# Using Simplify to perform substitutions [closed]

I have a matrix element $k_x$ and I want to substitute it as $-i\kappa$, I tried

Simplify[
Subscript[H, s],
{B == 0, Subscript[k, x] == -I κ, Subscript[k, y] == κ}] // MatrixForm


However, the result shows $k_x$ in some elements in the matrix and $-i\kappa$ some place. How do I substitute all of them? I did a quick test but found the result just gives $k_x$.

Simplify[
Subscript[k, x],
{B == 0, Subscript[k, x] == -I κ, Subscript[k, y] == κ}] // MatrixForm


I have played with ReplaceAll at the very beginning, but I actually want to do something in this way, is there any flexible way I can do a replacement?

Subscript[H, 01] =
FullSimplify[
ArrayFlatten[({
{Simplify[Subscript[H, S], {Subscript[k, x] == 0, Subscript[k, y] == 0}],
Simplify[
Subscript[H, S],
{B == 0, Subscript[k, x] == -I κ, Subscript[k, y] == κ}]},
{Simplify[
ConjugateTranspose[Subscript[H, S]],
{B == 0, Subscript[k, x] == I κ, Subscript[k, y] == κ}],
Simplify[
Subscript[H, S],
{Subscript[k, x] == 0, Subscript[k, y] == 0}] + ℏ ω IdentityMatrix[4]}
})],
{α, β, γ, κ, ζ, Subscript[k, x], Subscript[k, y]} ∈ Reals];


## closed as off-topic by MarcoB, m_goldberg, ilian, Bob Hanlon, ÖskåAug 17 '15 at 18:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – MarcoB, m_goldberg, ilian, Bob Hanlon, Öskå
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• Look up ReplaceAll. Also read this answer, mathematica.stackexchange.com/questions/18393/…, and pay attention to what it says about subscripts and MatrixForm, in case that is an issue. If you include an example for Subscript[H, s], folks will better be able to say what the precise issue is. – Michael E2 Aug 17 '15 at 3:03
• Good, thanks. I find out ReplaceAll [Subscript[H, s],k_x->-i [kappa]] work better than Simplify – Dhwister Aug 17 '15 at 8:48

Try this:

rule = Subscript[k, x] -> I*κ


Then any expression you may treat as follows: expression/.rule. For example:

expression = Subscript[k, x]^2 + Subscript[k, x]
expression /. rule


returns:

(*   Subscript[k, x] + \!$$\*SubsuperscriptBox[\(k$$, $$x$$, $$2$$]\)  *)

(*  I κ - κ^2 *)


Have fun!