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The column rank of a matrix is the dimension of its image as a linear map, equivalently the dimension of the span of its columns. Multiplication on the left by an invertible matrix preserves all column relations, and elementary column operations preserve their span. In row echelon form the pivot columns form a basis of the column space. Its dimension equals the number of nonzero rows, and hence the row rank.

The row rank of a matrix is the dimension of the linear span of its row vectors. Its column rank is the dimension of the linear span of its column vectors, equivalently the dimension of the image of the corresponding linear map. The equality of row rank and column rank states that these two numbers are equal for every matrix. Their common value is the rank of a matrix.

The row rank of $A$ is the dimension of the span of its rows in ${\mathbb{R}}^n$; its column rank is the dimension of the span of its columns in ${\mathbb{R}}^m$. Elementary row operations replace rows by invertible linear combinations, so preserve their span. They also preserve every column relation: $Ac=0$ if and only if $EAc=0$ for an invertible elementary matrix $E$. Reduce $A$ to row echelon form $R$, with $r$ nonzero rows. These rows are independent because their leading entries occur in distinct, successively later columns. The $r$ pivot columns are independent by their triangular pivot submatrix, while every column lies in the $r$-dimensional coordinate space supported on the nonzero rows. Thus both ranks of $R$, and hence both ranks of $A$, equal $r$. Row rank equals column rank.

Elementary column operations replace columns by invertible linear combinations and preserve their span. Equivalently, right multiplication by an invertible elementary matrix $F$ preserves row relations, since $c^{T}AF=0$ if and only if $c^{T}A=0$. Therefore either elementary row operations or elementary column operations preserve the matrix rank, and any finite sequence gives $\boxed{\mathrm{rank}(A^{\prime })=\mathrm{rank}(A)}$.

For the block matrices, put $M=\left(\begin{array}{cc}PQ& 0\\ Q& QR\end{array}\right)$. The explicit invertible multiplicationsfollow by first subtracting $P$ times the lower block row from the upper one and then subtracting the first block column times $R$ from the second. Changing the sign of the upper block row gives $N=\left(\begin{array}{cc}0& PQR\\ Q& 0\end{array}\right)$. These operations preserve matrix rank, proving that the two specified block matrices have equal rank. Their off-diagonal form gives $\mathrm{rank}(N)=\mathrm{rank}(Q)+\mathrm{rank}(PQR)$, since the two image components occupy independent coordinate blocks.

Choose $r_1=\mathrm{rank}(PQ)$ independent columns of $PQ$ and $r_2=\mathrm{rank}(QR)$ independent columns of $QR$. Select the corresponding $r_1$ columns from the first block column of $M$ and $r_2$ from the second. In a linear relation among these selected columns, the upper coordinate block forces every coefficient of the first selection to vanish; the lower block then forces every coefficient of the second to vanish. Consequently $\mathrm{rank}(M)\ge r_1+r_2$. Combining this with the equality of block ranks proves the Frobenius rank inequality:

A matrix is in row echelon form if zero rows follow all nonzero rows, each nonzero row has a first nonzero entry (pivot) to the right of the pivot in the preceding row, and entries below each pivot vanish. Elementary row operations bring every matrix to this form: choose a nonzero entry in the earliest nonzero column, swap its row to the top, eliminate entries below it, and continue on the remaining submatrix. The nonzero rows are independent by their successively positioned pivots; the pivot columns are independent and span the column space. Thus the number of pivots equals both row rank and column rank.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this space; multiplication on the right by an invertible matrix preserves all linear relations among the rows. In row echelon form the nonzero rows are linearly independent, so the row rank is the number of pivots. The pivot columns are also linearly independent and span the column space, proving equality with the column rank.