Let $A$ be a real matrix. If $A \hat x = r \hat y$, does $A^T \hat y = r \hat x$?

I believe it does, but am struggling to prove it.

Geometrically, I believe this is true, because $r$ captures the change of magnitude of $x$ under $A$, which should be shared by $A^T$, and $\hat y$ captures the change of direction of $x$ under $A$, which should be inverted under $A^T$. (Thinking of the polar decomposition of $A = UP$, the transpose of $A$ is often described as inverting the orthogonal matrix $U$ while preserving the diagonal matrix $P$.)

To prove this, I've experimented with the following approaches:

  1. Manipulations of transpose, such as $A^TA\hat x = A^T r \hat y = r A^T \hat y$.
  2. Definition of $A^T_{ij} = A_{ji}$. I didn't see how to use this.
  3. Finding the image of a basis under $A$. I didn't see how to go from $A\hat x$ to a full basis.

Is my conjecture true? If yes: How do I prove it? If not: What is wrong with my geometric argument?

  • $\begingroup$ This fails if $A$ is the zero matrix and $x$ is a non-zero vector. $\endgroup$
    – timon92
    Commented Dec 6, 2023 at 21:12
  • $\begingroup$ @timon92 If $A$ is the zero matrix, then $r = 0$, and so $A^Tv = rw$ for all $v,w$. $\endgroup$ Commented Dec 6, 2023 at 21:18
  • $\begingroup$ Not really. If $A$ is the zero matrix and $y$ is the zero vector, then $r$ can be arbitrary. $\endgroup$
    – timon92
    Commented Dec 6, 2023 at 21:25
  • $\begingroup$ I'll add another counterexample to the answer below. Suppose $A$ is an invertible square matrix that is not symmetric. Consider any of its eigenvectors $v$, where the relationship $Av = \lambda v$ holds. Because $A$ is not symmetric, $A\neq A^T \implies A^Tv \neq \lambda v$ $\endgroup$
    – Sam
    Commented Dec 6, 2023 at 21:28

1 Answer 1


The conjecture is false: here $A = \begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}$, $\hat x = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \hat y$

$$\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = 3\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$ $$\begin{pmatrix} 3 & 2 \\ 0 & 1 \end{pmatrix}^T \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} 3 & 0 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix} \neq 3 \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

The flaw in your geometric argument is that the "change in direction of $x$ under $A$ is not simply reversed by transposing $A$, and the "change in magnitude" is not simply preserved either. It's not entirely clear what you mean by this precisely, so I can't be more specific as to how it's flawed.

Rather than the transpose, you would want some operation that's more akin to the inverse of a matrix, but one that generalizes to non-square and non-invertible matrices.

  • $\begingroup$ Thanks. I added context to the OP. I think my error is in thinking that diagonal matrix $P$ can't change a vectors direction. $P$ has a basis of real eigenvectors, but since they have different eigenvalues, $P$ can change direction of a linear combination of these eigenvectors. $\endgroup$ Commented Dec 6, 2023 at 21:33

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