Yes, you have! The actual proof goes the other way: start by taking and going "up" your list of steps. But since it is clear that youcando that, what you have shown is enough. (This is sometimes called "synthetic" proof.)

There are NO "axes" in a general vector space. You may be thinking of basis vectors but there are an infinite number of bases for a vector space, just as you can have axes any where on a graph.A is singular when is a unit vector on one of the axis, and is a unit vector in the opposite direction of .For what and is singular:

Ex.

But I think you have a good starting idea. Try this in 2 dimensions first. If u= <a, b> and v= <c, d> then uv*= and so I+ uv*= . What is the determinant of that matrix? What has to be true of a, b, c, and d so that determinant is 0?

If singular, what is :

has rank m, and has rank 1.

If is singular has rank .

has then one free variable and its nullspace has rank 1.

It is a line in

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Just throwing around some ideas on the last two questions as I hate to post a question without attempt.

It seems that I can not get any further by my self,so I'm hoping some of you might spare a few moments.

Thanks.