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Suppose A is a 3x3 matrix whose columns are orthogonal and the length (two-norm) of each column equals 4. Then what is $A^T*A$?

How would I even start proving this? I have to remain general such that my reasoning applies to any matrix of that form. I just cannot pick a random 3x3 orthogonal column matrix to see the outcome.

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  • $\begingroup$ Try to find out what is the entry $\;1-1\;$ in the product...Can you see that it is the square of the norm of the first column of $\;A\;$ ? $\endgroup$ – DonAntonio Jan 18 '19 at 13:55
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How you define the element $\alpha_{ij}$ in the matrix $A^TA$? It's the scalar product of the $i$-th row of $A^T$ and the $j$-th column of $A$. But the $i$-th row of $A^T$ is the $i$-th column of $A$ (by definition of transpose). Thus $\alpha_{ij}=0$ if $i \neq j$ since the columns of $A$ are orthogonal, and $\alpha_{jj}=16$. Therefore $A^TA=16I$ where $I$ is the identity matrix of order $3$

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