Prove that if $A^t = A^{-1}$ then different rows of $A$ are orthogonal to each other

Question

Prove that if $A^t = A^{-1}$ then different rows of $A$ are orthogonal to each other

I saw this statement in my linear algebra book and I don't really know how to prove it, I hope someone can help me out.

Thank you in advance.

Answer 1

If $\tilde{\mathrm{a}}_i$ is the $i$-th row of $A$, then

$$A A^T = \begin{bmatrix} — \, \tilde{\mathrm{a}}_1^T \, —\\ — \, \tilde{\mathrm{a}}_2^T \, —\\ \vdots\\ — \, \tilde{\mathrm{a}}_n^T \, —\end{bmatrix} \begin{bmatrix} | & | & & |\\ \tilde{\mathrm{a}}_1 & \tilde{\mathrm{a}}_2 & \ldots & \tilde{\mathrm{a}}_n\\ | & | & & |\end{bmatrix} = \begin{bmatrix} \tilde{\mathrm{a}}_1^T \tilde{\mathrm{a}}_1 & \tilde{\mathrm{a}}_1^T \tilde{\mathrm{a}}_2 & \ldots & \tilde{\mathrm{a}}_1^T \tilde{\mathrm{a}}_n\\ \tilde{\mathrm{a}}_2^T \tilde{\mathrm{a}}_1 & \tilde{\mathrm{a}}_2^T \tilde{\mathrm{a}}_2 & \ldots & \tilde{\mathrm{a}}_2^T \tilde{\mathrm{a}}_n\\ \vdots & \vdots & \ddots & \vdots\\ \tilde{\mathrm{a}}_n^T \tilde{\mathrm{a}}_1 & \tilde{\mathrm{a}}_n^T \tilde{\mathrm{a}}_2 & \ldots & \tilde{\mathrm{a}}_n^T \tilde{\mathrm{a}}_n\end{bmatrix}$$

If $A^T = A^{-1}$, then $A A^T = I_n$ and, thus,

$$\tilde{\mathrm{a}}_i^T \tilde{\mathrm{a}}_j = \begin{cases} 1 & \text{if } i = j\\ 0 & \text{if } i \neq j\end{cases}$$

Hence, different rows of $A$ are orthogonal.

Answer 2

Let $\;R_i,\,R_j\;$ be two different rows in $\;A\;$, and observe that these two are different columns in $\;A^t\;$ , then:

$$A^t=A^{-1}\implies AA^t=I\implies R_i\cdot R_j=0\;\;\text{whenever}\;\;1\le i\neq j\le n$$

Answer 3

$A^t=A^{-1}$ is to say $AA^t=I$, then

$$\delta_{ij}=I_{ij}=(AA^t)_{ij}=\sum_{k=1}^{n}A_{ik}(A^t)_{kj}=\sum_{k=1}^{n}A_{ik}A_{jk}$$, where $\delta_{ij}$ is Kronecker delta.

Hence the different rows are orthogonal, in fact orthonormal.