@whuber I would like your guidance on this one. I realized after your comments that the idea of oversimplifying with drawings probably seemed happier than it is. It didn't take me too long to draw the pictures, and I don't want garbage cluttering the site. This being said, I wonder if including your comments as a caveat at the end of the post could turn a potentially misleading post into a post with good teaching points. I have been reading

this question on the math site, just to realize how much there is to explore on this topic. Incidentally, this answer sounds at this point so esoteric, that unless someone other than me upvotes either one of the answers, I have no idea which one is to be "accepted".

If neither of the answers clarifies the point to you, don't accept either!

But what the second answer is warning against (which they point out, at the very end, does not happen for symmetric matricies is this).

Suppose we have an eigenspace, but also a vector outside of the eigenspace that the matrix sends inside of the eigenspace.

When that happens, the matrix cannot be diagonalizable.

@MatthewDrury Thanks for your answer. Is this the "transversality" that you are describing? All I know about linear algebra I learned from Prof. Strang's free videos on youtube, but now I just came across a potential booster in these lectures. This answer was an eye opener...

@whuber I erased the post! No looking back... It was crap.

@AntoniParellada Transversality is a difficult concept in general, its foundational in differential topology.

In ths context its simpler though.

If you have two subspaces X and Y in a vector space, they are transverse when, if you take a basis for X and a basis for Y, and then union the basisies, the result is still linearly independent.

Said another way.

The intersection of two subspaces is always another subspace.

The two subspaces are transverse if the intersection subspace is {0}, only the zero vector.

A lot of effort in topology goes into reducing other definitions of transversality, for more complicated geometric spaces, to the definition that applies to linear spaces.

@AntoniParellada I don't think that post was crap. I could see what you were trying to do, even if the result didn't really work as intended.

No shame in taking a risk and trying to take a different perspective on something.

hey people :)

anybody have some feedback for me regarding this Q: stats.stackexchange.com/questions/231086/…

would be much appreciated :)

Opinion question: How bad is being a marginally informed statistical simpleton? I am reading about the evaluation of classification models and there are so many militant approaches (for example do not use 'accuracy' scores whatsoever to find your optimal model) and while I see (partially) the reasoning behind them the alternatives (for example, multiple F-tests between classifiers) are just a whole new can of worms themselves.

For instance I want a new binary classifier, will using Cohen's $k$ and AUC be so catastrophic? Yeah, sure I will definitely look at some TPR/TNR metrics as well as lift curves to make sure I am not riding on noise/overfitting but will it be such a disaster otherwise afterwards? I feel it is a bit like an $R^2$ debate. I wouldn't bet my life on it but it is somewhat informative metric after all.

(Just to be clear: I don't advocate picking a model using $R^2$ or AUC alone, I am just asking about reasonable lines in the sand :D )

@AntoniParellada I agree with @Matthew about your eigenvalue post.

As someone who doesnt get much exposure to applied statistics in an academic setting

Is this kind of thing common (second paragraph, with the bold assuming): stats.stackexchange.com/a/232333/74500

@MatthewDrury I guess it's common enough that I wouldn't be surprised to see in a psychology paper, but it is below even the usual abysmal statistical standards of psychology.

(I'm talking about psychology since I'm a psychologist.)

@AntoniP answered