Recent content by jonroberts74

so it goes both way "if ##T## is onto ##\Leftarrow\Rightarrow## the range is ##\mathbb{R^n}##" thanks I think I was misunderstanding what range meant in this sense.

Homework Statement I haven't learned kernel yet so if that's of use here I don't know it yet let ##T: \mathbb{R^3} \rightarrow \mathbb{R^2}## where ##T<x,y,z>=<2y,x+y+z>##[/B] prove that the range is ##\mathbb{R^2}## The Attempt at a Solution I know that T is not one-to-one, I checked that...

is that correct?

Homework Statement let ##T:\mathbb{R^3} \rightarrow \mathbb{R^3}## where ##T<x,y,z>=<x-2z,y+z,x+2y>## Is T one-to-one and is the range of T ##\mathbb{R^3}##? The Attempt at a Solution I took the standard matrix A ##\left[\begin{array}{cc}1&0&-2\\0&1&1\\1&2&0\end{array}\right]## det(A)=0 so...

okay perfect thanks! Yeah the more I know the better, I am a math major so I don't want to learn just the bare minimum.

figured this out another way Let A be an m x n matrix and ## s =\displaystyle\sum_{i=1}^{m}\sum_{j=1}^{n} a_{i,j}^2## ##A = \left[\begin{array}{cc}R_{1}\\R_{2}\\.\\.\\.\\R_{m}\end{array}\right] = \left[\begin{array}{cc}C_{1}&C_{2}&.&.&.&C_{n}\end{array}\right]## where ##R_{i},C_{i}## are rows...

oh yeah, I was squaring down the diagonal for whatever reason. I apologize, you are correct.

oh okay so to write the n x n matrix A ##[A]_{i,j} = \displaystyle\sum_{i,j=1}^{n} A_{i,j}##and to continue what you have ##tr(AA^T) = tr([AA^T]_{ij}) = tr(\sum_{k=1}^{m} A_{ik} A^T_{kj}) = (a_{i,1})^2+(a_{i,2})^2+...+(a_{i,n})^2+(a_{1,j})^2+(a_{2,j})^2+...+(a_{n,j})^2 = \sum_{j = 1}^{m}...

I tried find summation notation for a matrix the ones I found were in similar fashion also and ##\left[\begin{array}{cc} 1&4\\2&3\end{array}\right] \left[\begin{array}{cc} 1&2\\4&3\end{array}\right]##wont have the same trace as ##\left[\begin{array}{cc}...

the trace of the the product would be the sum of the squares along the diagonal and s is the entries of A (not the product) squared and summed. I know its true, I proved it for a 3x3 square matrix but it has to be symmetric. heres the 2x2 case Let ##A=\left[\begin{array}{cc} a_{11} & a_{12}...

Homework Statement Prove ##tr(AA^T)=tr(A^TA)=s## where ##s## is the sum of the squares of the entries of A I need help cleaning this up and I don't think my sigma notation is completely correct. The Attempt at a Solution I found the identity $$(AB)^T=B^TA^T$$then applying it to ##AA^T...

It's been a busy day but I worked it out on scratch paper ##\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{3}}{2}} \int_{\frac{1}{2}}^{\sqrt{1-r^2}} rdzdrd\theta## I got the same answer of ##\frac{5\pi}{24}## I want to try is solve it as if it set as ##W=\{(x,y,z)|\frac{1}{2} \le x...

I can see that, but I know the plane z=1 intersects the unit sphere at z=1 and phi is cos(0) at the z-axis so I used that, I did use \rho^2 = 1 on my scratch sheet I just didn't include it in my post. I'll be more careful in the future. I have to take care of some errands but I'll come back and...

##\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{\frac{sec\phi}{2}}^{1}\rho^2 sin\phi d\rho d\phi d\theta## ##\displaystyle \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \frac{\rho^3}{3} sin\phi \Bigg|_{sec\phi/2}^{1} d\phi d\theta####\displaystyle \int_{0}^{2\pi}...

yes the ##\phi## was a typo I had written something out but deleted it all except for that it appears. and I was going back to add in the ##sin\phi## I missed the equation in spherical is ##\rho^2 = 1 ## so ##\rho = 1## But I don't see what was wrong with the way I got 1 as the upper limit I...