# Eigenvalues of Gramian Matrix

if x is a column vector, then a matrix G = x*xT is a Gramian Matrix.
When I tried calculating the matrix G and its eigenvalues for cases when x = [x1 x2]' and [x1 x2 x3]'
by actually working out the algebra, it turned out (if I didn't do any mistakes) that the eigen values are all zeros except one which is equal to (x12+x22 OR x12 + x22 + x32) depending upon the case.
Is this a standard result for a Gramian Matrix to have a single non-zero eigenvalue? If, yes, is there a simpler proof?

Thank you.

Yes that is a standard and pretty simple fact. How did you compute the eigenvalues? Did you compute characteristic polynomials and then find their roots? If so, then yes, there is a much simpler proof.

Yes that is a standard and pretty simple fact. How did you compute the eigenvalues? Did you compute characteristic polynomials and then find their roots? If so, then yes, there is a much simpler proof.
Yes, I solved the roots of characteristic equation, and it was a nasty business for even a 3x3 matrix. :D Would love to know the simpler method.

Let ##\lambda \ne 0## be an eigenvalue, and ##\mathbf v\ne\mathbf 0## be he corresponding eigenvector. That means ##G\mathbf v =\lambda\mathbf v##. But $$G \mathbf v = \mathbf x (\mathbf x^T \mathbf v)$$ and ##(\mathbf x^T \mathbf v)## is a scalar. Therefore $$(\mathbf x^T \mathbf v) \mathbf x = \lambda \mathbf v,$$ so the eigenvector ##\mathbf v## must be a non-zero multiple of ##\mathbf x##. Substituting ##a\mathbf x## (where ##a\ne0## is a scalar) in the above equation, we get that indeed ## a\mathbf x ## is an eigenvector corresponding to ##\lambda= \mathbf x^T\mathbf x##.

I_am_learning
Let ##\lambda \ne 0## be an eigenvalue, and ##\mathbf v\ne\mathbf 0## be he corresponding eigenvector. That means ##G\mathbf v =\lambda\mathbf v##. But $$G \mathbf v = \mathbf x (\mathbf x^T \mathbf v)$$ and ##(\mathbf x^T \mathbf v)## is a scalar. Therefore $$(\mathbf x^T \mathbf v) \mathbf x = \lambda \mathbf v,$$ so the eigenvector ##\mathbf v## must be a non-zero multiple of ##\mathbf x##. Substituting ##a\mathbf x## (where ##a\ne0## is a scalar) in the above equation, we get that indeed ## a\mathbf x ## is an eigenvector corresponding to ##\lambda= \mathbf x^T\mathbf x##.
I cannot see why the bold part should be true?
But, doing the underlined part, i.e. substitution v=ax, I can see that it gives a solution, is that from this you infered that the bold part should hold?

I cannot see why the bold part should be true?
But, doing the underlined part, i.e. substitution v=ax, I can see that it gives a solution, is that from this you infered that the bold part should hold?