<p>No.
Consider block matrices $A = \left(\begin{matrix} 0 &amp; 0 \\ 0 &amp; X \end{matrix}\right)$ (with symmetric $X$) and $B = \left(\begin{matrix} 0 &amp; Y \\ 0 &amp; 0 \end{matrix} \right)$. Computing $AB^T$ from $A$, $B$ and $AB$ means computing $XY^T$ from $X$ and $Y$, since $AB = 0$</p>

CC BY-SA 3.0
2016-01-15T10:37:13.917Z
2016-01-15T10:37:13.917Z
Vladimir Lysikov
<p>Let $A,B\in\mathbb{F}^{n\times n}$ be two $n\times n$ matrices over
the underlying field $\mathbb{F}$. In addition, $A$ is guaranteed
to be a symmetric matrix, i.e, $A=A^{T}$. We assume complexity measure to be the number of field operations over $\mathbb{F}$. Now consider at the following problem:-</p>

<blockquote>
  <p>Given $A,B$ and $AB$ as input, can we compute $AB^{T}$ in $o(n^{\omega})$ time?</p>
</blockquote>

<p>Here $\omega$ is the exponent of matrix multiplication. Note that $AB^{T}=A^{T}B^{T}=(BA)^{T}$.
Thus computing $BA$ in $o(n^{\omega})$ time is also enough. </p>

<ds.algorithms><linear-algebra><matrices><matrix-product>
Matrix multiplication with transpose
Gorav Jindal
# Matrix multiplication with transpose

(Asked by: username_0 on 2016-01-14)

Let $A,B\in\mathbb{F}^{n\times n}$ be two $n\times n$ matrices over
the underlying field $\mathbb{F}$. In addition, $A$ is guaranteed
to be a symmetric matrix, i.e, $A=A^{T}$. We assume complexity measure to be the number of field operations over $\mathbb{F}$. Now consider at the following problem:-

> Given $A,B$ and $AB$ as input, can we compute $AB^{T}$ in $o(n^{\omega})$ time?

Here $\omega$ is the exponent of matrix multiplication. Note that $AB^{T}=A^{T}B^{T}=(BA)^{T}$.
Thus computing $BA$ in $o(n^{\omega})$ time is also enough.

**Tags:** ds.algorithms, linear-algebra, matrices, matrix-product

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## 1 answers

### Answer by username_1 on 2016-01-15. Score: 8 | Accepted answer

No.
Consider block matrices $A = \left(\begin{matrix} 0 & 0 \\ 0 & X \end{matrix}\right)$ (with symmetric $X$) and $B = \left(\begin{matrix} 0 & Y \\ 0 & 0 \end{matrix} \right)$. Computing $AB^T$ from $A$, $B$ and $AB$ means computing $XY^T$ from $X$ and $Y$, since $AB = 0$


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