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 "L_A(L_Q(F^n))=L_A(F^n)[/itex]&quot;. <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 &quot;terminology of images&quot; 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.

 

[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 "L_A(L_Q(F^n))=L_A(F^n)[/itex]&quot;. <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 &quot;terminology of images&quot; 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.

<br /> <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 "L_A(L_Q(F^n))=L_A(F^n)[/itex]&quot;.

<br /> <br /> <b>Question:</b> How do we justify &quot;L_A(L_Q(F^n))=L_A(F^n)[/itex]&amp;quot; since it was onto- sorry for asking such a silly question.&lt;br /&gt; &lt;br /&gt; Also for the proof of part (b.) to this theorem, I have the following outlined:&lt;br /&gt; &lt;br /&gt; 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)&lt;br /&gt; &lt;br /&gt; but the first equality apparently hinges on the result of the following problem:&lt;br /&gt; Let V and W be finite dimensional vector spaces and T: V--&amp;gt;W be an isomorphism. Let V_0 be a subspace of V:&lt;br /&gt; Prove that dim(V_0) = dim(T(V_0)).&lt;br /&gt; &lt;br /&gt; &lt;b&gt;Question:&lt;/b&gt; Is there anyway you could help me prove this new question?&lt;br /&gt; &lt;br /&gt; Thanks again

 

[quasar987]

Question: How do we justify "L_A(L_Q(F^n))=L_A(F^n)[/itex]&quot; since it was onto- sorry for asking such a silly question.

<br /> 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--&gt;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&#039;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)\}?

 

[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.