I was reviewing homogenous systems of first-order ODEs and was wondering how the identity matrix $I$ in


came about.

Here's some exposition:

Consider the homogenous first-order linear system



$$\textbf{x'}=\frac{d}{dt}\begin{pmatrix}x_1\\x_2\end{pmatrix}\quad\quad \bf{A}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\quad\quad\bf{x}=\begin{pmatrix}x_1\\x_2\end{pmatrix}.$$

We look for solutions of the from

$$\textbf{x}=\textbf{v}e^{rt}\quad (\star\star )$$

where $r$ and $\textbf{v}=(v_1\;v_2)^T$ are to be determined.

Plugging $(\star\star)$ back into $(\star)$ yields


since $e^{rt}$ is non-zero.

Furthermore we have that

$$(\textbf{A}-r\textbf{I})\textbf{v}=\textbf{0}\quad (\star\star\star)$$

where $I$ is the $2\times2$ identity matrix.

Therefore, to solve the system $(\star)$, we must solve the homogenous algebraic system $(\star\star\star)$ for $\bf v$.

This means that for a non-trivial solution of $(\star)$, $$\det(\textbf{A}-r\textbf{I})=\textbf{0}\quad (*)$$ must be satisfied.

Question: Where does the identity matrix $I$ come from? I know that without the $I$ the statement $(*)$ is meaningless, but what I'm asking is why is it this precise matrix and not any other matrix?


  • 5
    $\begingroup$ because $ \bf rv=rIv$, $\bf Av-rv=Av-rIv=(A-rI)v$ $\endgroup$ Jan 5 at 22:55
  • $\begingroup$ Wow, many thanks! $\endgroup$
    – whitenoise
    Jan 5 at 23:41

There are several errors in your question statement, so let me try to fix those as I answer your question.

We are given an equation $$\textbf{x}'=A\textbf{x},$$ where $\textbf{x}$ is a vector of length $n$, and $A$ is an $n\times n$ matrix. Suppose this has a solution of the form $\textbf{x}=e^{rt}\textbf{v}$, where $\textbf{v}$ is a constant vector and $r$ is a constant scalar. Since $(e^{rt}\textbf{v})'=re^{rt}\textbf{v}$, we obtain $$ re^{rt}\textbf{v}=A(e^{rt}\textbf{v}) $$ and hence $$ r(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0}. $$ Here we would like to factor out $e^{rt}\textbf{v}$, but there is a problem: Since $r$ is a scalar and $A$ a matrix, the expression $(r-A)$ makes no sense---unless we stipulate that $r$ is really $rI$, where $I$ is the $n\times n$ identity matrix. This works because $$ r(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0} \;\;\;\Leftrightarrow\;\;\; rI(e^{rt}\textbf{v})-A(e^{rt}\textbf{v})=\textbf{0} $$ and hence $$ (r-A)(e^{rt}\textbf{v})=\textbf{0} \;\;\;\Leftrightarrow\;\;\; (rI-A)(e^{rt}\textbf{v})=\textbf{0}. $$

  • 2
    $\begingroup$ Remember: matrix transformation is linear. So, $A(e^{rt}v) = e^{rt}(Av)$. Now because $e^{rt}$ cannot be zero, we can divide by it on both sides to get $Av = rv$, and hence $(A-rI)v = 0$. So the OP had it right. $\endgroup$ Jan 5 at 23:30
  • $\begingroup$ @StephenDonovan the OP has numerous errors in his question. For instance, he uses the notation $\textbf{r}$ instead of $r$. $\endgroup$
    – Ben W
    Jan 6 at 0:20
  • $\begingroup$ Yes, but given the context it should be clear it's a scalar. It is misleading notation but if you know about the system he's talking about everything is still communicated rather clearly. $\endgroup$ Jan 6 at 4:49
  • $\begingroup$ Thanks for clearing things up for me @Steven Donovan. Yes, $r$ is supposed to be a scalar. I will change it as bold r was merely a typsetting error on my part. $\endgroup$
    – whitenoise
    Jan 6 at 4:53

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