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Suppose $A$ is an $n \times n$ matrix. Show that $AA^{*}=I$ if and only if the rows of $A$ form an orthonormal basis.

So far the only thing that I have done with this problem is knowing that $(AA^{*})_{ij}=\langle v_i, v_j\rangle$ for all $i$ and $j$. But I do not know how to get that this in fact equals $0$ and how to show that the norm of each row is $1$. Any help is appreciated. Thanks in advance.

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It's hard to know where your problem lies. What is for you the definition of $I$? –  Alex B. Dec 4 '12 at 17:12
    
Think about the entries of both sides of $AA^*=I$ when $i=j$ and when $i\ne j$. –  anon Dec 4 '12 at 17:12
    
$A_{ii}A^{*}_{ii}=1$ and whenever $i \neq j$ is $0$. –  tk2 Dec 4 '12 at 17:25
    
@AlexB. $I$ is the identity matrix. –  tk2 Dec 4 '12 at 17:25
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Your second to last comment is incorrect. The matrix identity says that $(AA^*)_{ii}=1$, i.e. $\langle v_i,v_i\rangle=1$, and not $A_{ii}A^*_{ii}$. I hope that clears up your confusion. –  Alex B. Dec 4 '12 at 17:28
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As you say: $(AA^{\top})_{i,j} = \langle {\bf v}_i,{\bf v}_j \rangle$. Since $(I)_{i,j} = 1$ for all $i=j$ and $(I)_{i,j} = 0$ for all $i\neq j$, it follows that $AA^{\top} = I$ if and only if $$ \langle {\bf v}_i,{\bf v}_j \rangle = \left\{ \begin{array}{ccc} 1 & : & i = j \\ 0 & : & i \neq j \end{array}\right.$$ Thus $||{\bf v}_i|| = 1$ for all $i$ and ${\bf v}_i \perp {\bf v}_j$ for all $i \neq j.$

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$AA^*=I$. Because of $\det A = \det A^*$ and $\det I = 1$, hence $\det A\neq0$ and rows of matrix $A$ are linearly independent. So rows form basis. And it's orthonormal because row $i$ multiplied by transposed itself equals $1$, and row $i$ multiplied by transposed row $j$ ($i\neq j$) equals $0$

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