Please verify my proof for this. I try to keep it short.

Let $A\in\mathbb{R}^{n\times n}$ be positive-definite and symmetric. Show that $\langle x,y \rangle_A:=\langle Ax,y \rangle$ defines an inner product.

First note that $\langle x,y \rangle=\langle Ax,y \rangle = (Ax)^Ty=x^TA^Ty=x(Ay)$.

  1. Symmetry follows over the commutativity of the inner product.

$\langle x,y \rangle = x(Ay)=(Ay)x=(Ay)^Tx=y^TA^Tx=y^TAx=\langle y,x \rangle$.

  1. Linearity. Let $x,y,z$ be vectors and $r$ a real number. Then we have

$\langle rx,y \rangle=(rx)^TAy=rx^TAy=r\langle x,y \rangle$


$\langle x+y,z \rangle = (x+y)^TAz=(x^T+y^T)Az=x^TAz+y^TAz=\langle x,z \rangle + \langle y,z \rangle$.

  1. Positive-definiteness of the function.

Since $A$ is positive-definite we have $\langle x,x \rangle = x^TAx>0$ for a non-zero vector $x$. Furthermore we have $\langle 0,0 \rangle = 0^TA0=0$. Thus it follows that $\langle x,x \rangle \geq 0$ for any vector $x$.

Now suppose that $\langle x,x \rangle = 0$ and thus $\langle x,x \rangle = x^TAx=0$. Since $A$ is positive-definite this is the case iff $x=0$. Thus $\langle x,x \rangle=0 \Leftrightarrow x=0$.

Follow up question: Does this hold if $A$ is only positive semidefinite?

Can anybody explain to me what this would change? Your help is much appreciated.

  • 3
    $\begingroup$ If $A$ is semidefinite, there will be nonzero $x$ with $\left<x,x\right>=0$. $\endgroup$ – Lord Shark the Unknown Jun 29 '18 at 12:59

If $A$ is strictly positive semidefinite (i.e. there exists $x \neq 0$ such that $x^TAx = 0$) then $\langle x,x\rangle = 0$ although $x \neq 0$. Thus, $\langle x,x \rangle = 0 \iff x = 0$ is not true since the forward direction is not true, although the other direction is.

Note that positive definiteness means that $x^TAx \ \mathbf{ > } \ 0$ for any $x \neq 0$. Positive semidefiniteness is a relaxation of this to an inequality, and as you can see a positive semidefinite matrix does not always yield an inner product in this manner.


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