$A$ is a square matrix of order $2$ with real entries and ${\rm Tr}(A)+|A|=2$. Show that $|A^2+|A|\cdot A+{\rm Tr}(A)I_2|\geq 4$

Question

$A$ is a square matrix of order $2$ with real entries and ${\rm Tr}(A)+|A|=2$. Show that

$$|A^2+|A|\cdot A+{\rm Tr}(A)I_2|\geq 4$$

My Attempt

I could observe that ${\rm Tr}(A)+|A|=2\Rightarrow 1+a+d+ad-bc=3 \Rightarrow|A+I|=3$

By Cayley-Hamilton theorem for a $2\times 2$ matrix we have

$A^2-{\rm Tr}(A)\cdot A+|A|\cdot I_2=0$.

$A^2={\rm Tr}(A)\cdot A-|A|\cdot I_2$

So $A^2+|A|\cdot A+{\rm Tr}(A)I_2={\rm Tr}(A)\cdot A-|A|\cdot I_2+|A|\cdot A+{\rm Tr}(A)I_2$

$=({\rm Tr}(A)+|A|)A+({\rm Tr}(A)-|A|)I_2$

$=2A+({\rm Tr}(A)-|A|)I_2=2A+2(1-|A|)I_2$

$|A^2+|A|\cdot A+{\rm Tr}(A)I_2|=4|A+(1-|A|)I_2|$

Not able to proceed from here. I am wondering whether Cayley Hamilton was the right approach or not.

Answer 1

Maybe not the shortest solution, but it works :

By Cayley Hamilton, one has $$A^2 = \mathrm{Tr}(A)A-|A|I_2$$

One deduce that \begin{align*} A^2+|A|.A+\mathrm{Tr}(A)I_2 &= \mathrm{Tr}(A)A-|A|I_2+|A|.A+\mathrm{Tr}(A)I_2 \\ &= (\mathrm{Tr}(A)+|A|)A +(\mathrm{Tr}(A)-|A|)I_2 \quad \quad \quad (*) \end{align*}

By assumption, one has also $\mathrm{Tr}(A)+|A|=2$, so $(*)$ becomes \begin{align*} A^2+|A|.A+\mathrm{Tr}(A)I_2 &= 2A + 2(\mathrm{Tr}(A)-1)I_2 \end{align*}

In particular, using the fact that $|\lambda I_2 + A|=|A|+\lambda\mathrm{Tr}(A)+\lambda^2$, you get \begin{align*}\big| A^2+|A|.A+\mathrm{Tr}(A)I_2 \big| &= 4\big|(\mathrm{Tr}(A)-1)I_2 + A \big| \\ & =4\left(|A| + (\mathrm{Tr}(A)-1)\mathrm{Tr}(A) + (\mathrm{Tr}(A)-1)^2 \right) \\ & =4\left(|A| + 2\mathrm{Tr}(A)^2-3\mathrm{Tr}(A)+1 \right) \\ & = 4\left(2\mathrm{Tr}(A)^2-4\mathrm{Tr}(A)+3 \right) \quad \quad \textit{using again }\mathrm{Tr}(A)+|A|=2\\ & = 4\left(1 + 2(\mathrm{Tr}(A)-1)^2 \right) \\ & \geq 4 \end{align*}