How to find counterexamples, linear algebra prove/disprove

Question

I have to determine if the following statements are true/false.

Let $ A \in M_n({\mathbb{R}})$.

1. if $ A^2 $ is orthogonal then $ A $ is orthogonal

2. if $ A^2 $ is normal then $ A $ is normal

Well, none of those statements are correct, they all false. But I had difficulties finding counter examples.

for the first statement I could not find a counter example. If any one can find one and help me understand what should be the way of thinking in order to find the counterexample, that would be great.

for the other statement I did found a counter example but I basically gussed it. Is there a way to find a counterexample without guessing and trying things untill something will work?

Thanks in advance.

Answer 1

There is no general recipe to find counterexamples, but for matrices you would usually start by trying the easiest $2 \times 2$ matrices. For example, the simplest orthogonal $2 \times 2$ matrix is undoubtably $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. So can you find a non-orthogonal matrix $A$ such that $A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$?

Answer 2

Part 1:

$A^2$ is orthogonal, that is $AAA^TA^T=D$ where $D$ is diagonal. Chose a $D$ that has some negative value in its diagonal then if $D=CC^T$ where $C$ is also diagonal, then $C$ contains some complex value.

Part 2:

$A^2$ is normal, that is $AAA^*A^*=A^*A^*AA$. Choose a diagonal $A=\sqrt{-1}I$ then $A^*=-A$ and it is not normal.

Answer 3

as far as generating counterexamples, part of my own mental checklist is:

1.) triangular matrices are easy to work with (whether for nilpotence or diagonalization)
2.) symmetric matrices tend to be easy to work with
3.) Idempotence: $A^2 = A$
4.) Involutions: $A^2 = I$

For your particular problem, combining (1) and (4) will do it. So pick an upper triangular diagonalizable matrix that has +1 and -1 as eigenvalues (if you make this 2x2 the eigenvalues are distinct so it is automatically diagonalizable), and make sure the matrix has some non-zero strictly upper triangular element. From here you call on the inequality of Schur: $\big\Vert A\big\Vert_F^2 \geq \sum_{k=1}^n \vert \lambda_k\vert^2$ with equality iff $A$ is normal. The Frobenius norm is easy to calculate and $\sum_{k=1}^n \vert \lambda_k\vert^2$ is easy to calculate for Triangular matrices.