Jodles
Reputation
306
Top tag
Next privilege 500 Rep.
Access review queues
 Oct 20 comment Different results for row reduction in Matlab @J.M. I am using MATLAB 7.11.0 (R2010b). Oct 20 revised Different results for row reduction in Matlab Added screenshot from MATLAB Oct 20 comment Different results for row reduction in Matlab @J.M. and Agusti, I think I might have narrowed it down to a rounding error, causing the top and bottom row to be slightly different (but not shown due to lack of decimals shown in Matlab). As you are both more experienced here than me, do you think I should answer my own post or delete it? Thank you! Oct 20 revised Different results for row reduction in Matlab Possible solution Oct 20 asked Different results for row reduction in Matlab Oct 19 accepted Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set Oct 10 comment Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set Thank you for your answer, Arturo! I think my problem might stem from a misunderstanding of what linear independence requires. I thought for functions to be linearly independent, they need to be so for all $\mathbb{R}$; however, it seems like it's the other way around: i.e. if a set of functions are not linearly dependent for all $\mathbb{R}$, then they are linearly independent. So if we can show one example of linear independence, then we're done. Is that the gist of what you were saying? Oct 10 comment Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set Thank you for your answer, André! Forgive me for asking, but it is not obvious to me why the fact that $\cos t$ can take on more than 6 different values makes a contradiction? Oct 8 asked Showing that $\{ 1, \cos t, \cos^2 t, \dots, \cos^6 t \}$ is a linearly independent set Oct 8 comment Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set Thank you for the counterexample! Oct 8 accepted Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set Oct 8 asked Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set Sep 18 awarded Enthusiast Sep 4 comment Showing that $\mathbf{u}=\mathbf{0}$ for $c\mathbf{u}=\mathbf{0}$, $c$ nonzero Good point, thanks! Sep 4 accepted Showing that $\mathbf{u}=\mathbf{0}$ for $c\mathbf{u}=\mathbf{0}$, $c$ nonzero Sep 4 comment Showing that $\mathbf{u}=\mathbf{0}$ for $c\mathbf{u}=\mathbf{0}$, $c$ nonzero @Srivatsan: Good to know, thanks! Sep 4 comment Showing that $\mathbf{u}=\mathbf{0}$ for $c\mathbf{u}=\mathbf{0}$, $c$ nonzero @Chandrasekhar: Thanks! Sep 4 asked Showing that $\mathbf{u}=\mathbf{0}$ for $c\mathbf{u}=\mathbf{0}$, $c$ nonzero Sep 3 revised Rotation of a vector distribution to align with a normal vector LaTeX for arctan, arccos. Sep 3 suggested approved edit on Rotation of a vector distribution to align with a normal vector