Prove $\langle x_1 y_1^T, x_2 y_2^T \rangle = (x_1^T x_2)(y_1^T y_2)$

Question

I am trying to prove $$\langle x_1 y_1^T, x_2 y_2^T \rangle = (x_1^T x_2)(y_1^T y_2)$$ given $x_1, x_2, y_1, y_2 \in \mathbb{R}^n$

I tried to apply $$\langle x, y \rangle= y*x = y_1 x_1 + y_2 x_2 + \cdots + y_n x_n$$ but I don't know how to use trace of $y_1$ and $y_2$.

I appreciate any help!

Answer 1

We need to assume a convention, let's say that $x_1, x_2, y_1, y_2$ are all column vectors ($n$ rows by $1$ column). Then $<x,y>=x^T y$ is an inner product.

$x_1 y_1^T$ and $x_2 y_2^T$ make sense, these are $n$ by $n$ matrices!

but

$<x_1 y_1^T, x_2 y_2^T > = y_1x_1^Tx_2y_2^T=y_1 (x_1^T x_2) y_2^T = (x_1^Tx_2)y_1y_2^T$, an $n$ by $n$ matrix!

We have $$ \boldsymbol{y}_1 \boldsymbol{y}_2^T = \pmatrix{y^1_1 \\y^1_2\\ \vdots \\y^1_n} \pmatrix{y^2_1 & y^2_2 & \cdots & y^2_n} = \pmatrix{y^1_1 y^2_1 & y^1_1y^2_2 &\cdots& y^1_2 y^2_n \\y^1_2 y^2_1 & y^1_2y^2_2 &\cdots& y^1_1 y^2_n \\ \vdots & \vdots & \ddots \\ y^1_n y^2_1 & y^1_n y^2_2 &\cdots & y^1_n y^2_n}$$

So that $$\text{trace} <\boldsymbol{x}_1 \boldsymbol{y}_1^T , \boldsymbol{x}_2 \boldsymbol{y}_2^T> = (\boldsymbol{x}_1^T \boldsymbol{x}_2 )(\boldsymbol{y}_1^T \boldsymbol{y}_2).$$