Transpose matrices and vectors - rules

Question

Given vectors $x, d \in \mathbb{R}^{1\times N}$ and matrix $A \in \mathbb{R}^{N\times N}$, why does this hold truth?

$$\frac{1}{2}x^TAd + \frac{1}{2}d^TAx = x^TAd$$

Answer 1

It isn't true if $A$ is not symmetric: $$ \frac12\pmatrix{1&0}\pmatrix{0&1\\ 0&0}\pmatrix{0\\ 1} +\frac12\pmatrix{0&1}\pmatrix{0&1\\ 0&0}\pmatrix{1\\ 0} \ne\pmatrix{1&0}\pmatrix{0&1\\ 0&0}\pmatrix{0\\ 1}. $$ The statement is true when $A$ is symmetric, because every $1\times1$ matrix is equal to its transpose.

Answer 2

I take $\Bbb R^{1 \times N}$ to be the space of $N$-rowed column vectors; otherwise expressions such as $Ad$ etc. make no sense.

I further assume that, in light of the comments to the question itself, that $A$ is a symmatric matrix, viz.

$A = A^T. \tag 0$

We note that both

$d^TAx, x^TAd \in \Bbb R; \tag 1$

that is, $d^TAx$ and $x^TAd$ are $1 \times 1$ real matrices. As such, we have

$(d^TAx)^T = d^TAx; \tag 2$

by virtue of (0),

$(d^TAx)^T = x^TA^T(d^T)^T = x^TAd; \tag 3$

combining (2) and (3) we thus find

$x^TAd = (d^TAx)^T = d^TAx, \tag 4$

whence

$\dfrac{1}{2}d^TAx = \dfrac{1}{2}x^TAd; \tag 5$

if we now add $\dfrac{1}{2}x^TAd$ to each side we obtain

$\dfrac{1}{2}x^TAd + \dfrac{1}{2}d^TAx = x^TAd, \tag 6$

$OE\Delta.$

Answer 3

Here is a long winded solution with the actual computation and a small finessing of sums if you are like me and don't realize this is just a $1 \times 1$ matrix and its transpose.

Let $x$ and $d$ be column vectors with $N$ rows and $A$ be an $N \times N$ symmetric matrix, and let $x_i, d_i,$ and $a_{i,j}$ denote the elements of $x, d,$ and $A$ respectively. Then the product

\begin{align} x^TAd &= \begin{pmatrix} x_1 & x_2 & \cdots \end{pmatrix} \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots\\ a_{2,1} & \ddots & \\ \vdots & & \\ \end{pmatrix} \begin{pmatrix} d_1\\ d_2\\ \vdots \end{pmatrix}\\ &=\begin{pmatrix} \sum_{\ell=1}^{n} x_{\ell} a_{\ell,1} & \sum_{\ell=1}^{n} x_{\ell} a_{\ell,2} & \cdots \end{pmatrix} \begin{pmatrix} d_1\\ d_2\\ \vdots \end{pmatrix}\\ &=\sum_{k=1}^{n}\sum_{\ell=1}^{n} d_{k} x_{\ell} a_{\ell,k}, \tag{1} \end{align}

and similarly

\begin{align} d^TAx = \sum_{k=1}^{n}\sum_{\ell=1}^{n} x_{k} d_{\ell} a_{\ell,k}. \tag{2} \end{align}

If $A$ is symmetric then $a_{i,j}=a_{j,i}$ for all $a_{i,j} \in A$. Using this fact and switching the sums around in $(2)$ yields \begin{align} d^TAx &= \sum_{\ell=1}^{n} \sum_{k=1}^{n} x_{\ell} d_{k} a_{k,\ell}\\ &= \sum_{\ell=1}^{n} \sum_{k=1}^{n} x_{\ell} d_{k} a_{\ell,k}. \end{align}

So, in this case, $$x^TAd = d^TAx.$$