mathematical proof

[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.

[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}
[tex]= 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.

[hgfalling]

You'll have a hard time proving this statement because it's not true.

Consider the following matrix:


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.

[HallsofIvy]

You'll have a hard time proving this statement because it's not true.

Consider the following matrix:


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.
Well done.