The conjugate of a scalar is the same scalar in a matrix-scalar multiplication?

Question

I am reading the book, Applied Linear Algebra and Matrix Analysis.
When I was doing the exercise of Section 2.4, Page 97, I thought the equation is right as followed:

$(cA)^{*} = cA^{*}$

But answer tells me it is false.
So I am wondering it and searching the definition of a scalar in Wikipedia.
And I find it.

In linear algebra, real numbers or other elements of a field are called scalars and relate to vectors in a vector space through the operation of scalar multiplication, in which a vector can be multiplied by a number to produce another vector...

SO I wonder that in this equation $c$ can be a complex number, so the equation is not right.

If not mind, anyone could help me and give some tutorials of it?
I will appreciate it.

Answer 1

[Just sharing my thoughts]

Whenever we define a matrix we always mention the corresponding field, from where the elements are taken, otherwise, the term matrix is not well defined.

In a usual sense, when we define a matrix we consider $\Bbb R$ as a field, but it is a good habit to mention field every time a matrix is considered.

Your equation $(cA)^*=cA^*$ is not true for $c\in\Bbb C$ but it is true for $c\in\Bbb R;$ when $c\in \Bbb C$ we have $(cA)^*=\overline c A^*$ where $\bar c$ is the conjugate of $c$

Answer 2

The term "scalar" is used to mean some element of a field, usually clear from context. Here, the field is clearly $\mathbb{C}$, and hence $c$ must not be real, so the statement is false since $c$ can be complex. For example, $c=i$ and $A=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ provides a counter-example (verify that this is indeed a counter-example). If $c$ is real, the statement is true.

It is true, however, that $(cA)^*=\bar{c} A^*$, where $\bar{c}$ is the complex conjugate of $c$ (that is, if $c=a+bi$, then $\bar{c}=a-bi$).