91Ó°ÊÓ

Here, it is given that \(A\) is an \(m \times n\) matrix. So, \({A^T}A\) be an \(n \times n\) matrix since \({A^T}A\) is invertible. Therefore, the rank of \({A^T}A\) is \(n\).

Now, the dimension of \({\rm{Nul}}\,{A^T}A\)will be \(\dim \left( {Nul\,{A^T}A} \right) = n - n = 0\).

It is known that if \(A\) is a \(m \times n\) matrix, then a vector \(x\) in \({\mathbb{R}^n}\) satisfies \(Ax = 0\) if and only if \({A^T}Ax = 0\). Hence,

\(\begin{aligned}{}{\rm{Nul}}\,A &= {\rm{Nul}}\,{A^T}A\\\dim \left( {{\rm{Nul}}\,A} \right) &= \dim \left( {{\rm{Nul}}\,{A^T}A} \right)\\\dim \left( {{\rm{Nul}}\,A} \right) &= 0\end{aligned}\)

The dimension of column space of \(A\) is given by:

\(\begin{aligned}{}\dim \left( {{\rm{Col}}\,A} \right) & = {\rm{rank}}\,A\\ &= n - 0\\ &= n\end{aligned}\)

Therefore, the above calculation shows that the columns of \(A\) are linearly independent.