Theorem: Rank of a Matrix: Proof & Questions

[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.

 

[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 "[tex]L_A(L_Q(F^n))=L_A(F^n)[/itex]". <br /> 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 $rank(A) = dim(R(L_A ))$.<br /> <br /> 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 <u>image </u>of the map f.[/tex]

 

[jeff1evesque]

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 "[tex]L_A(L_Q(F^n))=L_A(F^n)[/itex]". <br /> 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 $rank(A) = dim(R(L_A ))$.<br /> <br /> 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 <u>image </u>of the map f.[/tex]

$$<br /> Oh yeah, i should have remembered the idea of an invertible matrix having properties that equivalent. And the answer for part C is simple once parts a and b is established. I think your explanation for (#1) is pretty good, however, for me it is still a little fuzzy- could you try to explain it to me in another way?$$

 

[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.

 

[quasar987]

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?

Which part is fuzzy to you?

 

[jeff1evesque]

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 "[tex]L_A(L_Q(F^n))=L_A(F^n)[/itex]".[/tex]

[tex] <br /> <b>Question:</b> 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.<br /> <br /> Also for the proof of part (b.) to this theorem, I have the following outlined:<br /> <br /> $$dim(R(L_A))$$ = $$dim(L_PR(L_A))$$ = $$dim((L_P(L_A(F^n)))$$ = $$dim(R(L_PL_A))$$ = $$dim(R(L_P_A))$$ = $$rank(PA)$$<br /> <br /> but the first equality apparently hinges on the result of the following problem:<br /> Let V and W be finite dimensional vector spaces and T: V-->W be an isomorphism. Let $$V_0$$ be a subspace of V:<br /> Prove that $$dim(V_0)$$ = $$dim(T(V_0)).$$<br /> <br /> <b>Question:</b> Is there anyway you could help me prove this new question?<br /> <br /> Thanks again[/tex][/tex]

 

[quasar987]

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.[/tex]

[tex] Ask yourself what does it mean that $L_Q$ is onto. It means precisely that $L_Q(F^n)=F^n$.<br /> <br /> <blockquote data-attributes="" data-quote="jeff1evesque" data-source="post: 2163397" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> jeff1evesque said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Also for the proof of part (b.) to this theorem, I have the following outlined:<br /> <br /> $$dim(R(L_A))$$ = $$dim(L_PR(L_A))$$ = $$dim((L_P(L_A(F^n)))$$ = $$dim(R(L_PL_A))$$ = $$dim(R(L_P_A))$$ = $$rank(PA)$$<br /> <br /> but the first equality apparently hinges on the result of the following problem:<br /> Let V and W be finite dimensional vector spaces and T: V-->W be an isomorphism. Let $$V_0$$ be a subspace of V:<br /> Prove that $$dim(V_0)$$ = $$dim(T(V_0)).$$<br /> <br /> <b>Question:</b> Is there anyway you could help me prove this new question? </div> </div> </blockquote>That's very good work. Indeed, if you could just prove this new question, then (b) would be solved.<br /> <br /> Recall that by definition, the vector space $V_0$ has <i>dimension d</i> if it admits a set of d linearly independent vectors that span $V_0$ (i.e. a <i>basis </i>of d elements). So, suppose $\{e_1,...,e_d\}$ is a basis for $V_0$. What can you say about the sets $\{T(e_1),...,T(e_d)\}$?[/tex]

 

[jeff1evesque]

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 $\{T(e_1),...,T(e_d)\}$?

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

 

[quasar987]

Not really, no.