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$A$ is a matrix that is $k$ x $n$ and is $\in\mathbb{R}$, where n is greater than k. Also, the set of row vectors is orthonormal w.r.t. dot product.

Show $(A^{T}A)^{2}=A^{T}A$.

I know bits and pieces, but can't seem to connect the dots. Here's what I know. Let $\alpha:V$$\rightarrow W$, where V is the vector space defined by $\mathbb{R}^{n}$ and W is the vector space defined by $\mathbb{R}^{k}$ Let B be the basis for V, $B={{v_{1}...v_{n}}}$ and let D be the basis for W, $D={w_{1}...w_{k}}$ Then A is a representation matrix for $\alpha$ and $A^{T}$ is the representation matrix for $\alpha^{*}$, the adjoint operator for $\alpha$. I don't see the connection for a zero dot product making the product matrix squared equal to itself.

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    But what are the entries of that matrix? Use the fact that the rows of $A$ form an orthonormal set.2012-11-10

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Using the orthonormality of the rows of $A$, along with the fact that $n\geq k$, you need only show what $AA^T$ is the $k\times k$ identity matrix.

To see how, write $A=\left[\begin{array}{c}r_1\\\vdots\\r_k\end{array}\right],$ where the $r_j$ are $n$-component row vectors. Orthonormality tells us that for $i,j\in\{1,...,k\}$, we have $r_ir_j^T=\begin{cases}1 & i=j\\0 & i\neq j.\end{cases}$ What does that tell you about the $i$th row, $j$th column entry of the $k\times k$ matrix $AA^T=\left[\begin{array}{c}r_1\\\vdots\\r_k\end{array}\right]\left[\begin{array}{ccc}r_1^T & \cdots & r_k^T\end{array}\right]?$

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    Thank you, could you elaborate as to why?2012-11-10
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If $x_1,\ldots,x_n$ are an orthonormal set, and $ A=\begin{bmatrix}x_1^T\\ \vdots \\ x_n^T\end{bmatrix}, $ then $(AA^T)_{kj}=x_k^Tx_j$. The orthonormality then gives $AA^T=I$. So $(A^TA)^2=A^TAA^TA=A^TIA=A^TA$.

When $A$ is square, the condition $AA^T=I$ already implies $A^TA=I$ (from where of course it follows in particular that $(A^TA)^2=A^TA$).

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    You right. I just never think about non-square matrices! I'll edit accordingly.2012-11-10