# The rank of a matrix

1. Apr 15, 2009

### jeff1evesque

Theorem: Let A be an m x n matrix. If P and Q are invertible m x m and n x n matrices, respectively, then
(a.) rank(AQ) = rank(A)
(b.) rank(PA) = rank(A)
(c.) rank(PAQ) = rank(A)

Proof:
$$R(L_A_Q)$$ = $$R(L_AL_Q)$$ = $$L_AL_Q(F^n)$$ = $$L_A(L_Q(F^n))$$= $$L_A(F^n)$$ = $$R(L_A)$$

since $$L_Q$$ is onto. Therefore,
rank(AQ) = dim(R($$L_A_Q$$)) = dim(R($$L_A$$)) = rank(A). (#1)

Question1: How is $$L_Q$$ onto?
Question2:How does the onto-ness imply (#1)?
Question3:Can anyone help me/supply ideas for the proof for parts (b.) and (c.) of the theorem?

NOTE: the symbol R denotes the terminology of images.

Last edited: Apr 15, 2009
2. Apr 15, 2009

### quasar987

Answer 1: Q is an invertible matrix. This is the same as saying that $L_Q$ is an invertible linear map. And this is the same as saying that $L_Q$ is 1-1 and onto.

Answer 2:In the proof, they said " [...] since $L_Q$ is onto." to justify the step "$$L_A(L_Q(F^n))=L_A(F^n)[/itex]". Now that you have $R(L_A_Q)=R(L_A)$, it follows in particular that $dim(R(L_A_Q))=dim(R(L_A))$. But by definition of the rank of a matrix, we have $rank(AQ) = dim(R(L_{AQ} ))$ and $rank(A) = dim(R(L_A ))$. Btw - the "terminology of images" is not a recognized term in mathematics and nor is the symbol R for it. You should use Im(f) instead of R(f) and this is called the image of the map f. 3. Apr 15, 2009 ### jeff1evesque Oh yeah, i should have remembered the idea of an invertible matrix having properties that equivalent. And the answer for part C is simple once parts a and b is established. I think your explanation for (#1) is pretty good, however, for me it is still a little fuzzy- could you try to explain it to me in another way? 4. Apr 16, 2009 ### sanctifier Notations: L(V,W) stands for a vector space of linear transformations form vector space V to W. L(V) stands for a vector space of linear transformations form vector space V to itself. rk(?) stands for the rank of "?". ker(?) stands for the kernel of a linear transformation "?". im(?) stands for the image of "?". inv(?) stands for the inverse of a linear transformation "?". Answer 1: Think about the kernel of a linear transformation, if the inverse of a linear transfomation exists, then its kernel is {0}, i.e., the zero vector. In other words, the only way that makes 2 distinct vectors map to a same vector is that these two vectors belong to the kernel of the linear transfomation and their same mapping can only be {0}, since: σu=σv ←→ σ(u-v)=0 ←→ ker(σ) σ∈L(V) and u,v∈V Answer 2: Your first two questions are identical to: σ,τ,inv(τ)∈L(V), rk(στ) = rk(τσ) = rk(σ) Let φ=στ, according to dim(ker(φ)) + rk(φ)=dim(V), rk(φ)=dim(V) - dim(ker(φ)), dim(V) is a fixed number, the only thing need be considered is ker(φ). The mapping process of φ can be decomposed to two steps: 1, mapping a vector to im(τ) by τ; 2, mapping the result of 1 to im(σ) by σ. As inv(τ)∈L(V), namely, ker(τ) = {0}, so step 1 will map V to V, ker(φ)=ker(τ)={0} for now, but inv(σ)∈L(V) is unknow, so the decisive factor is ker(σ) and after step 2, ker(φ)=ker(σ). You can analyse τσ in a similar way. Answer 3: Let μ=τστ, based on answer 2, rk(στ)=rk(φ)=rk(σ), so μ=τφ, it's the same question mentioned in answer 2. 5. Apr 16, 2009 ### quasar987 Which part is fuzzy to you? 6. Apr 16, 2009 ### jeff1evesque Question: How do we justify "[tex]L_A(L_Q(F^n))=L_A(F^n)[/itex]" since it was onto- sorry for asking such a silly question. Also for the proof of part (b.) to this theorem, I have the following outlined: [tex]dim(R(L_A))$$ = $$dim(L_PR(L_A))$$ = $$dim((L_P(L_A(F^n)))$$ = $$dim(R(L_PL_A))$$ = $$dim(R(L_P_A))$$ = $$rank(PA)$$

but the first equality apparently hinges on the result of the following problem:
Let V and W be finite dimensional vector spaces and T: V-->W be an isomorphism. Let $$V_0$$ be a subspace of V:
Prove that $$dim(V_0)$$ = $$dim(T(V_0)).$$

Question: Is there anyway you could help me prove this new question?

Thanks again

7. Apr 16, 2009

### quasar987

Ask yourself what does it mean that $L_Q$ is onto. It means precisely that $L_Q(F^n)=F^n$.

That's very good work. Indeed, if you could just prove this new question, then (b) would be solved.

Recall that by definition, the vector space $V_0$ has dimension d if it admits a set of d linearly independent vectors that span $V_0$ (i.e. a basis of d elements). So, suppose $\{e_1,...,e_d\}$ is a basis for $V_0$. What can you say about the sets $\{T(e_1),...,T(e_d)\}$?

8. Apr 17, 2009

### jeff1evesque

Is there any way to prove this problem without your specified definitions above, with a more emphasis on the idea of "isomorphism"? The reason I ask about this is because it isn't until the next theorem that wee know ...the rank of a matrix is the dimension of the subspace generated by its columns- in particular rank(A) = $$dim(R(L_A))$$ = dim( $$span({a_1, a_2, .., a_n})$$, where $$a_n$$ are the jth column of A.

Thanks,

JL

9. Apr 17, 2009

### quasar987

Not really, no.