Find All Values of a which Will Guarantee that A Has Eigenvalues 0, 3, and -3.

Let A$A$ be the matrix given by

A=⎡⎣⎢−2−5403−21a−1⎤⎦⎥$$A=\left[\begin{array}{ccc}-2& 0& 1\\ -5& 3& a\\ 4& -2& -1\end{array}\right]$$

for some variable a$a$. Find all values of a$a$ which will guarantee that A$A$ has eigenvalues 0$0$, 3$3$, and −3$-3$.

Let p(t)$p(t)$ be the characteristic polynomial of A$A$, i.e. let p(t)=det(A−tI)=0$p(t)=det(A-tI)=0$. By expanding along the second column of A−tI$A-tI$, we can obtain the equation

p(t)=det⎛⎝⎜⎡⎣⎢−2−5403−21a−1⎤⎦⎥–⎡⎣⎢t000t000t⎤⎦⎥⎞⎠⎟=∣∣∣∣−2−t−5403−t−21a−1−t∣∣∣∣=(3−t)∣∣∣−2−t41−1−t∣∣∣+2∣∣∣−2−t−51a∣∣∣=(3−t)[(−2−t)(−1−t)−4]+2[(−2−t)a+5]=(3−t)(2+t+2t+t2−4)+2(−2a−ta+5)=(3−t)(t2+3t−2)+(−4a−2ta+10)=3t2+9t−6−t3−3t2+2t−4a−2ta+10=−t3+11t−2ta+4−4a=−t3+(11−2a)t+4−4a.$$\begin{array}{rl}p(t)& =det\left(\left[\begin{array}{ccc}-2& 0& 1\\ -5& 3& a\\ 4& -2& -1\end{array}\right]–\left[\begin{array}{ccc}t& 0& 0\\ 0& t& 0\\ 0& 0& t\end{array}\right]\right)\\ & =\left|\begin{array}{ccc}-2-t& 0& 1\\ -5& 3-t& a\\ 4& -2& -1-t\end{array}\right|\\ & =(3-t)\left|\begin{array}{cc}-2-t& 1\\ 4& -1-t\end{array}\right|+2\left|\begin{array}{cc}-2-t& 1\\ -5& a\end{array}\right|\\ & =(3-t)\left[(-2-t)(-1-t)-4\right]+2\left[(-2-t)a+5\right]\\ & =(3-t)(2+t+2t+t^2-4)+2(-2a-ta+5)\\ & =(3-t)(t^2+3t-2)+(-4a-2ta+10)\\ & =3t^2+9t-6-t^3-3t^2+2t-4a-2ta+10\\ & =-t^3+11t-2ta+4-4a\\ & =-t^3+(11-2a)t+4-4a.\end{array}$$

For the eigenvalues of A$A$ to be 0$0$, 3$3$, and −3$-3$, the characteristic polynomial p(t)$p(t)$ must have roots at t=0,3,−3$t=0,3,-3$. This implies

p(t)=–t(t−3)(t+3)=–t(t2−9)=–t3+9t.$$\begin{array}{r}p(t)=–t(t-3)(t+3)=–t(t^2-9)=–t^3+9t.\end{array}$$

Therefore,

For this condition to hold, the consistent terms on the left and right hand sides of the above condition must be equivalent. This implies 4−4a=0, which infers a=1. Consequently A has eigenvalues 0,3,−3 accurately when a=1.−t3+(11−2a)t+4−4a=−t3+9t.