Linear Algebra, orthogonal columns and length

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.

• 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 $\;A\;$ ? – DonAntonio Jan 18 '19 at 13:55

1 Answer

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 $$3$$