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 $T$.

Is $r$ the rank of $M$ ?

  • $\begingroup$ 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. $\endgroup$ – Adrien Boulanger Mar 6 '12 at 14:16
  • $\begingroup$ Rank of a matrix is the order of a largest submatrix with non zero determinant. $\endgroup$ – user12290 Mar 6 '12 at 14:34

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 $s=r$.


T: V-> R(Where R is the subspace of W to which T maps V ).

R has a dimension of r[Given] and Let V has a dimension m.

  1. You know Matrix Representation of T wrt to a Fixed basis Of R(say B) and that of V(say A).

Proof That any Linear Transformation from V-> R requires at it first step that their basis Are fixed.

So, Now M(m*r)B(r*1)=T(A)(m*1) where M,B, and T(A) are represented as Matrices and T(A) means the matrix of the Transformed Basis Elements.

MB=T" where the notations denote the respective Matrices.

Now, B is of rank r, T" is of rank m(As T" is the matrix of a Linear Transformation of the BASIS of V)

Now p(T")<=min(p(M),p(B))=min(p(M),r) and By Sylvester's Inequality, Sylvester rank inequality

p(T")>=p(M) => p(T")=p(M)=m

Also, as r is the dimension of R. if m=r it holds.

NOTE: If I have done Wrong ,I will be glad if You Point out my mistake.

COUNTER Example: Take V= $\ E^2 \ $ and R=$\ E^3 \ $ where E is the Euclidian Space.

So m=2,r=3, Here take T(A)= $ \ (e_1,e_2)\ $, B= $ \ (e_1,e_2,e_3) \ $

and Hence Dimension of M = 2.(CHECK!!)

Hence We can Tell it is Equal to r, iff m>=r


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