Problem of characteristic and minimal polynomials.

Question

Let $A$ be a non-diagonal $2\times 2$ matrix with complex entries such that $A=A^{-1}.$ Write down its characteristic and minimal polynomials.

Solution

$A=A^{-1}\implies A^2=I\implies Ch_A(x)=x^2-1=(x-1)(x+1)$

Since Characteristic polynomial and Minimal polynomial have the same roots so,$Mp_A(x)=x^2-1$

The answer in answer key is same as I computed,but I'm getting vibes that something is wrong during transition from $2^{nd}$ step to $3^{rd}$ step.

Please give suggestions about my solution...

Thank You!!

Answer 1

The solution you've written down is incorrect, and you can guess that this has to be the case, because you didn't make use of the fact that $A$ is a non-diagonal matrix. Another way to see your proof is incorrect is by your comment: in your proof you didn't use the fact that $A$ is a $2 \times 2$ matrix (but in your comment you correctly pointed out that it is the second to third implication which is erroneous becase the characteristic polynomial for a $3 \times 3$ matrix would have to be cubic).

Now, the correct approach is the following:

• Since $A = A^{-1}$, it follows that $A^2 = I$.
• Since $A^2 = I$, this means the polynomial $\alpha(x) = x^2-1 = (x-1)(x+1)$ is such that $\alpha(A) = 0$. In other words, $\alpha(x)$ is an annihilating polynomial for the matrix $A$. Therefore, the minimal polynomial $\mu_A(x)$ has to divide the annihilating polynomial $\alpha(x)$.
• From the above, we can deduce that there are $3$ possibilities for the minimal polynomial: either $\mu_A(x) = x-1$, or $\mu_A(x) = x+1$ or $\mu_A(x) = x^2 - 1$. However, the minimal polynomial has to be $x^2-1$, because if it was $x \pm 1$, then the matrix $A$ would satisfy $A \pm I = 0$, and hence $A = \pm I$. But this is a diagonal matrix, and hence would contradict the assumption that $A$ is a non-diagonal matrix. Therefore, the minimal polynomial is $\mu_A(x) = x^2-1$.
• Lastly, we know that the minimal polynomial divides the characteristic polynomial (Cayley-Hamilton theorem), so the characteristic polynomial can be written as \begin{align} C_A(x) &= \phi(x) \cdot \mu_A(x) \end{align} for some polynomial $\phi(x)$. But now, we make use of the fact that $A$ is a $2 \times 2$ matrix. This means $C_A(x)$ has to be a degree 2 polynomial. But notice also that $\mu_A(x) = x^2 - 1$ is already degree 2. Therefore, $\phi(x)$ must be a constant, and in fact it is equal to $1$, because the leading coefficient of the characteristic polynomial of a $2 \times 2$ matrix is $1$.

This entire argument shows that \begin{align} \mu_A(x) = C_A(x) = x^2 -1. \end{align}