# Finding satisfying matrix for a particular condition

1. Mar 13, 2014

### Seydlitz

Hello guys,

I'm trying rather hard to find a real square matrix that will satisfy this $AA^T=-I$. The first thing that comes to my mind is of course orthogonal matrix. But clearly it isn't. In fact, $A^T=-A^{-1}$ in order for it to work. The condition is pretty strict if you also consider that only even square matrix can satisfy this condition for real determinant and hence real matrix. (The reason is $-I$ will have a determinant value of $-1$ if the matrix is odd.)

Trivial matrix and symmetric matrix will not even work even though $I(-I)=-I$ because $-I$ is not the transpose of $I$.

Moreover, I'm not sure if square matrix will work because clearly the diagonal of $AA^T$ cannot be $-1$ if $A$ is real matrix.

$iI$ works, but then, it's not real.

2. Mar 13, 2014

### jbunniii

There is no such matrix. The reason is that $AA^T$ is positive semidefinite, whereas $-I$ is negative definite.

To see this, let $x$ be any nonzero real vector of the appropriate dimension. Then $x^T(AA^T)x = (x^TA)(A^Tx) = (A^T x)^T (A^Tx) \geq 0$. On the other hand, $x^T (-I) x = -(x^T x) < 0$.

3. Mar 14, 2014

### Seydlitz

Thanks jbunnii for your explanation. I was already quite convinced that it is, seeing the case using 2x2 matrix.