Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $r$?

share|improve this question
    
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
add comment

1 Answer 1

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:

$$M(v_1,\ldots,v_n)^T=v_1Me_1+\cdots+v_nMe_n=v_1r_1+\cdots+v_nr_n$$

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 $s=r$.

share|improve this answer
    
Thank you very much. –  user12290 Mar 6 '12 at 16:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.