How to quickly calculate $\det(f(A))$, where $f(x)=2x^3+3x$ and A is a 3x3 matrix?

Question

Given $$A=\left( \begin{matrix} 1 & 0 & 1 \\ 1 & -2 & -3 \\ 1 & 2 &3\end{matrix} \right)$$ how would one calculate $\det(f(A))$ most efficiently besides first raising $A$ to the 3rd power, adding the matrices and calculating the determinant (which is quite ugly in itself)? Or, better asked, is there any way to skip steps? Any observations that could help here?

Answer 1

As the characteristic polynomial of $A$ is

$$\lambda^3-2 \lambda^2-4,$$

Cayley Hamilton theorem: gives $A^3=2A^2+4I$.

Therefore $f(A)=2 A^3+3A=4A^2+3A+8I.$

Then one has to end the calculation by hand...

Answer 2

You can do the following: first you can compute $h(z):=\det(A-zI)$, now note that $f(z)=2z(z-\alpha )(z-\bar \alpha )$ for $\alpha =i\sqrt{3/2}$, then using the fact that $\det(M_1M_2)=\det(M_1)\det(M_2)$ for any two matrices $M_1$ and $M_2$ you have that

$$ \begin{align*} \det(f(A))&=2^3\cdot h(0)\cdot h(\alpha )\cdot h(\bar \alpha )\\ &=8\cdot h(0)\cdot h(\alpha )\cdot \overline{h(\alpha )}\\ &=8\cdot h(0)\cdot |h(\alpha )|^2 \end{align*} $$

where the second equality follows because $h$ is a polynomial. I'm not sure about how efficient is this procedure, however it seems useful in some cases.

Answer 3

As computed by Jean Marie, we know that the characteristic polynomial of $A$ is $\chi_A(X)=X^3-2X^2-4=X^3-aX^2+bX-c$ with $(a,b,c)=(2,0,4)$. Now, naming $\alpha,\beta,\gamma$ the three roots (in $\mathbb{C}$) of $\chi_A$, we have the relations between roots and coefficients of $\chi_A$: $$a=\alpha+\beta+\gamma;\quad b=\alpha\beta+\beta\gamma+\gamma\alpha;\quad c=\alpha\beta\gamma.$$

We further observe that $$\renewcommand{\arraycolsep}{0pt} \begin{array}{rcccl} \hfill\alpha^2+\beta^2+\gamma^2&{}={}&(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)&{}={}&a^2-2b,\\[3pt] \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2&{}={}&(\alpha\beta+\beta\gamma+\gamma\alpha)^2 -2(\alpha+\beta+\gamma)\alpha\beta\gamma&{}={}&b^2-2ac. \end{array}$$

Also $A$ is trigonalizable: there si some invertible $M\in\mathcal{M}_{3,3}(\mathbb{C})$ such that $A=M^{-1} \begin{pmatrix} \alpha&*&*\\ 0&\beta&*\\ 0&0&\gamma \end{pmatrix}M$. For any complex polynomial $P$, we then have $P(A)=M^{-1} \begin{pmatrix} P(\alpha)&\star&\star\\ 0&P(\beta)&\star\\ 0&0&P(\gamma) \end{pmatrix}M$, and $\det(P(A))=P(\alpha)P(\beta)P(\gamma)$.

Now let's proceed with $P(X)=2X^3+3X$. We have: \begin{eqnarray*} \det(P(A))&=&(2\alpha^3+3\alpha)\times(2\beta^3+3\beta)\times(2\gamma^3+3\gamma)\\ &=&\alpha\beta\gamma\times(2\alpha^2+3)\times(2\beta^2+3)\times(2\gamma^2+3)\\ &=&\alpha\beta\gamma\times\big[8(\alpha\beta\gamma)^2+12(\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)\\ &&\hskip8em{}+{}18(\alpha^2+\beta^2+\gamma^2) + 27\big]\\[3pt] &=&c\big[8c^2+12(b^2-2ac)+18(a^2-2b)+27\big]\\ &=&140. \end{eqnarray*} The advantage of this method is that we calculated no power of $A$ and no determinant either (at least in the “usual” way…).