# The connection between the rank of a matrix and its zero-mode eigenvectors

I would be most thankful if you could help me prove that if an arbitrary n by n matrix has rank m < n, then the matrix has (n-m) linearly independent eigenvectors corresponding to the eigenvalue zero. Thank you!

if rank of $A_{n*n}$ is $m <n$ then Ax=0 will have non trivel answer as $x_1$and so 0 will be a $eigen value$ $Ax_1=0x_1$.
Now we try to find eigen vectors of zero egien value: (A-0I)x=0 Ax=0 and so $X_1$ and other non trivel answersof Ax=0 will be eigen vector correspondence with zero eigen value too since Ax=0 has n-m independence answer so number of independ eigen vectors corresponding to 0 eigen value is n-m too!:)