# Let a be an n×1 matrix of real constants. How do you know a′a ≥ 0?

## Expert Answers I may have deleted something by mistake or made a typo that didn't render in math mode. It should read:

The product `a'a` is then

`[[a_(11),a_(21),...,a_(n1)]][[a_(11)],[a_(21)],[.],[.],[.],[a_(n1)]]=[a_(11)^2+a_(21)^2+...+a_(n1)^2]`

Approved by eNotes Editorial Team Let `a=[[a_(11)],[a_(21)],[.],[.],[.],[a_(n1)]].` The product `a'a` is then

`[[a_(11)],[a_(21)],[.],[.],[.],[a_(n1)]]=[a_(11)^2+a_(21)^2+...+a_(n1)^2],`

which is a `1xx1` matrix, which usually is just considered a real number. Since the sum of squares of real numbers is always greater than or equal to zero, the proof is complete.

Note that essentially what we did was compute the dot product of `a`(considered as a column vector) with itself, which is always greater than or equal to zero.

Approved by eNotes Editorial Team

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