Question 1036149
If the set { {{{v[1]}}}, {{{v[2]}}} } is a linearly independent set in {{{R^2}}}, then it is also a basis for {{{R^2}}}.

==> for any non-zero vector v in {{{R^2}}}, 

{{{v = c[1]*v[1] + c[2]*v[2]}}} for some constants {{{ c[1]}}} and {{{c[2]}}} both not necessarily zero.  

Left multiplying the previous equation with A, we get


{{{Av = c[1]*Av[1] + c[2]*Av[2] =  c[1]*THETA + c[2]*THETA = THETA}}}, where {{{THETA}}} is the zero-vector in {{{R^2}}}.

In other words, the effect of left-multiplying A with any non-zero vector in {{{R^2}}} is to turn it into the zero vector.

Only the 2x2 zero matrix has this effect in {{{R^2}}}, and therefore {{{A = (matrix(2,2,0,0,0,0))}}}.