# Mathematical proof

1. Aug 22, 2010

### megaman

I am looking for a rather simple proof that if the matrix A has eigenvalues>0, then (x^t)Ax>0 for any vector x not 0.

My first tought was if the eigenvalues are bigger than 0, then (x^t)Ax=(x^t)"eigenvaulue"x="eigenvalue"(x^t)x>0, if x is nonzero and eigenvalues is bigger than 0.

Is this proof good enough? I am a little unsure, becuase I think I have not proven it for all x, just the eigenvectors.

2. Aug 22, 2010

### HallsofIvy

I'm not sure what you mean by '(x^t)"eigenvaulue" x'. which eigenvalue?

It is true that if A is diagonalizable, then it is similar to a diagonal matrix having the eigenvalues on the diagonal.

For example, if A itself is
$$A= \begin{bmatrix}a_1 & 0 \\ 0 & a_2\end{bmatrix}$$
where $a_1$ and $a_2$ are both positive, we have
$$x^*Ax= \begin{bmatrix}x_1 & x_2 \end{bmatrix}\begin{bmatrix}a_1 & 0 \\ 0 & a_2\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}$$
$$= a_1x_1^2+ a_2x_2^2[/itex] which is postive because it is the sum of two positive numbers (or one positive number and 0). But not all matrices are diagonalizable. 3. Aug 22, 2010 ### hgfalling You'll have a hard time proving this statement because it's not true. Consider the following matrix: [tex] A= \begin{bmatrix}9 & 5.5 \\ 1 & 1\end{bmatrix}$$

The eigenvalues of A are both positive (verify).

Let x = (1,-3.1)T

Then
$$x^*Ax= \begin{bmatrix} 1 & -3.1 \end{bmatrix}\begin{bmatrix}9 & 5.5 \\ 1 & 1\end{bmatrix} \begin{bmatrix}1 \\ -3.1\end{bmatrix} = -1.54$$

The relationship you describe is true a) if A is symmetric or b) if (A+A*)/2 is positive definite.

4. Aug 23, 2010

Well done.