<s>[INST] Show that the pseudoinverse of $T$ is $1/4T$. If  $T$ is a selfadjoint matrix then, by wiki, the pseudo inverse of $T$ is given by
$$T^{+}= \lim_{t \to 0}(T^2+tI)^{-1} T. \tag{1} $$

  
*
  
*I don't understand under which norms we have
  $$\lim_{t \to 0}\|T^{+}-(T^2+tI)^{-1} T \|=0?$$
  
*Also if $T=\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$. I want to use the formula $(1)$ in order to show that 
  $$T^+=\frac{1}{4}T.$$

If I don't made mistakes, I find
$$T^2+tI=\left(\begin{array}{cc}2+t&2\\2&2+t\end{array}\right)$$
and so
$$(T^2+tI)^{-1}=\frac{1}{(2+t)^2-t}\left(\begin{array}{cc}2+t&-2\\-2&2+t\end{array}\right).$$
Also
$$(T^2+tI)^{-1} T=\frac{1}{(2+t)^2-t}\left(\begin{array}{cc}t&t\\t&t\end{array}\right).$$
 [/INST] The norm is irrelevant, as all norms on $M_2(\mathbb{C})$ are equivalent, being a finite dimensional vector space.
The determinant of $T^2+tI$ is $(2+t)^2\color{red}{-4}=t^2+4t$. So you need to compute the limit of
$$
\frac{1}{t^2+4t}\begin{pmatrix}t & t \\ t & t \end{pmatrix}=
\frac{1}{t+4}\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}
$$
With a different technique: the LU decomposition of $T$ is
$$
T=
\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}
$$
so a full rank decomposition $T=AB$ is
$$
T=
\begin{pmatrix} 1 \\ 1 \end{pmatrix}
\begin{pmatrix} 1 & 1 \end{pmatrix}
$$
Then
$$
A^+=A(A^HA)^{-1}=\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}
\qquad
B^+=(BB^H)^{-1}B=\begin{pmatrix} 1/2 & 1/2 \end{pmatrix}
$$
and
$$
T^+=B^+A^+=
\begin{pmatrix} 1/2 & 1/2 \end{pmatrix}
\begin{pmatrix} 1/2 \\ 1/2 \end{pmatrix}
=
\begin{pmatrix} 1/4 & 1/4 \\ 1/4 & 1/4 \end{pmatrix}
$$
</s>