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Suppose $L: M(4,3) \to R^7$ is linear and onto.

  1. Determine the rank L.
  2. Determine the nullity L.

I know that rank is equal to the leading 1's in a reduced row echelon matrix, and the nullity is equal to the number of columns corresponding to free variables. So if I was given an actual matrix I could easily find the rank and nullity, however, I'm not given an exact matrix, only M(4,3). Does that mean I should assume the matrix looks like this:

1 0 0
0 1 0
0 0 1
0 0 0

Therefore the rank is 3 and the nullity is 0? Thus the dimension of M is 3 since:

rank L + nullity L = dim M ?

Thank you.

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up vote 0 down vote accepted
  1. $L$ is onto$\implies Im(L)=\mathbb R^7\implies \text{rank}(L)=7.$

  2. $\text{nullity}(L)=\dim M_{4\times 3}(\mathbb R)-7=12-7=5.$

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How did you know that the rank(L) is the codomain (range) of L? Not the domain? And how is dim M(4,3) 28? – Goose Mar 25 '13 at 1:53
oui. Looks good now. – ncmathsadist Mar 25 '13 at 1:55
@Goose: rank of a linear transformation is the dimension of the image space. $M(4,3)$ is of dim $12$ (not 28, my fault) since $\{E_{ij}=(e_{mn})_{4\times 3}:i=1(1)4,j=1(1)3\}$ forms a basis of $M(4,3)$ where $$e_{mn}=\begin{cases}1\text{ if m=i,n=j }\\0\text{ otherwise}\end{cases}.$$ – Sugata Adhya Mar 25 '13 at 2:01
I thought the dimension of a matrix is the number of leading 1s in the row reduced form? – Goose Mar 25 '13 at 2:06
$M(4,3)$ doesn't denote a unique matrix here. It's the space of all the $4\times 3$ matrices over $\mathbb R.$ – Sugata Adhya Mar 25 '13 at 2:08

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