# Thread: Re: Linear Algebra-Transition Matrices

1. ## Re: Linear Algebra-Transition Matrices

how to post the problem? Please, guide me

2. ## Re: Linear Algebra-Transition Matrices

Originally Posted by Loser66
how to post the problem? Please, guide me
Just the same way as you posted this. If you are trying to post a picture or an equation we can help you there. Just say the word.

-Dan

3. ## Re: 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)}
It should be in the new post, but I don't know how to do.

4. ## Re: Linear Algebra-Transition Matrices

Originally Posted by Loser66
Let A in L(U,V), B in L(V,W)
a) Prove that rank (A) <= min{dim(U) , dim(V)}
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 $\text{rank}(A)\le\dim(V)$.

Originally Posted by Loser66
b) Prove that rank (BA) <= min{rank (A), rank(B)}
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),
$\text{rank}(BA)=\dim(\text{Im}(BA))\le \dim(\text{Im}(A))=\text{rank}(A).$

5. ## Re: Linear Algebra-Transition Matrices

Thanks for reply. My work is Problem 1.pdf
at problem1, if you have time, please, check the next one and guide me more if I am not right. Thanks in advance. This is my practice review for final. I desperate need help.

6. ## Re: 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.

Let x ∈ R^p $\to$ x ∈ Col BA, then BA(x) = B(Ax) ∈ Col B
I believe you mean, "If $x\in\text{Col}(BA)$, then $x=BAy$ for some $y$, so $x=B(Ay)$, which means $y\in\text{Col}(B)$".

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

Problem 6.
Let a =(a1,a2,a3) and v = (v1,v2,v3) then, a x v = a2v3 – a3v2 –a1v3+a3v1 +a1v2-a2v1
Is $a$ a fixed vector? This should be said explicitly. The right-hand side of the last equality is a number while it should be a vector, i.e., three numbers. The matrix of A is not Av, and it should not depend on v1, v2, v3 (the first matrix). I am not sure how you determine the matrix of $A$. In my book, its columns are images of the basis vectors. For example, (1, 0, 0) is mapped to $(0, a_3,-a_2)$, so that would be the first column. Overall, the matrix of $A$ is
$\begin{pmatrix} 0&-a_3&a_2\\ a_3&0&-a_1\\ -a_2&a_1&0 \end{pmatrix}$