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comment Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
@GitGud, Thanks for the additional info!
Jul
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accepted Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
Jul
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comment Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
Great answer, Thanks!
Jul
9
comment Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
@GitGud rank$(BA) \le n$? But $BA_{n \times n}$ so how does that help?
Jul
9
comment Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
@GitGud, Ok, so for $A=0$ it contridicts AB, how about BA then?
Jul
9
revised Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
added 216 characters in body
Jul
9
comment Given $A_{m\times n}$ and $B_{n \times m} (m<n)$. prove that AB is not singular and BA is singular
@DavidButlerUofA: Let me try rephrase, maybe I made a mistake translating the question: Given $A_{m\times n}$ and $B_{n \times m} (m<n)$.1) Is AB Singular?2) Is BA Singular? My answers say that 1 is false and 2 is true.
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