I would like to build a vector with the values obtained from computation of the diagonal terms of the following Table:

 Table[Table[l + j, {l, 0, 10, 1}], {j, 0, 0.1, 0.01}] // MatrixForm

My real Table is much more complicated and time consuming, therefore I am looking for a fast way to avoid all computations of terms out of diagonal. Is there any function or way?

I would like to underline that my real Table has the following shape, which is forcing me to find a smart way to compute the diagonal terms:

vectorj=Table[j,{j, 0, 0.1, 0.01}]
Table[j*f[x][[j]],{j,1,Length[vectorj]}] // MatrixForm

The problem is that I need in the last table j to be an integer to get the [[j]] term of f[x], but the multypling factor j should be actually the value in the j position, say in position 2, that should be 0.01. I want to avoid building too many tables so I thought there might be a way to avoid useless computations.


2 Answers 2


To talk about a diagonal of the matrix it must be square. So in general make each of your two vectors (which are equal length due to squareness):


Then apply your matrix function f[i,j] to these two vectors as f[t,t] along the diagonal:

  • $\begingroup$ a couple of things: first, isn't there syntax error in the definition of ivector and jvector? secondly, i don't get anything after running your code $\endgroup$
    – Andrea G
    Commented May 24, 2017 at 9:56
  • 1
    $\begingroup$ Yes there was a typo in my two vectors. This is now fixed. This error was what was causing there to be no output. $\endgroup$
    – Ian Miller
    Commented May 24, 2017 at 10:25

First, I'd rewrite your code above

Table[Table[l + j, {l, 0, 10, 1}], {j, 0, 0.1, 0.01}] // MatrixForm


Table[l + j/100, {j, 0, 10}, {l, 0, 10}] // MatrixForm

Now the indices are comparable, and you can just write the following to get the diagonal elements:

Table[j + j/100, {j, 0, 10}]

In general, if you are making j do double duty as both an argument to a function and an index to a table, better to treat it fundamentally as an integer index and compute the argument.

  • $\begingroup$ ok this is fine if we can make the indices comparable. In this case we were lucky because of that 1/100 factor. And in a general case where this is not possible? $\endgroup$
    – Andrea G
    Commented May 23, 2017 at 15:13
  • $\begingroup$ I think it will always be possible to rewrite into an integer index form, since it is just a linear transformation from the original sequence to the sequence {0,1,2,...}. Can you think of a counter-example? $\endgroup$
    – MikeY
    Commented May 23, 2017 at 15:17
  • $\begingroup$ Values that I want the function to assume: 0, 0.1, 0.6, 100, 1023.5; Indices of the vector: 1, 2, 3, 4, 5. What do you think of such a case? $\endgroup$
    – Andrea G
    Commented May 23, 2017 at 15:26

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