# Why the identity matrix in first-order linear systems?

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

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

Here's some exposition:

Consider the homogenous first-order linear system

$$\bf{x'=Ax}\quad(\star)$$

where

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

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

$$\textbf{Av}=r\textbf{v},$$

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?

Thanks.

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

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 $$(r-A)(e^{rt}\textbf{v})=\textbf{0} \;\;\;\Leftrightarrow\;\;\; (rI-A)(e^{rt}\textbf{v})=\textbf{0}.$$
• 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. Jan 5 at 23:30
• @StephenDonovan the OP has numerous errors in his question. For instance, he uses the notation $\textbf{r}$ instead of $r$. Jan 6 at 0:20
• 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. Jan 6 at 4:53