Given a $3\times3$ real matrix $A$ with eigenvalues $0,1,2$, find real constants $a,b,c$ such that $aI+bA+cA^2$ has eigenvalues 0,1,3

Question

Suppose $A$ is a $3\times 3$ real matrix with three distinct eigenvalues $0,1,2$. Find real constants $a,b,c$ such that the matrix $aI+bA+cA^2$ has eigenvalues $0,1,3$.

My only initial thought on how to approach this problem is to use the Cayley-Hamilton theorem. i.e. $$A^3+3A^2+2A=0=(cA^2+bA+aI)^3-4(cA^2+bA+aI)^2+3(cA^2+bA+aI)$$ but this is a pretty ugly system (I didn't bother to show the algebraic steps of expanding the characteristic polynomials, but they are $C_p(A)=x(x-1)(x-2)$ and $C_p(aI+bA+cA^2)=x(x-1)(x-3)$). Thanks in advance for any assistance.

Answer 1

Since $A$ has distinct eigen-values, we know it is diagonalizable, so can be written as $A=P^{—1}DP$ for some invertible matrix $P$ and where $D=\pmatrix{0&0&0\\0&1&0\\0&0&2}$.

Note that $aI+bA+cA^2=P^{-1}(aI+bD+cD^2)P$, so we are trying to find $a,b,c$ such that $aI+bD+cD^2$is equivalent with $\pmatrix{0&0&0\\0&1&0\\0&0&3}$. But $aI+bD+cD^2=\pmatrix{a&0&0\\0&a+b+c&0\\0&0&a+2b+4c}$.

Thus $a=0,\,b=c=\frac12$ is a solution.

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Hope this helps.

Answer 2

Continued from my comment.

We now know that the eigenvalues of matrix $$ a \boldsymbol I + b \boldsymbol A+c \boldsymbol A^2 $$ could be $$ a\cdot 1 + b \cdot 0 + c \cdot 0, a + b + c, a+2b + 4c. $$ Let them equal to $0,1,3$ respectively then solve for $a,b,c$. The result might not be unique.

Also, what we doing here is interpolation, given the data $\{p(0), p(1),p(2)\} = \{0,1,3\} $. So if possible, the Lagrange interpolation formula is applicable here, maybe much more direct to the question.