I have two column vectors $X$ and $Y$. Now the Equation is
$$ \frac{1}{2}(X Y^T)^2 - X^TXY^TY $$
Where $X^T$ is the transpose of $X$.
I need to solve the equation basically get something like $-\frac1{2}(X^TY)^2$
Edited Post
The full equation is like this.
$ \iota(A,\tau) = \frac{1}{2} (A-A_{t})^{2} + \tau(1-X^{T}.A.Y) $
Where $\tau $ is the Langrangian Multiplier. We take the derivate w.r.t A and set the langrangian to 0. This yields
$ A=A_{t} + \tau(X.Y^{T}) $
To find the value of $\tau$ we put value of A into the First Equation.
$ \iota(A,\tau) = \frac{1}{2}(\tau X.Y^{T})^2 + \tau -\tau(X^{T}.A_{t}.Y)-\tau^2(X^{T}.X.Y^{T}.Y)$
Now i need to simplify this so that i can diffrentiate it wrt to $\tau$ and finally to put the value of $\tau$ in the second equation. A is a square matrix
The given formula is an expression rather than an equation and the author wants to give it a different form. I assume that X(Y^T) is meant to be the inner product of X and Y. Let a be the angle between the two. Then the expression becomes (1/2)(|X|^2)(|Y|^2){cos(a)}^2- (|X|^2)(|Y|^2) and it can be reformulated using trigonometric identities. But, in view of the factor 1/2, the outcome will not be the form the author suggests. One possible form is -(1/2){(|X|^2)(|Y|^2)+(XxY)^2} where x indicates the vector product.