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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.

An elementary column operation interchanges two columns of a matrix, scales one by a nonzero scalar, or adds a multiple of another column. It is an elementary row operation on the transpose. Hence it has the same three effects on the determinant: changing sign, multiplying by the scale factor, or leaving the determinant unchanged.

An elementary matrix is obtained by applying one elementary row operation to the identity matrix. Multiplying $A$ by it on the left performs the same row operation on $A$. Such a matrix is invertible: reverse a row swap, reverse a nonzero row scaling by its reciprocal, or reverse adding a multiple of another row by adding its negative. Right multiplication gives an elementary column operation. In particular these operations preserve matrix rank.

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: