Question

A matrix Ais said to be nilpotent if there exists some positive integer msuch that A^m=0. Verify that

$A=\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)$

is nilpotent. Discuss why it is relatively easy to compute e^Atwhen Ais nilpotent. Compute e^Atand then use (1)to solve the system X'=AX.

Short answer

When matrix A is solved A³=0, so A^m=0. It is verified. When A is nilpotent it is easy to compute e^Atbecause it has a finite number of terms. The value of e^Atis $\left(\begin{array}{ccc}1-t-\frac{t^2}{2}& t& t+\frac{t^2}{2}\\ -t& 1& t\\ -t-\frac{t^2}{2}& t& 1+t+\frac{t^2}{2}\end{array}\right)$. The general solution for the system X'=AX is $X\left(t\right)=\left(\begin{array}{c}{c}_1(1-t-\frac{t^2}{2})+c_{2}t+c_{3}t+\frac{t^2}{2}\\ -c_{1}t+c_2+c_{3}t\\ c_1(-t-\frac{t^2}{2})+c_{2}t+c_3\left(\begin{array}{cc}t& 1+t+\frac{t^2}{2}\end{array}\right)\end{array}\right)$.

Step-by-step solution

Define matrix exponential.

Consider a square matrix A of size n*n. This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

where I is the unit matrix of order n.

Therefore the infinite matrix power series is

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression e^At. It is given by the formula,

Find whether the matrix A is nilpotent.

It is given that,

$A=\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)$

So,

$A^2=\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)=\left(\begin{array}{ccc}-1& 0& 1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right)$
$$\begin{gathered} A^{3=}{A}^2.A=\left(\begin{array}{ccc}-1& 0& 1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right)\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \\ {A}^m=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right) \end{gathered}$$

Since A³=0, then A^m=0.

Hence it is verified that A is nilpotent.

Find the value of eAt.

Now we have to compute e^At.

$$\begin{gathered} e^{At}=I+At+A^2\frac{t^2}{2!}+A^3\frac{t^3}{3!}+.... \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)+\left(\begin{array}{ccc}-1& 1& 1\\ -1& 0& 1\\ -1& 1& 1\end{array}\right)t+\left(\begin{array}{ccc}-1& 0& 1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right)\frac{t^2}{2!}+0+.... \\ =\left(\begin{array}{ccc}1-t-\frac{t^2}{2}& t& t+\frac{t^2}{2}\\ -t& 1& t\\ -t-\frac{t^2}{2}& t& 1+t+\frac{t^2}{2}\end{array}\right) \end{gathered}$$

We also notice that it is relatively easy to compute e^Atwhen A is nilpotent because we end up with a finite number of terms.

Hence the value of e^Atis $\left(\begin{array}{ccc}1-t-\frac{t^2}{2}& t& t+\frac{t^2}{2}\\ -t& 1& t\\ -t-\frac{t^2}{2}& t& 1+t+\frac{t^2}{2}\end{array}\right)$.

Find the general solution of the system.

The general solution of the system is calculated using,

X(t) = e^AtC

$=\left(\begin{array}{c}1-t-\frac{t^2}{2}tt+\frac{t^2}{2}\\ -t1t\\ -t-\frac{t^2}{2}t\begin{array}{cc}t& 1+t+\frac{t^2}{2}\end{array}\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\\ c_3\end{array}\right)$

$=\left(\begin{array}{c}{c}_1(1-t-\frac{t^2}{2})+c_{2}t+c_{3}t+\frac{t^2}{2}\\ -c_{1}t+c_2+c_{3}t\\ c_1(-t-\frac{t^2}{2})+c_{2}t+c_3\left(\begin{array}{cc}t& 1+t+\frac{t^2}{2}\end{array}\right)\end{array}\right)$

Hence the solution for the given system is $X\left(t\right)=\left(\begin{array}{c}{c}_1(1-t-\frac{t^2}{2})+c_{2}t+c_{3}t+\frac{t^2}{2}\\ -c_{1}t+c_2+c_{3}t\\ c_1(-t-\frac{t^2}{2})+c_{2}t+c_3\left(\begin{array}{cc}t& 1+t+\frac{t^2}{2}\end{array}\right)\end{array}\right)$