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There are two vectors:

$\boldsymbol{\hat{y}} = (\hat{y_1}, \hat{y_2}, \dots, \hat{y_n})$

$\boldsymbol{{y}} = ({y_1}, {y_2}, \dots, {y_n})$

The vectors $\boldsymbol{\hat{y}}$ and $\boldsymbol{{y}}$ have some matches, but also some values that do not match. From these vectors I would like to create a matrix, which contains like the identity matrix in the diagonal ones, but only if the values match, otherwise zero.

For example:

$\boldsymbol{\hat{y}} = (1.0, 2.0, 3.0, 4.0)$

$\boldsymbol{{y}} = (1.0, 2.2, 3.0, 4.0)$

From these vectors results:

$$ M= \left(\begin{matrix}1&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&1\end{matrix}\right)$$

Is there a way to describe the definition of $M$ mathematically?

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[The] matrix, which contains like the identity matrix in the diagonal ones, but only if the values match, otherwise zero.

is a very mathematical description. Don't confuse lack of formulas for lack of mathematics. Personally, I would maybe write it a little differently, but that's mostly just to remove any potential for misinterpretation. For instance, saying that the matrix is diagonal, and explicitly stating that the non-zero entries are $1$ is a bit clearer than saying "contains like the identity matrix". At least to people who are used to reading mathematical texts.

Here is my suggestion:

The diagonal $n\times n$ matrix with $1$ in the diagonal entries corresponding to where $y$ and $\hat y$ are equal, and $0$ otherwise.

Alternately, if you really want some formulas, one can use matrix algebra to say

The maximal rank diagonal matrix $A$ with only $0$ and $1$ as entries for which $Ay = A\hat y$

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  • $\begingroup$ Many thanks for the effort! $\endgroup$ – Anne Bierhoff Oct 16 '19 at 6:40
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An alternate way is as follows:

Let $\hat y=(\hat y_1,\hat y_2,...,\hat y_n)$ and $ y=( y_1, y_2,..., y_n)$.

Define $M=diag(a_{11}, a_{22},....,a_{nn})$ as $a_{ii}=1$ if $\hat y_i=y_i$ and $a_{ii}=0$, otherwise.

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  • $\begingroup$ You could even use the $\delta$-function (or at least one of the variations on it) and let $a_{ii} = 1-\delta(\hat y_i-y_i)$. If you really want to fill in with formulas. $\endgroup$ – Arthur Oct 16 '19 at 7:06
  • $\begingroup$ @Arthur...yes. I highly agree. Thanks a lot for pointing out, $\endgroup$ – Nitin Uniyal Oct 16 '19 at 11:03

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