I am from physics with humble mathematical background so apologies if you find this trivial. Given a vector $v=\begin{pmatrix} a \\ \alpha \\ b \\ \beta \end{pmatrix}$that lives in $\mathbb{R}^{4\times 1}$, I define a projector $P$ with $P_{11}=P_{33}=1$ and all other elements zero, such that $u:=Pv = \begin{pmatrix} a \\ 0\\ b \\ 0 \end{pmatrix}$. My question is whether the following replacement is true:
$$\begin{pmatrix} a \\ 0\\ b \\ 0 \end{pmatrix} \overset{?}{\equiv} \begin{pmatrix} a \\ b \end{pmatrix}=:w$$
Since $u$ and $w$ belong to different spaces, to me it seems that this equivalence is not true. I "invented" the word truncated for the vector $w$, only to help me frame the question title.
Also, can a similar thing be said about the matrices, e.g.
$$ P \begin{pmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\\ a_{41} & a_{42} & a_{43} & a_{44} \end{pmatrix}P = \begin{pmatrix} a_{11} & 0 & a_{13} & 0 \\ 0 & 0 & 0 & 0\\ a_{31} & 0 & a_{33} & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \overset{?}{\equiv} \begin{pmatrix} a_{11} & a_{13} \\ a_{31} & a_{33} \end{pmatrix}$$
EDIT: Thanks to the comment by @ronno and answer by @Nils. What I mean by the "equivalence" is the that expectation values should be the same in the two representations.
It depends on your definition of equality. If you say that $u=w$ when both of the object are not from the same set, it means that you consider equality up to an isomorphism. It is often the case in physics. However be aware that the application that you wrote is not an isomorphism from $ \mathbb{M}_{4\times 4} \to \mathbb{M}_{2\times 2} $ for any four by four matrix but only from the susbet $$S \subset \mathbb{M}_{4\times 4} = \{M \in \mathbb{M}_{4\times 4}|M = \begin{pmatrix}a_{11}&0&a_{13}& 0\\ 0&0&0&0\\a_{31}&0&a_{33}&0\\0&0&0&0 \end{pmatrix} \}$$ It this case there is indeed a linear application $f:M\in S \to \mathbb{M}_{2\times 2}$ such that $$f(M) = \begin{pmatrix}a_{11}&a_{13}\\a_{31}&a_{33}\end{pmatrix}$$ which is a bijection. Furthermore, you can write the matrix $P\in \mathbb{M}_{4\times 4}$ such as $$P = \begin{pmatrix}1&0&0&0\\ 0&0&0&0\\ 0&0&1&0\\0&0&0&0 \end{pmatrix}$$ and by applying $f$ you get $$ f(P) = \begin{pmatrix}1&0 \\0&1 \end{pmatrix}$$ which is the identity matrix of $\mathbb{M}_{2\times 2}$ which was expected.
Edit: The expectations value of a physical observable are the set of eigenvalues of our operator in a given basis (the diagonality is assured by the hermiticity of the operator). In these case, if I understand the question, you are asking for the expectation value of $M$ with respect to the vector $u$. Basically, we're looking for $$ u^T M u = a^2 a_{11}+aba_{31}+aba_{13}+b^2a_{33}= \left< u \right| M\left|u\right>_{\mathbb{M}_{4\times4}} $$ (I'm keeping the notation you introduced) Doing the same calculations with their $\mathbb{M}_{2\times2}$ representation, you came up with the same expecation value with $u_{\mathbb{R}^2} = \begin{pmatrix}a\\b\end{pmatrix}$ and $M_{\mathbb{M}_{2\times2}} = \begin{pmatrix}a_{11}&a_{13} \\a_{31}&a_{33} \end{pmatrix}$ .$$\left< u \right| M\left|u\right>_{\mathbb{M}_{2\times2}}=a^2 a_{11}+aba_{31}+aba_{13}+b^2a_{33} = \left<u\right|M\left|u\right>_{\mathbb{M}_{4\times4}}$$