2
$\begingroup$

I'm having trouble with some maths regarding the expression of the matrix quadratic form (i.e. $x^TAx$) and the proof that, where the eigenvalues $\lambda_1,\lambda_2,...,\lambda_n$ are all positive, the quadratic form is positive definite.

My understanding is that the definition of positive-definiteness is when $x^TAx>0$ for all x where at least one element of $x \neq 0$.

My textbook produces the following proof, but I don't understand the last line:

Where $s_i$ = a normalized eigenvector of A, $\lambda_i$ = the corresponding eigenvalue, and $C = [s_1|s_2|...|s_n]$, $C^{-1}AC=D=$ the "diagonalization" of A.

Since the eigenvectors of A are orthonormal, $C^{-1} = C^T$.

Suppose we define a transformation $x=CX$. Then the equation becomes: $x^TAx = (CX)^{T}ACX = X^TDX = \lambda_1X_1^2 + \lambda_2X_2^2+...+\lambda_nX_n^2$.

It follows from this that a quadratic form is positive-definite if and only if all its eigenvalues are positive.

So, in summary, I don't understand why the following derivation true:

$x^TAx = \lambda_1X_1^2 + \lambda_2X_2^2+...+\lambda_nX_n^2$ therefore, for $x^TAx > 0$ for all x where at least one element of $x_i \neq 0$ to be true, $\lambda_1, \lambda_2, ..., \lambda_n > 0$.

Can someone please help with the derivation of the last step?a

Thanks in advance for your help.

  • 1
    I think your definition of positive definiteness may be the source of your confusion. The correct definition is that $A$ is positive definite if $x^TAx>0$ for **all** vectors $x$ other than the zero vector.2012-04-30
  • 0
    @WillOrrick that is what I mean by at least one element of x is not zero. To restate: for all x except x = [0;0]. Thanks for you comment though.2012-04-30
  • 0
    "...therefore, for $x^TAx\gt0$ with at least one element of $x_i\ne0$ to be true, $λ_1$, $λ_2$,...,$λ_n\gt0$." --- I still think you really need to change this to "...for $x^TAx\gt0$ to be true **for all** $x$ having at least one element $x_i\ne0$..." The "for all $x$" allows you to choose an $x$ that will help you reach the conclusion. Since $x^TAx\gt0$ has to be true for any $x$, you can choose $x$ in such a way that only one of the eigenvalues contributes. Then you can conclude that that eigenvalue must be positive. Repeat for every eigenvalue.2012-04-30
  • 0
    Thanks for adding that. I think you also need to correct it in the second gray box. This may help clarify the logic of the proof.2012-04-30
  • 0
    @WillOrrick no problem and fixed in the second gray box too :)2012-05-01
  • 0
    The statement in the second gray box is becoming a little bit hard to parse. I hope this is a fair phrasing: "From $x^TAx=\lambda_1X_1^2+\ldots+\lambda_nX_n^2$, it follows that the statement '$x^TAx\gt0$ for all $x$ with at least one element $x_i\ne0$' implies the statement '$\lambda_1,\ \lambda_2\ ,\ldots,\ \lambda_n\gt0$'." The contrapositive of this is '$\lambda_i\le0$ for at least one $i$ implies that $x^TAx$ is zero or negative for some $x$.' It may be easier to see why the contrapositive is true, which is essentially what leonbloy's answer shows.2012-05-02
  • 0
    correction: ...zero or negative for some **non-zero** $x$.2012-05-02

3 Answers 3