1
$\begingroup$

Suppose $V$ is the space of all $n×n$ matrices with real elements. Define $T:V→V$ by $T(A)=AB−BA, \ A∈V,$ where $B∈V$ is a fixed matrix. Let $f$ be the trace function, which is $T^tf$?

We know that $f(T(A)) = f(AB) - f(BA) = 0$. It's kind of obvious that $T^tf = 0$. How to formalize this idea?

  • 2
    You're using some strange notation here. Is $T^t$ supposed to be $T^\dagger$, which is to say the adjoint? Is the "dash function" supposed to be the [trace](https://en.wikipedia.org/wiki/Trace_(linear_algebra))? – 2017-01-16
  • 0
    Yes, in fact it is the trace function. And $T^t$ is the transposed of a linear transformation, is something wrong in that too? – 2017-01-17
  • 0
    The transpose makes sense if your linear transformation is a **matrix**. That is, the transpose operation is defined for linear transformations from $\Bbb R^n$ to $\Bbb R^m$. For any other kind of linear transformation, you have to explain what exactly you mean by the transpose. – 2017-01-17
  • 0
    $T^t$ is a linear transformation from $W* \rightarrow V*$ , where $T:V \rightarrow W$. – 2017-01-17
  • 0
    Right. Usually, that's called the **adjoint** as opposed to the transpose. The transpose refers specifically to something that you do to matrices, usually to the matrix of a linear transformation from $\Bbb R^n$ to $\Bbb R^m$. – 2017-01-17
  • 0
    All right, the book I'm following is written as transposed. But I did! – 2017-01-17
  • 0
    Interesting. As you can see from the [wiki page](https://en.wikipedia.org/wiki/Transpose), the usual definition of a transpose (at least the only definition I had ever seen) is much more narrow. What book are you using, by the way? – 2017-01-17
  • 0
    Let $U$ and $V$ be vector spaces over the field $\mathbb K$ and let $T: U \rightarrow V$ be a linear transformation. Note that if $f \in V^{\ast}$, then the compound $f \circ T: U \rightarrow \mathbb K$ is an element of $U^{\ast}$ (we recall that the composite of linear transformations is linear). Therefore, we can define the following function \begin{align} T^{t}: & V^{\ast} \rightarrow U^{\ast} \\ & f ↦ T^{t}(f) : U \rightarrow \mathbb K \\ && u ↦ T^{t}(f)(u) = (f \circ T)(u). \end{align} Where $T^t$ is linear. – 2017-01-17
  • 0
    Definition. The linear transformation $T^t$ defined above is called the transpose of $T$. This is the complete definition. I have put this definition because in the book it also has the definition of "adjunct operator ($T^{*}$)". I do not know if it's the same thing we're talking about. The book is a Brazilian by author Flávio Ulhoa Coelho, however most Brazilian books approach so, so I want to know if it's the same thing we're talking about. – 2017-01-17
  • 0
    Yes, we were talking about the same thing. – 2017-01-17

1 Answers 1

0

You've done everything that you're supposed to do. If I wanted to write a formal answer, I'd write

For every $A \in V$, we have $$ T^*f(A) = f(T(A)) = f(AB - BA) = f(AB) - f(BA) = f(AB) - f(AB) = 0 $$ Thus, $T^*f = 0$.

By definition, saying that a linear transformation is $0$ is the same as saying that it's the map $x \mapsto 0$ (for all $x$).

I would say that the equality $f(AB) = f(BA)$ is not inherently obvious and may merit an explicit mention/justification.