Question 356159
Consider the linear combination:
(a1)(y1)+(a2)(y2)+(a3)(y3)+....+(a(n-1))(y(n-1))+(an)(yn)= 0, a1, a2, a3,...an are scalar coefficients.  
For the purpose of contradiction, suppose that {y1, y2, y3,...,yn} is a linearly dependent set.  Therefore not all of a1, a2, a3, ...an are equal to zero, by definition.  Since A is nonsingular, {{{A^-1}}} exists.
Now 
(a1)(y1)+(a2)(y2)+(a3)(y3)+....+(a(n-1))(y(n-1))+(an)(yn)= 
(a1)(Ax1)+(a2)(Ax2)+(a3)(Ax3)+....+(a(n-1))(Ax(n-1))+(an)(Axn)= 
A((a1)(x1)+(a2)(x2)+(a3)(x3)+....+(a(n-1))(x(n-1))+(an)(xn))= 0.
Since A is nonsingular, this means that {{{A^1}}} exists.  Left-multiply the equation
A((a1)(x1)+(a2)(x2)+(a3)(x3)+....+(a(n-1))(x(n-1))+(an)(xn))= 0
by {{{A^-1}}}.  This means that 
(a1)(x1)+(a2)(x2)+(a3)(x3)+....+(a(n-1))(x(n-1))+(an)(xn)= 0,and 
not all a1, a2, a3, ...an are equal to zero, CONTRADICTION, because {x1, x2, x3, ...xn} is a linearly independent set.  Therefore 
{y1, y2, y3,...,yn} has to be a linearly independent set.