Proof that (x^t)Ax>0 if Eigenvalues of Matrix A > 0

[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.

 

[blue_raver22]

Your proof is on the right track, but it is not complete. You have correctly identified that if x is an eigenvector of A with eigenvalue λ, then (x^t)Ax = λ(x^t)x > 0 since λ and (x^t)x are both positive.

However, this only proves that (x^t)Ax > 0 for eigenvectors of A. To prove it for all vectors x, you need to use the fact that any vector x can be written as a linear combination of eigenvectors of A. This is because eigenvectors of A form a basis for the vector space.

So, let x be any nonzero vector. Then, we can write x as a linear combination of eigenvectors of A, say x = c1v1 + c2v2 + ... + cnvn, where c1, c2, ..., cn are scalars and v1, v2, ..., vn are eigenvectors of A.

Substituting this in (x^t)Ax, we get:

(x^t)Ax = (c1v1 + c2v2 + ... + cnvn)^t A (c1v1 + c2v2 + ... + cnvn)

= (c1v1)^t A (c1v1) + (c2v2)^t A (c2v2) + ... + (cnvn)^t A (cnvn)

= c1^2 (v1^t A v1) + c2^2 (v2^t A v2) + ... + cn^2 (vn^t A vn)

Now, since v1, v2, ..., vn are eigenvectors of A, we know that (v1^t A v1), (v2^t A v2), ..., (vn^t A vn) are all positive (as shown in your initial proof). And since c1^2, c2^2, ..., cn^2 are also positive (since they are squares of real numbers), we can conclude that each term in the above expression is positive.

Therefore, (x^t)Ax is a sum of positive terms and hence, it is positive. This proves that (x^t)Ax > 0 for all nonzero vectors x, given that the eigenvalues of A are positive.

In summary, your proof was correct for eig