Exercício 5 Para cada transformação linear T a seguir, determine \operatorname{Ker}(T), \operatorname{Im}(T) e as dimensões de \operatorname{Ker}(T) e \operatorname{Im}(T).
T: M_{2 \times 2}(\mathbb{R}) \rightarrow M_{2 \times 2}(\mathbb{R}), T(X)=M X-X M, onde \left[\begin{array}{ll}1 & 2 \\\\ 0 & 1\end{array}\right].

Passo 1

O enunciado define a transformação linear
T\colon M_{2\times2}(\mathbb{R})\to M_{2\times2}(\mathbb{R}),\quad T(X)=MX - XM,
onde
M=\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}.
Precisamos achar \operatorname{Ker}(T), \operatorname{Im}(T) e as dimensões desses subespaços.

Passo 2

Escrevemos uma matriz genérica
X=\begin{pmatrix}a & b\c & d\end{pmatrix}
e calculamos
MX = \begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}a & b\c & d\end{pmatrix} =\begin{pmatrix}a+2c & b+2d\c & d\end{pmatrix},
XM = \begin{pmatrix}a & b\c & d\end{pmatrix}\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix} =\begin{pmatrix}a & 2a+b\c & 2c+d\end{pmatrix}.
Logo
T(X)=MX - XM =\begin{pmatrix}2c & 2(d-a)\\\ 0 & -2c\end{pmatrix}.

Passo 3

Para \operatorname{Ker}(T) impomos T(X)=0, isto é
2c=0,\quad 2(d-a)=0,\quad -2c=0
que dão c=0 e d=a. Assim
\operatorname{Ker}(T) =\Bigl\{\begin{pmatrix}a & b\\\ 0 & a\end{pmatrix}:a,b\in\mathbb{R}\Bigr\},
portanto
\dim\bigl(\operatorname{Ker}(T)\bigr)=2.

Passo 4

Um elemento genérico de \operatorname{Im}(T) é
T(X)=\begin{pmatrix}2c & 2(d-a)\\\ 0 & -2c\end{pmatrix}.
Se definirmos u=2c e v=2(d-a), obtemos
\operatorname{Im}(T) =\Bigl\{\begin{pmatrix}u & v\\\ 0 & -u\end{pmatrix}:u,v\in\mathbb{R}\Bigr\},
logo
\dim\bigl(\operatorname{Im}(T)\bigr)=2.

Resposta

\operatorname{Ker}(T) = \Bigl\{ \begin{pmatrix} a & b \\\\ 0 & a \end{pmatrix} \colon a,b \in \mathbb{R} \Bigr\}, \quad \dim(\operatorname{Ker}(T)) = 2.
\operatorname{Im}(T) = \Bigl\{ \begin{pmatrix} u & v \\\\ 0 & -u \end{pmatrix} \colon u,v \in \mathbb{R} \Bigr\}, \quad \dim(\operatorname{Im}(T)) = 2.

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