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We know linear transformation $T$ over two vector spaces $V,W$ and the rank $r$ (dimension of image $T$) of $T$ . We also know matrix representation $M$ of $T$. Why rank of $M$ is $r$?

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How do you define the rank of M ? Because my definition of the rank of a matrix is just the rank of the linear application which is symbolised by this matrix in caconical base. – Adrien Boulanger Mar 6 '12 at 14:16
Rank of a matrix is the order of a largest submatrix with non zero determinant. – user12290 Mar 6 '12 at 14:34
up vote 2 down vote accepted

Using your definition of rank as the order of the largest submatrix with non-zero determinant (call this $s$), a proof is as follows. Write $e_i$ for the basis vector of $V$ represented by the column vector with $0$s everywhere except for a $1$ in the $i$-th position, and (abusing notation), also write $e_i$ for this column.

First we show $r\geq s$. If the largest submatrix with non-zero determinant arises from taking rows $r_{i_1},\ldots,r_{i_s}$ (and then some $s$ columns) then in particular the rows $r_{i_1},\ldots,r_{i_s}$ are linearly independent, so the vectors $T(e_{i_j})$ corresponding to $Me_{i_j}=r_{i_j}^T$ (for $j=1,\ldots,s$) are linearly independent, and the dimension of the image is at least $s$.

To show $s\geq r$, note that the rows of $M$ span an $r$-dimensional subspace of $K^n$, where $K$ is your field. This is because:


so by varying $v_1,\ldots,v_n$, we find that the span of rows of $M$ is (or rather represents) the image of $T$, which has dimension $r$. So in particular (by sifting) there must be $r$ linearly independent rows of $M$. Any submatrix of $M$ formed from these $r$ rows has non-zero determinant, since its rows are linearly independent, and so the maximal size of such a submatrix is at least $r$.

So $r\geq s$ and $s\geq r$, and hence $s=r$.

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Thank you very much. – user12290 Mar 6 '12 at 16:29

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