Answer
Every square in the matrix will not have $n^{}$ linearly independent eigenvectors because the null space and column space have the potential to overlap, in which case $x^{}$ would be in both. Additionally, there may not be $r^{}$ independent eigenvectors in the column space.
Work Step by Step
Every square in the matrix will not have $n^{}$ linearly independent eigenvectors because the null space and column space have the potential to overlap, in which case $x^{}$ would be in both. Additionally, there may not be $r^{}$ independent eigenvectors in the column space.
