7
$\begingroup$

Now, I'm aware of the threads existing about this question such as:

How to enter matrices in block matrix format?

How to form a block-diagonal Matrix from a list of matrices?

But I wasn't able to find the solution to my problem there. I need to make a matrix $$ \begin{pmatrix} C_1 &I\\ 0 & C_2 \end{pmatrix} $$

My problem is that $C_1$ and $C_2$ are of different sizes! $C_1$ is of size 30 and $C_2$ is of size 48. So one would expect the following code to work:

MatrixForm[ArrayFlatten[{{c1, IdentityMatrix[30], 0}, {0, c2}}]]

or maybe

MatrixForm[ArrayFlatten[{{c1, IdentityMatrix[30], ConstantArray[0,{18,18}]}, 
{ConstantArray[0,{30,30}], c2}}]]

But neither actually works!

Edit

I ended up using J.M's answer:

ArrayFlatten[{{c1, PadRight[IdentityMatrix[30], {Automatic, 48}]}, {0, c2}}]]
Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ Why not ArrayFlatten[{{c1, IdentityMatrix[30]}, {0, c2}}]? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Nov 23, 2015 at 13:59
  • $\begingroup$ this doesn't work either :( $\endgroup$
    – Sertii
    Commented Nov 23, 2015 at 14:03
  • 2
    $\begingroup$ Try ArrayFlatten[{{c1, PadRight[IdentityMatrix[30], {Automatic, 48}]}, {0, c2}}]] then. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Nov 23, 2015 at 14:09
  • $\begingroup$ Awesome! That worked! $\endgroup$
    – Sertii
    Commented Nov 23, 2015 at 14:10
  • $\begingroup$ Yes they are both square. C1 and C2 are companion matrices and i'm trying to make the Jordan rational normal form. I don't think there is a problem, I might be wrong though $\endgroup$
    – Sertii
    Commented Nov 23, 2015 at 14:15

1 Answer 1

Reset to default
3
$\begingroup$ 
m30 = ConstantArray[3, {30, 30}];
i30 = IdentityMatrix[30];
m48 = ConstantArray[4, {48, 48}];
m12 = ConstantArray[0, {48, 12}];

Transpose@Join[m30, i30]~Join~Join[m12, m48, 2]
Improve this answer
$\endgroup$
2
  • $\begingroup$ I ended up using an anwer provided above but thank you for taking the time anyway! $\endgroup$
    – Sertii
    Commented Nov 23, 2015 at 14:16
  • 1
    $\begingroup$ They are different answers, J.M.'s follows the idea that C1 and C2 have no columns in common - which can be inferred from your sketch above. Eldo's answer maintains the fact that I is an identity matrix. You can't have both: if I is an identity matrix, then the shape described doesn't make sense $\endgroup$
    – Jason B.
    Commented Nov 23, 2015 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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