Do Gramian Matrices Have Only One Non-Zero Eigenvalue?

[I_am_learning]

if x is a column vector, then a matrix G = x*x^T 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.

 

[Hawkeye18]

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.

 

[I_am_learning]

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.

 

[Hawkeye18]

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?

Thank you for your help.

 

[Hawkeye18]

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?

Thank you for your help.

No, the "bold" part is true independently of the "underlined" part, they both prove different parts of the statement.

For the "bold" part: we know that $(\mathbf x^T\mathbf v)$ is a number, let call it $\beta$. Then the equation is rewritten as $\beta\mathbf x = \lambda\mathbf v$, and solving it for $\mathbf v$ gives us $\mathbf v = (\beta/\lambda) \mathbf x$.

Now, the constant $\beta$ depends on the unknown $\mathbf v$, and we do not know what $\lambda$ is, so we cannot say from here that $\mathbf v = (\beta/\lambda) \mathbf x= a\mathbf x$ is an eigenvector. But what we can say is that if $\mathbf v$ is an eigenvector corresponding to a non-zero eigenvalue $\lambda$, then it must be a non-zero multiple of $\mathbf x$.

Substituting then $\mathbf v=a\mathbf x$ we get that it is indeed an eigenvector and find $\lambda$. So the "underlined" part give you that $\mathbf v=a\mathbf x$ is an eigenvector, and gives the corresponding eigenvalue. The "bold"" part shown that there are no other eigenvectors corresponding to a non-zero eigenvalue.