# Problem with dot product and outer product of vectors.

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

• 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. – tom Sep 19 '13 at 10:36
• 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? – user1551 Sep 19 '13 at 10:49