Linked Questions

46
votes
8answers
18k views

How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?

Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$ I proved $\operatorname{rank}(AB) \...
15
votes
3answers
34k views

Rank product of matrix compared to individual matrices. [duplicate]

Possible Duplicate: How to prove $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))$? If $A$ is an $m\times n$ matrix and $B$ is a $n \times r$ matrix, prove that the rank of matrix $...
1
vote
1answer
68 views

Rank of matrix products is never greater than what [duplicate]

If I have a product of matrices of rank 1, the product is not going to have rank greater than 1. why?
1
vote
0answers
57 views

Rank of product of two matrices [duplicate]

I want to show that $\text {rank} ( AB)\le \min(\text{rank} B, \text{rank} A) $ and when the equality occurs? Please help me with this problem by giving hints, solving, or suggesting a book. Thanks
0
votes
1answer
44 views

What can be said about $\text{Rank}(C)?$ [duplicate]

Assume that $A\in\mathbb{R}^{n\times k}, \ B\in\mathbb{R}^{k\times m}$ and that $\text{Rank}(A)=r, \ \text{Rank}(B)=s.$ What can be said about $\text{rank}(C),$ where $C=AB?$ The only thing I can ...
10
votes
3answers
437 views

Given a square matrix A of order n, prove $\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$

Given $A\in F^{n \times n}$ prove: $$\operatorname{rank}(A^n) = \operatorname{rank}(A^{n+1})$$ $\operatorname{rank}(A^{n+1}) \leq \operatorname{rank}(A^n)$ is easy, just from: How to prove $\text{...
4
votes
3answers
231 views

If rank$(A)=r$, show that rank$(A^\top A)=r$

Let $A$ be $m\times n$ matrix with rank $r=\min(m,n)$. How do we show that rank$(A^T A)$ is $r$.
0
votes
3answers
124 views

Finding the rank of a certain general matrix

If I have an $m \times n$ matrix $A$ and an $n \times m$ matrix $B$ such that $AB=I_m$, how do I go about calculating the rank of $A$ and the rank of $B$? Any clues would be much appreciated!
1
vote
1answer
185 views

Proving $\displaystyle rang(AB) \le \inf(rang(A),rang(B))$

Supposing $\displaystyle A\in \mathbb{M}_{np}(\mathbb{R})$ and $B\in\mathbb{M}_{pq}(\mathbb{R})$: How can prove that: $\displaystyle rang(AB) \le \inf(rang(A),rang(B))$
1
vote
1answer
111 views

Let $A,B$ be $m \times n$ and $n \times m$ matrices, respectively. Prove that if $m > n$, $AB$ is not invertible

We haven't done anything about rank or dimensions or linear dependence / basis or determinants. Possible related facts : A matrix is invertible iff it is bijective as a linear transformation. An ...
0
votes
0answers
175 views

$\det(AB) = 0$ when $A$ has more rows than $B$

Someone posted here that when $A$ has more rows than $B$, $\det(AB)=0$. How?
1
vote
1answer
62 views

Find a matrix A that is diagonalized by matrix P and its transpose to be the identity matrix

If I have a matrix $P_{M\times N}$ where $M>N$, can I find a matrix $A_{M\times M}$ so that it satisfies this equation? $P^{T}AP=I$ where $I$ is identity matrix.
0
votes
1answer
60 views

Rank of linear transformations from vector space to same vector space

I have a question about a homework question so don't expect a full solution. Just if someone could tell me how I should approach this question. I'm not really sure what kind of manipulation is allowed....
1
vote
2answers
54 views

$\text{rank}(BA)\le \text{rank}(A)$ proof

$A:V\to W, B:W\to Z$ be linear map on finite dimensional vector spaces. I need to show $\text{rank}(BA)\le \text{rank}(A)$. I thought like this: suppose $\text{rank}(A)=m$, suppose $\text{rank}(BA)=...
0
votes
2answers
68 views

Let $p(t), q(t) ∈ \mathbb C[t]$ be relatively prime, $A ∈ M_n(\mathbb{C})$. Show that $\operatorname{rank}(p(A))+\operatorname{rank}(q(A)) ≥ n$.

Let $p(t), q(t) ∈ \mathbb C[t]$ be relatively prime, $A ∈ M_n(\mathbb{C})$. Show that $\operatorname{rank}(p(A))+\operatorname{rank}(q(A)) ≥ n$. I have been stumped on this question for quite awhile. ...

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