I have no idea about how to prove this statement hence require a proof. The only definition I have is that rank is the highest order square submatrix of a matrix which has non-zero determinant.Our professor has given the statement in question as a definition without any prrof hence asking for help.Thanks!
[Math] Why is the number of non zero rows in row echelon form of the matrix the rank of the matrix
matrix-rank
Related Solutions
Let $J= \operatorname{diag}(J_1,...,J_k)$ be the Jordan normal form.
Note that $J^2= \operatorname{diag}(J_1^2,...,J_k^2)$
The rank of $J$ is given by the sum of the ranks of the blocks that is, $\operatorname{rk} J = \sum_k \operatorname{rk} J_k$. Similarly, $\operatorname{rk} J^2 = \sum_k \operatorname{rk} J_k^2$.
It follows from the hypothesis that the rank of the Jordan block corresponding to the zero eigenvalue is zero. That is, the Jordan block is identically zero (and so is the square of the Jordan block).
For the blocks $J_k$ corresponding to non zero eigenvalues, we have $\operatorname{rk} J_k = \operatorname{rk} J_k^2$.
It follows that $\operatorname{rk} J = \operatorname{rk} J^2$.
Alternative:
Let $N = \ker A$. We see that $z=\dim N$ is the number of zero eigenvalues of $A$. Let $b_1,..,b_z$ be a basis for $N$ and complete the basis with $b_{z+1},...,b_n$. Note that $N$ is $A$ invariant and so in this basis, $A$ has the form $\begin{bmatrix} 0 & A_{12} \\ 0 & A_{22}\end{bmatrix}$. Furthermore, we must have $\det A_{22} \neq 0$ otherwise $A$ would have more that $ z$ zero eigenvalues.
Then we want to show that $\ker A^2 = N$. Note that $A^2$ has the form $\begin{bmatrix} 0 & A_{12}A_{22} \\ 0 & A_{22}^2\end{bmatrix}$ and it follows from this that if $x \in \ker A^2$ then $x \in \ker A$.
If you agree that row operations don't affect the rank of the matrix, then this follows from one of the definitions of the rank of a matrix, which is the dimension of the row space of the matrix.
The rows of that matrix span a vector space, and the dimension of that vector space is equal to the number of non-zero vectors that span it.
The dimensionality is clearly not greater than the number of rows. But its also not less: there are no redundant vectors, because every row contains an entry where it is the only non-zero element in its column, so it is impossible to build that row from a linear combination of the other rows.
Best Answer
The rank of a matrix also tells you how many linearly independent vectors it has i.e. if rank is 2 then there will be 2 linearly independent vectors in the matrix. And these vectors will be the columns (given your vectors are the columns of the matrix) that contain leading 1's in the rref of said matrix. And if you observe, you will see that the number of leading 1's will always be equal to the number of non-zero rows in the rref of a matrix. Thus rank = # of non-zero rows in the rref of a matrix.