how to post the problem? Please, guide me

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- Apr 27th 2014, 01:39 PMLoser66Re: Linear Algebra-Transition Matrices
how to post the problem? Please, guide me

- Apr 27th 2014, 07:36 PMtopsquarkRe: Linear Algebra-Transition Matrices
- May 1st 2014, 03:21 PMLoser66Re: Linear Algebra-Transition Matrices
Let A in L(U,V), B in L(V,W)

a) Prove that rank (A) <= min{dim(U) , dim(V)}

b) Prove that rank (BA) <= min{rank (A), rank(B)}

Please, help

It should be in the new post, but I don't know how to do. - May 1st 2014, 04:52 PMemakarovRe: Linear Algebra-Transition Matrices
Let $(e_1,\dots,e_n)$ be a basis of $U$. Then $\text{Im}(A)=\text{span}(Ae_1,\dots,Ae_n)$ (why?), so $\text{rank}(A)=\dim(\text{Im}(A))\le n$. I'll leave it to you to show that $\text{rank}(A)\le\dim(V)$. (Smile)

I assume $A$ is applied first in $BA$. Obviously, $\text{Im}(BA)\subseteq\text{Im}(B)$, so

\[

\text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(B))=\text{rank}(B).

\]

In addition, $\text{Im}(BA)=B(\text{Im}(A))$, so using statement a),

\[

\text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(A))=\text{rank}(A).

\] - May 5th 2014, 04:03 AMLoser66Re: Linear Algebra-Transition Matrices
Thanks for reply. My work is Attachment 30840

at problem1, if you have time, please, check the next one and guide me more if I am not right. Thanks in advance(Shake). This is my practice review for final. I desperate need help.(Headbang) - May 5th 2014, 11:33 AMemakarovRe: Linear Algebra-Transition Matrices
I think problem 1 (a) and (b) are largely correct. If I could make a general advice, it would be to not hesitate to use more English text in complete sentences. It would make the role of the grader, or any reader, much easier.

I assume Col(A) means the column space of A. But in the beginning of problem 1, A and B are defined as linear operators, not matrices. You need to say something about a basis.

Quote:

Let x ∈ R^p $\to$ x ∈ Col BA, then BA(x) = B(Ax) ∈ Col B

Note the the solution in post #4 does not use the rank-nullity theorem.

Problem 6.

Quote:

Let a =(a1,a2,a3) and v = (v1,v2,v3) then, a x v = a2v3 – a3v2 –a1v3+a3v1 +a1v2-a2v1

\[

\begin{pmatrix}

0&-a_3&a_2\\

a_3&0&-a_1\\

-a_2&a_1&0

\end{pmatrix}

\]