Matrices with complex entries

[compliant]

Homework Statement

Let B be an m×n matrix with complex entries. Then by B* we denote the n×m matrix that is obtained by forming the transpose of B followed by taking the complex conjugate of each entry. For an n × n matrix A with complex entries, prove that if u*Au = 0 for all n × 1 column vectors u with complex entries, then A is a zero matrix.

The Attempt at a Solution

Let $$u = \left[ z_{1}, z_{2}, z_{3},..., z_{n} \right]^{T}$$

and $$u^{*} = \left[ \overline{z_{1}}, \overline{z_{2}}, \overline{z_{3}},...,\overline{z_{n}} \right]$$

Then
$$u^{*} A = \left[ {{\sum{\overline{z_{i}}a_{i1}}}},{{\sum{\overline{z_{i}}a_{i2}}}},...,{{\sum{\overline{z_{i}} a_{in}}}} \right]$$

And that's as far as I got, because I have no idea how to make $$a_{ij} = 0$$

 

[mighty2000]

for all i and j.

Thank you for bringing up this interesting problem. I am always eager to tackle new challenges and explore new ideas.

After looking at your attempt at a solution, I believe I can help you take it a step further. Let's start by considering the given condition: u*Au = 0 for all n × 1 column vectors u with complex entries. This means that the inner product of any vector u with the matrix A is equal to 0. In other words, u and Au are orthogonal for all possible vectors u.

Now, let's take a closer look at the matrix A. We know that u and Au are orthogonal, meaning that their inner product is equal to 0. But this also means that the dot product of u and each column of A is equal to 0. In other words, each column of A is orthogonal to u.

Now, let's consider the first column of A. Since it is orthogonal to u, it must be a zero vector (i.e. all its entries are equal to 0). Similarly, we can apply this logic to all the other columns of A. This means that all the columns of A are zero vectors, and hence A is a zero matrix.

I hope this helps you understand the problem better. Good luck with your studies!