Why Does a Rank One Matrix with a=1 Still Have Solutions?

[Carbon273]

TL;DR

What conditions will there be no solutions, a unique solution and a p-parameter of solutions? If possible may you express in particular and homogenous vectors. I'm a bit rusty so bear with me, trying to reestablish my understanding here.

 

[StoneTemplePython]

I'd strongly suggest multiplying by a permutation matrix $\mathbf P$ so that the LHS matrix is

$\mathbf {11}^T + (a-1)\mathbf I$
and the RHS has the natural progression of $1, a, a^2$

assuming this is in reals, you should be able to confirm that any $a \neq 1, -2$ implies an invertible matrix. And $a=1$ still has (at least one) solution because... and as for $a=-2$ well...

 

[Carbon273]

Hmm why is that you used that expression for the LHS. I am curious, where does the 11^T comes from?

 

[StoneTemplePython]

Hmm why is that you used that expression for the LHS. I am curious, where does the 11^T comes from?

because $\mathbf {11}^T$ is the ones matrix and it is easy to work with. The (scaled) Identity matrix is even easier to work with.

and the RHS is most natural as $1, a, a^2$ -- this is called the moment curve.

 

[mathman]

No solution when the determinant of the matrix is 0.

 

[StoneTemplePython]

No solution when the determinant of the matrix is 0.

This is not correct.

The matrix is rank one when $a=1$ and yet there are $\gt 0$ solutions. e.g. any of the standard basis vectors will do for x here.

I was trying to get OP to come to this conclusion when I said

And $a=1$ still has (at least one) solution because...