Homework Help: Proof x.(My) = (M*x).y

1. Jun 24, 2008

Nick89

1. The problem statement, all variables and given/known data
This was a question in a recent exam and I would like to know if the answer I gave is correct since I not 100% sure...

If $\mathbf{x}$ and $\mathbf{y}$ are complex vectors in C^n (complex) and $M$ is a square (n x n)-matrix (also in C^n), prove that:
$$\mathbf{x} \cdot (M \mathbf{y}) = (M^* \mathbf{x}) \cdot \mathbf{y}$$
(where M* denotes the complex conjugate + the transpose of M: $M^* = \overline{M}^T$
(The dot denotes the standard complex dot-product)

3. The attempt at a solution
I did the following:

$$\mathbf{x} \cdot \mathbf{y} = \mathbf{x}^T \mathbf{\overline{y}}$$
So
$$\mathbf{x} \cdot (M \mathbf{y}) = \mathbf{x}^T \overline{M \mathbf{y}} = \mathbf{x}^T \overline{M} \overline{\mathbf{y}}$$

Now:
$$\mathbf{x}^T \overline{M} = (M^* \mathbf{x})^T$$ because $$(M^* \mathbf{x})^T = \mathbf{x}^T (M^*)^T = \mathbf{x}^T \overline{M}$$

So, now we have:
$$\mathbf{x} \cdot (M \mathbf{y}) = (M^* \mathbf{x})^T \overline{\mathbf{y}} = (M^* \mathbf{x}) \cdot \mathbf{y}$$

Is this solution correct? Or did I make an error somewhere? (I'm not entirely sure of the very first statement for example...)

Thanks.

2. Jun 24, 2008

Dick

Yes, that's basically correct. But if you aren't sure of the first statement, look up how you defined the inner product. You could also have written your definition x.y=(y*)x. An equally valid definition of inner product is x.y=(x*)y. The first inner product is antilinear in y and the second one is antilinear in x. The choice of one or the other is a matter of convention.

3. Jun 24, 2008

Nick89

Well that was really the problem. The book we are using first defines inner products on the real numbers, it defines the standard dot product as:
$$\vec{x} \cdot \vec{y} = \sum_{i=1}^n x_i y_i$$
and it is then shown that x.y = x^T y.

Then it goes on later about the complex numbers. It does define the standard dot product as:
$$\vec{x} \cdot \vec{y} = \sum_{i=1}^n x_i \overline{y_i}$$
but it does NOT explicitly show that the dotproduct can also be written as $$\mathbf{x}^T \overline{\mathbf{y}}$$.
I can easily verify that this works but I wasn't sure whether it is true for every condition...

So it is correct? Yay :)

4. Jun 24, 2008

Dick

That's the same as (y*)x, yes. Which is the same as your expression.