Does this special identity matrix have a name?












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 $2times 2$ 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?






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 7




    $begingroup$
    Unless I am mistaken, your example is $I_6$.
    $endgroup$
    – Mindlack
    Jan 21 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 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 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 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 at 17:29
















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 $2times 2$ 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?






linear-algebra matrices







share|cite|improve this question









$endgroup$








  • 7




    $begingroup$
    Unless I am mistaken, your example is $I_6$.
    $endgroup$
    – Mindlack
    Jan 21 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 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 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 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 at 17:29














0












0








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 $2times 2$ 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?






linear-algebra matrices







share|cite|improve this question









$endgroup$




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 $2times 2$ 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?






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 17:02









Eduardo W.Eduardo W.

1




1








  • 7




    $begingroup$
    Unless I am mistaken, your example is $I_6$.
    $endgroup$
    – Mindlack
    Jan 21 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 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 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 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 at 17:29














  • 7




    $begingroup$
    Unless I am mistaken, your example is $I_6$.
    $endgroup$
    – Mindlack
    Jan 21 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 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 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 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 at 17:29








7




7




$begingroup$
Unless I am mistaken, your example is $I_6$.
$endgroup$
– Mindlack
Jan 21 at 17:07




$begingroup$
Unless I am mistaken, your example is $I_6$.
$endgroup$
– Mindlack
Jan 21 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 at 17:24




$begingroup$
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 at 17:24




1




1




$begingroup$
You mean, it is not a block notation, but a matrix of which the entries are matrices?
$endgroup$
– Mindlack
Jan 21 at 17:26




$begingroup$
You mean, it is not a block notation, but a matrix of which the entries are matrices?
$endgroup$
– Mindlack
Jan 21 at 17:26




1




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 at 17:27




$begingroup$
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 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 at 17:29




$begingroup$
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 at 17:29










1 Answer
1






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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 17:28











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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 17:28
















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






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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 at 17:28














0












0








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






share|cite|improve this answer









$endgroup$



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







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 17:22









user458276user458276

554211




554211












  • $begingroup$
    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 at 17:28


















  • $begingroup$
    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 at 17:28
















$begingroup$
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 at 17:28




$begingroup$
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 at 17:28


















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