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recently I stumbled upon the problem of defining a diagonal matrix whose elements are identity matrices of $dim = n$, where $n$ is the column/row index. For example, for $n=3$:

$\mathbb{I}_3 = \left[{\begin{array}{ccc} I_1 & 0 & 0 \\ 0 & I_2 & 0 \\ 0 & 0 & I_3 \end{array} }\right]$,

and the subscript indicates the size of the matrix, i.e., $I_2$ is a $2\times 2$ identity matrix and so on.

This definition may look silly, but I need a matrix with this property to explicitly define the direct sum of matrices with a notation that's more usual than $\bigoplus_i^n$.

So, $\mathbb{I}_n$ does have a special name?

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    $\begingroup$ Unless I am mistaken, your example is $I_6$. $\endgroup$
    – Aphelli
    Commented Jan 21, 2019 at 17:07
  • $\begingroup$ It looks like, but $\mathbb{I}_3$ is actually a 3x3 matrix, where the elements of the diagonal are progressively larger identity matrices. This shape is neccessary for the mathematical properties I need. $\endgroup$
    – Eduardo W.
    Commented Jan 21, 2019 at 17:24
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    $\begingroup$ You mean, it is not a block notation, but a matrix of which the entries are matrices? $\endgroup$
    – Aphelli
    Commented Jan 21, 2019 at 17:26
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    $\begingroup$ Precisely! The 1st element is a 1x1 identity matrix. The 2nd element on the diagonal is a 2x2 identity. The 3rd is a 3x3 identity and so on. So, expanding the elements back into full matrices, the full dimension of $\mathbb{I}_n$ is $n!$. $\endgroup$
    – Eduardo W.
    Commented Jan 21, 2019 at 17:27
  • $\begingroup$ I don't think what you're asking for has/needs a special name. Generally you talk about matrices built out of smaller matrices as being "block matrices" and then you use your arguments about how to compute with the blocks. $\endgroup$
    – rschwieb
    Commented Jan 21, 2019 at 17:29

1 Answer 1

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There’s not a special unique name, but in general, $\mathbb{I}_{n} = I_{\frac{n(n+1)}{2}}$.

In your example, $n=3$ so $\dfrac{n(n+1)}{2} = 6$

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  • $\begingroup$ I see... I thought this couldn't be new, probably some Russian or French mathematician would already had defined this and it could be referred by its name. But thanks! $\endgroup$
    – Eduardo W.
    Commented Jan 21, 2019 at 17:28

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