# How can we determine the numbers $k_1$ and $k_2$ ?

Suppose that the matrix $A=\begin{pmatrix}1+ci & w_1 \\ 2+i & z_2\end{pmatrix}$ with $c\in \mathbb{R}$ is hermitian, of order $1$ and the vector $(k_1, k_2)\in \mathbb{C}^2$ with $k_2$ positive real number, belongs to the orthogonal complement of the row space of $A$ and has norm $\sqrt{3}$.

How can we determine the numbers $k_1$ and $k_2$ ?



We know that $A$ is hermitian, so it is equal to its own conjugate transpose. So we have $A=\overline{A^T}$, i.e. $$\begin{pmatrix}1+ci & w_1 \\ 2+i & z_2\end{pmatrix}=\begin{pmatrix}1-ci & 2-i \\ \overline{w_1} & \overline{z_2}\end{pmatrix}$$ From that we get that $c=0$, $w_1=2-i$ and $z_2=\overline{z_2}$ and so $z_2$ is real.

Are the information correct so far?

Then we have that $A$ has the order 1, that means that $A^1=I_2$, or not?

Now we have to calculate the orthogonal complement of the row space of $A$, or not? How can we do that?

The last information is that the vector $(k_1, k_2)$ has the norm $\sqrt{3}$ and so we get $\sqrt{k_1^2+k_2^2}=\sqrt{3}\Rightarrow k_1^2+k_2^2=3$, right?

• I agree with the form of $A$ you have so far. You may try to prove it yourself, but the orthogonal complement of the row space of $A$ is the kernel of $A$. I don't know what does it mean that the order of the matrix is 1. – Javi Sep 5 '18 at 17:43
• so it looks like (0,0) is a vector in the kernel of $A$, but could it be the only vector? (no) why not? what is the value of $z_2$ that makes other vectors exist in the kernel? and finally, which vector or vectors in the kernel satisfy the requirements given for $(k_1,k_2)$? – Javi Sep 5 '18 at 18:18
• well that is what I've just asked you... read the statement of the problem, it actually says (with other words) that there is a non-zero vector in the kernel, which one is it? – Javi Sep 5 '18 at 18:29
• because $k_2$ is positive, yes, or also because the norm of $(k_1, k_2)$ is not zero. I agree with the expression you got for the kernel, but not with your selection of $(k_1, k_2)$. You indeed selected a vector of the kernel, but not one with norm $\sqrt{3}$ – Javi Sep 5 '18 at 18:38
• you are mistaking the calculation of the norm with complex components. It should be $\sqrt{(|-(2-i)| x_2)^2 + x_2^2}$ – Javi Sep 5 '18 at 18:49

You may make use of the fact that the orthogonal complement of the row space of $A$ is the kernel of $A$ (that is easy to prove.)
Then, $z_2$ may be determined, keeping in mind that there is at least 1 non-zero vector in the kernel of $A$ (which vector?). Then find what the kernel is.
Finally, you may find one (or more maybe?) vectors $(k_1, k_2)$ in the kernel of $A$ which satisfies the requirements given for $k_2$ and $\|(k_1, k_2)\|$.