0
$\begingroup$

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?

$\endgroup$
5
  • 7
    $\begingroup$ Unless I am mistaken, your example is $I_6$. $\endgroup$ – Mindlack Jan 21 '19 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. Jan 21 '19 at 17:24
  • 1
    $\begingroup$ You mean, it is not a block notation, but a matrix of which the entries are matrices? $\endgroup$ – Mindlack Jan 21 '19 at 17:26
  • 1
    $\begingroup$ 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. Jan 21 '19 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 Jan 21 '19 at 17:29
0
$\begingroup$

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 $\dfrac{n(n+1)}{2} = 6$

$\endgroup$
1
  • $\begingroup$ 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. Jan 21 '19 at 17:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.