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

  • 1
    $\begingroup$ Well dimensions in your question does not match. $(X Y^T)^2 $ is matrix and $ X^TXY^TY = \|X\|^2 \|Y\|^2$ is scalar. So you cant add those together. $\endgroup$ – tom Sep 19 '13 at 10:36
  • $\begingroup$ I concur with tom's comment. And where is your equation? I only see an expression. An equation, by definition, equates one thing to another. Do you mean setting the displayed expression to zero? $\endgroup$ – user1551 Sep 19 '13 at 10:49

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.

  • $\begingroup$ Welcome to Math.SE. Please consider formatting your answers in LateX in future. You may learn by seeing some other answers. $\endgroup$ – Vishal Gupta Sep 19 '13 at 11:23
  • $\begingroup$ Thank you for the quick response. I have added the full equation so the problem becomes clearer $\endgroup$ – Ali Shahzad Sep 19 '13 at 12:40

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