what does it mean when a Matrix is multiplied by it's transpose to give $3I$(identity)

Question

if $A$ is $2\times 2$ Matrix such that $AA^T$ = $3I$ what does this say about $A$? $3I$ is a symmetrical matrix does that mean that $A$ is symmetrical therefore $AA^T$ is $A^2$?

Answer 1

$$\pmatrix{a&b\\c&d}\times \pmatrix{a&c\\b&d}=\pmatrix{a^2+b^2&ac+bd\\ac+bd&c^2+d^2}=\pmatrix{3&0\\0&3}$$

is satisfied by, for example $a=d=\sqrt{2}$ , $b=1$ , $c=-1$ , so $A$ need not be symmetric.

Answer 2

If an $n \times n$ matrix $A$ satisfies $AA^T = \alpha I$ for some real $\alpha \neq 0$, that means that the rows of $A$ are orthogonal (but not necessarily orthonormal) and have the same norm.

This can be seen by considering the element at position $i,j$ in the product $AA^T$, which is the dot product of the $i$th row, $a_i$, of $A$ and the $j$:th column of $A^T$, which is the $j$th row, $a_j$, of $A$. Since $AA^T = \alpha I$, we see that this product is zero if $i \neq j$ and $\alpha$ otherwise: $$\langle a_i , a_j \rangle = \begin{cases} 0 & i \neq j \\ \alpha & i = j \end{cases}$$ since $\langle a_i, a_i \rangle = \|a_i\|_2^2$, we also see that all the vectors have the same norm.

For the special case of $\alpha = 1$ we get an orthogonal matrix.

We can consider what a matrix $A$ with $AA^T = \alpha I$ would do when applied to both sides of a dot product of a vector. We get: $$\langle Av, Au \rangle = (Av)^TAu = v^T A^T A u = \alpha v^Tu = \alpha \langle v, u \rangle$$ so they are mappings where the image is always $\alpha$ times larger than the element which was mapped.

Matrices $A$ which satisfy $AA^T = \alpha I$ for any $\alpha \neq 0$, i.e. the set $$\{A \mid \exists \alpha \neq 0: AA^T = \alpha I \}$$ form a group under multiplication. The identity is, as usual, $I$. If $A$ and $B$ are matrices such that $AA^T = \alpha I$ and $BB^T = \beta I$ we get that the product $AB$ satifies: $$AB(AB^T) = ABB^TA^T = A(\beta I)A^T = \beta AA^T = \alpha \beta I.$$

We know that the the rows form an orthogonal basis, so $A$ has full rank, and therefore has an inverse $A^{-1}$. We can see that the inverse must be $$A^{-1} = \frac{1}{\alpha} A^T$$ so we get $$AA^{-1} = \frac{1}{\alpha} AA^T = \frac{1}{\alpha} \alpha I = I.$$