Let $A$ be $2 \times 2$ nonzero real matrix.which of the following is true?
$(A)$ trace of $A^2$ is positive$(B)$ $A$ has non zero eigenvalue.
$(C)$ All entries of $A^2$ can't be negative.
$(D)$$A^2$ has at least one positive entry.
I tried to find examples to counter these statements.
I took $A$ as $\begin{pmatrix}1&2\\-3&2\end{pmatrix}$ and $A^2$ is $\begin{pmatrix}-5&6\\-9&-2\end{pmatrix}$
This cancels out option $A,C$
Now I am not sure how to figure out option $(B)$ and $(D)$ I tried to change numbers of $A$ to find the example that counters $(D)$ but it is time-consuming. Is there any fact that I am missing for $(B)$ and $(D)$? I think $A$ can have zero eigenvalues because in my experience I never saw any statement saying a matrix must have zero value to have an eigenvalue zero. So (D) is my last option to tick. What could be another way to solve this problem quickly?
Hint If $$A= \begin{bmatrix}0&1 \\ 0 &0 \end{bmatrix}$$ what is $A^2$?
Note that if $A= \begin{bmatrix}a&b \\ c &d\end{bmatrix}$ then $$A^2= \begin{bmatrix}a^2+bc& ab+bd \\ ac+cd &d^2+bc \end{bmatrix}= \begin{bmatrix}a^2+bc& b(a+d) \\ c(a+d) &d^2+bc \end{bmatrix}$$
If $bc \geq 0$ then the (1,1) entry of $A^2$ cannot be negative.
If $bc <0$ then $b,c$ have opposite signs. Then one of $b$ or $c$ has the same sign as $(a+d)$ making the (1,2) or (2,1) entry non-negative.