Assume $x\in\mathbb{R}^n$. We denote $2$-norm of a vector as $||x||_2$ meaning: $$||x||_2=\left(\sum_{i=1}^n{|x_i|^2}\right)^{\frac{1}{2}}.$$ Also assume $A\in \mathbb{R}^{n\times n}$. Denote $2$-norm of a matrix as $||A||_2$ meaning: $$||A||_2=\underset{||x||_2=1}{\,\,\text{ max }\,\,} ||Ax||_2=\sqrt{\rho(A^TA)}=\sqrt{\rho(AA^T)}=\sigma_1(A),$$ Which $\rho(A^TA)$ is the maximum eigenvalue of $A^TA$ and its square root is the maximum singular value of $A$ or $A^T$.
Now for $x,y\in \mathbb{R}^n$ show $$||yx^T||_2=||y||_2||x||_2.$$
My try: $$ yx^T= \begin{bmatrix} y_1\\y_2\\\vdots\\y_n \end{bmatrix} \begin{bmatrix} x_1 & x_2 & \dots & x_n \end{bmatrix} =\begin{bmatrix} y_1x_1 & y_1x_2 & \dots & y_1x_n\\ y_2x_1 & y_2x_2 & \dots & y_2x_n\\ \vdots & \vdots & \ddots & \vdots\\ y_nx_1 & y_nx_2 & \dots & y_nx_n\\ \end{bmatrix}.$$ So we should find $2$-norm of this matrix. Also: $$||y||_2||x||_2=\sqrt{(y_1^2+y_2^2+\dots y_n^2)(x_1^2+x_2^2+\dots x_n^2)}.$$ But I couldn't find out how to connect these. I tried to calculate $||yx^T||_2$ by multiplying it to a vector with $2$-norm equal to $1$ and finding the maximum $2$-norm, but it wasn't helpful.
Any help or hint is immensely valuable to me!