Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

\begin{align} \tag{1} \dot x&=Ax\\ \tag{2} sX(s)-x(0)&=AX(s) \\ \tag{3} (sI-A)X(s)&=x(0) \end{align} Considering these equations, how can we go from $(2)$ to $(3)$?

My question is why was the identity matrix $I$ introduced here? My concern is about introducing $I$ when collecting $X(s)$.


share|cite|improve this question
Thx Tong for fixing my question. – OOzy Pal Apr 28 '13 at 4:26
up vote 0 down vote accepted

$$ sX(s)-x(0)=AX(s) \\ sX(s)-AX(s)=x(0) \\ $$

In the second equation, notice that there is a factor of $X(s)$ on the right-hand side of both $sX(s)$ and $AX(s)$, so we can take it out like so:


But this wouldn't make any sense, because $A$ is a matrix and $s$ is a scalar. To make it so that the equation makes sense, we make $s$ into a matrix by multiplying by the identity.


This is just to ensure that it doesn't matter which way round we do things. We could do $sI-A$ first, and then multiply the result by $X(s)$ on the right, or we can multiply through by $X(s)$ to get


which is what you started with.

Adding the identity $I$ doesn't change the value of the equation, it's just so that it "makes sense" if we were to do the subtraction before the matrix multiplication.

share|cite|improve this answer

$$\begin{array}{rcl} \dot x&=Ax \\ sX(s)-x(0)&=AX(s) \\ sX(s)-AX(s)&=x(0) \\ sIX(s)-AX(s)&=x(0) \\ (sI-A)X(s)&=x(0) \end{array}$$

$I$ was already there but it wasn't written, because writing it doesn't change the meaning of the expression. Consider this analogical example from scalar equations:

$$\begin{array}{rcl} ab-a&=c \\ a(b-\underset{?}{\underset{\downarrow}{1}})&=c \end{array}$$

How did that $1$ appear there? It wasn't there, was it? Yes it was there, but it wasn't written, it was hidden, because it is the identity element of multiplication, just like $I$ is in matrix case.

$$\begin{array}{rcl} ab-a&=c \\ ab-a1&=c \\ a(b-1)&=c \end{array}$$

The hidden identity element comes into appearance when it is needed. In your case, it has to appear, because when you take the common factor $X(s)$ out of the parenthesis, you left $s-A$ inside. The dimensions must match but they don't. It was originally a difference of matrices $(s(X(s)-AX(s))$, but now a difference between a scalar and a matrix $(s-A)$ which is an undefined operation.

share|cite|improve this answer
Thank you hkBattousai for answering my question. Both Answers made sense. Thx – OOzy Pal Apr 28 '13 at 4:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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