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Let $n > k$ be positive integers and let $A \in M_{k×n}(\mathbb{R})$ satisfy the condition that its set of rows is orthonormal with respect to the dot product. Show that $(A^T A)^2 = A^T A$.

I'm not sure if my proof is correct or there is any better proof for this problem. I appreciate any help.

Let $0\neq v\in\mathbb{R}^n$.

Since the rows of $A$ are orthonormal, then we get $AA^T=I_{k\times k}$, hence $\langle A^TAv, A^TAv\rangle=\langle v,(A^TA)^* A^TAv\rangle=\langle v,A^TAA^TAv\rangle=\langle v, A^TAv \rangle $. Note that $(A^T)^*=A.$ As, $v\in \mathbb{R}^n$ is arbitrary, we get $(A^T A)^2 = A^T A$.

Parisina
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2 Answers2

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HINT: $(A^T A)(A^T A) = A^T (A A^T) A$.

Show that the fact that $A$ has orthonormal rows means that the middle matrix is the identity. That's almost immediate from the definition of the matrix product and the word "orthonormal".

John Hughes
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  • Yes, It's kind of trivial to show the middle one is the identity. Thanks. I spent one hour to solve this but the answer is so short! – Parisina Dec 27 '16 at 22:48
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    Lesson: when an exercise in a class or book takes more than an hour, it may be because you're approaching it wrong, and should step back and rethink things for a few minutes. I often find reviewing all relevant definitions helps clear my head. Like "What does it mean for rows to be orthonormal? It means that $a_i \cdot a_j = 0$ or $1$, depending on whether $i \ne j$ or $i = j$ (where $a_i$ is the $i$th row of $A$)." – John Hughes Dec 27 '16 at 22:50
  • So, If $k=n$, I can say that $A$ is invertible and $A^{-1}=A^T$, since we have $AA^T=I$. Hence, $A^TA=I$ which implies columns of A are linearly independent. Right? – Parisina Dec 28 '16 at 04:40
  • I mean if $k=n$, having orthonormal rows implies having orthonormal columns. – Parisina Dec 28 '16 at 04:42
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    Yes, that's correct. – John Hughes Dec 28 '16 at 04:47
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The set of rows of $A$ being orthonormal, means the set of columns of $A^\top$ is orthonormal as well. So we have $$ A^T = (b_1, \dotsc, b_k) \in \mathbb{R}^{n\times k} $$ with $$ b_i \cdot b_j = b_i^T b_j = \delta_{ij} $$ which means $$ A A^T = I $$ so we have $$ (A^T A)^2 = (A^T A)(A^T A) = A^T(A A^T) A = A^T I A = A^T A $$

mvw
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