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let n x n matrix A be a nonsingular
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show that , if
\[\left\{ V _{1},V _{2},...,V _{n} \right\}\]
is the basis for \[\mathbb{R} ^{n}\]
then \[\left\{ AV _{1},AV _{2},....AV _{n} \right\}\] is also the basis for \[\mathbb{R} ^{n}\]
(i.e the linear system Ax=0 has only the trivial solution x=0)
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One way to do this is to proceed by contradiction. Assume that:\[AV_1,AV_2,AV_3,\ldots ,AV_n \]is not a basis (is not linearly independent), and show that contradicts\[V_1,V_2,\ldots , V_n\]being linearly independent.
c1V1+...cnVn=X belonging Rn so AX=Y span Rn also because X=A^-1Y exsist.
Then c1AV1+...cnAVn=Y is also a basis of R^n
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