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Lin Alg Proofs and Counterexamples

I have compiled 57 prove or disprove lin alg questions for my final; however, these may be useful to all.

Contributors to some of the solutions are HallsofIvy, Failure, Tikoloshe, jakncoke, tonio, and Defunkt.

If you discover any errors in one of the solutions, then feel free to reply with the number and correction.

**Moderator Edit:**

1. If you want to thank dwsmith, please click on the Thanks button (do NOT post replies here unless you have a suggestion or erratum).

2. The original thread can be viewed at http://www.mathhelpforum.com/math-he...tml#post511378.

Re: Lin Alg Proofs and Counterexamples

Re: Lin Alg Proofs and Counterexamples

I have a question which seems to be little elementary. But if someone gives me a proof or explanations I would be happy and indebted to them for that.

Though it seems to be elementary I hope you would clear the doubt if your precious time permits.

My doubt is regarding the rank of a matrix. Let A be a matrix of order mXn and B be a matrix of order nXp with entries from a field of charactersitic zero. If A is a full row rank matrix (need not be a square matrix) then is it true that rank(AB)=rank(B)? Does it depend on the field? I ask this question because in the field of complex numbers I have a counter example.

Take $A=\begin{pmatrix} 2 & \dfrac{1+\sqrt(3/5)i}{2} & \dfrac{1-\sqrt(3/5)i}{2}\\ 1 & 1 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1/2 & 1\\ \dfrac{2}{1+\sqrt(3/5)i} & 1\\ \dfrac{2}{1-\sqrt(3/5)i} & 1 \end{pmatrix}$. See that rank(A)=rank(B)=2, But rank(AB)=1. Is the result not true for $F=\mathbb{C}$?

Re: Lin Alg Proofs and Counterexamples

Quote:

Originally Posted by

**shahul** I have a question which seems to be little elementary. But if someone gives me a proof or explanations I would be happy and indebted to them for that.

Though it seems to be elementary I hope you would clear the doubt if your precious time permits.

My doubt is regarding the rank of a matrix. Let A be a matrix of order mXn and B be a matrix of order nXp with entries from a field of charactersitic zero. If A is a full row rank matrix (need not be a square matrix) then is it true that rank(AB)=rank(B)? Does it depend on the field? I ask this question because in the field of complex numbers I have a counter example.

Take $A=\begin{pmatrix} 2 & \dfrac{1+\sqrt(3/5)i}{2} & \dfrac{1-\sqrt(3/5)i}{2}\\ 1 & 1 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 1/2 & 1\\ \dfrac{2}{1+\sqrt(3/5)i} & 1\\ \dfrac{2}{1-\sqrt(3/5)i} & 1 \end{pmatrix}$. See that rank(A)=rank(B)=2, But rank(AB)=1. Is the result not true for $F=\mathbb{C}$?

another example

$A=\begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$

$B=\begin{pmatrix} 1 & 0\\ 0 & 1\\ 0 & 1 \end{pmatrix}$

over the real numbers.

In general rank(AB) $\displaystyle \leq $ min{rank A, rank B}

Re: Lin Alg Proofs and Counterexamples

Yes #Idea. But would anybody help me to see what are the conditions for the matrix $A$ to have rank(AB)=rank(B)?

Re: Lin Alg Proofs and Counterexamples

Actually my question is to prove a conjecture. It is as follows:

Given a matrix $A$ of order $m\times n$ with entries from a field of characteristic zero. Define $A^\theta$ as the transpose of the matrix obtained from $A$ by replacing each of its non-zero elements by their inverse and leave zeros as such. Show that $Rank(AA^\theta)=Rank(A^\theta A)=min\{Rank(A),Rank(A^\theta)\}$.

As per my example It is found to be false for the field of complex numbers. Will it be true for the field of real numbers?

Re: Lin Alg Proofs and Counterexamples

Try the matrix

$A=\begin{pmatrix} 0 & 0 & 2 & 1\\ 2 & 2 & 2 & 2\\ 1 & 2 & 0 & 1\end{pmatrix}$

Re: Lin Alg Proofs and Counterexamples

Thank you so much for this eye-opener.