What is the kernel of a number?

Question

tr: R^2×2 → R

trace \begin{bmatrix}a&b\\c&d\end{bmatrix}

Is a linear transformation.

What is ker(tr)?

I am confused with this question, how can I find the trace of a a real number. All the videos and material I see is finding the trace of a matrix.

Edit: As people have pointed out I meant to say find the kernel of a real number. Which is also inaccurate because I am finding the kernel of the transformation which is a matrix.

Answer 1

Let $A$ and $B$ be groups and let $f$ be a homomorphism. The kernel of $f : A \longrightarrow B$ is the set

$$\ker f \enspace := \enspace \{ \, a \in A \, | \, f(a) = e_B \in B \, \} \quad ,$$

where $e_B$ is the neutral element of $B$.

${}$

Example 1: If $A, B$ are vector spaces, then the neutral element is the zero vector. So $\ker f$ is the set of all vectors in $A$ which map to the zero vector $0_B$ in $B$.

Example 2: Let $A = B = \mathbb{R}^{2\times 2}$ be the set of $(2 \times 2)$-matrices and let the group operation be the matrix-multiplication. Then the neutral element of $B$ is the identity matrix and so the kernel is the set of all matrices $a \in A$ for which $f(a) = 1_{2\times2}$.

Here: In your specific case $A = \mathbb{R}^{2\times2}$ is the space of real $(2\times 2)$-matrices and $f = \operatorname{tr} : \mathbb{R}^{2\times2} \longrightarrow \mathbb{R}$, and so $B = \mathbb{R}$. The kernel of $\operatorname{tr}$ is therefore the set of all matrices which get mapped to the neutral element of $\mathbb{R}$, which is $0$.

So given a matrix of the form

$$a \enspace = \enspace \left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right]$$

the trace is

$$\operatorname{tr}(a) \enspace = \enspace a_{11} + a_{22} \quad .$$

In order for the trace to vanish, you must have $a_{11} = -a_{22}$, and so the kernel is the set

$$\ker \operatorname{tr} \enspace = \enspace \left\{ \; \left[ \begin{matrix} a_{11} & a_{12} \\ a_{21} & -a_{11} \end{matrix} \right] \; \Bigg| \; a_{11}, a_{12}, a_{21} \in \mathbb{R} \; \right\} \quad .$$

As you can see, one can choose three components of the matrices in the kernel freely and so the dimension of the kernel is 3. This can also be seen by the rank-theorem

$$\underbrace{\dim A}_{= \, 4} \enspace = \enspace \underbrace{\dim \operatorname{im} \operatorname{tr}}_{= 1} + \dim \ker \operatorname{tr} \quad \Longrightarrow \quad \dim \ker \operatorname{tr} \enspace = \enspace 3 \quad .$$