# Eigenvalues of Gramian Matrix

1. Jan 24, 2015

### I_am_learning

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.

2. Jan 24, 2015

### 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.

3. Jan 24, 2015

### I_am_learning

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.

4. Jan 24, 2015

### 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 $\lambda= \mathbf x^T\mathbf x$.

5. Jan 24, 2015

### I_am_learning

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.

6. Jan 25, 2015

### Hawkeye18

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.

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