Proving (My) = (M*x).y | Complex Vectors and Matrices

[Nick89]

Homework Statement

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)

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.

 

[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.

 

[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 :)

 

[Dick]

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