right? I think the key to whether this works is in how the blocks are developed, as Greg Martin indicates above. For example, there may be a notion of "maximal" blocks, in which case the 2 x 2 case may be one block. Or two columns. Or two rows. Etc. I feel like there is a way to make this work, or, at least, that I have to play with it more to convince myself it doesn't. Nov 26, 2023 at 16:38
• You last claim is that some matrix $P$ which is (somehow) assembled from the $P_i$s will suffice to transpose $A$. In my example case, the only $2\times2$ matrix you can build is $P = \pmatrix{1 & 1 \\ 1 & 1}$, which does not in fact transpose my matrix $A$. So at least the last paragraph of your answer is wrong. It's true that you only need to show the existence of a matrix $P$ rather than exhibiting it...but I'll be honest: I don't think you're going to be able to show the existence, either (and not because of you: I suspect that in general no such $P$ exists). Nov 26, 2023 at 21:29