Exercise 8, Section 82 from PR Halmos's Finite-Dimensional Vector Spaces, 2nd Edition

If $A$ is a positive semidefinite operator, and if $\langle Ax, x\rangle = 0$ for some vector $x$, show that $Ax = 0$. The underlying inner product space is not specified as finite-dimensional. The scalar field is not specified as real or complex.

I am able to establish the assertion assuming that the inner product space is finite-dimensional. Struggling with extending the argument to infinite-dimensional spaces however.

My argument for the finite-dimensional case goes as follows. Section 82 ("Functions of Transformations") of the book argues that every positive operator on a finite-dimensional inner product has a positive square root (function) associated. Thus, we observe that $0 = \langle Ax, x\rangle $ $= \langle \sqrt A \sqrt Ax, x\rangle$ $= \langle \sqrt Ax, {\sqrt A}^*x\rangle$ $= \langle \sqrt Ax, \sqrt Ax\rangle$ $= \Vert \sqrt Ax \Vert^2$ $\implies$ $\sqrt Ax = 0$ $\implies \sqrt A \sqrt Ax = 0$ $\implies Ax = 0$.

Unclear on how to extend this argument to the infinite-dimensional case. Would appreciate an advice. Thanks.

  • $\begingroup$ Does Halmos' definition of positive (semi)definite include the assumption that $A$ is self-adjoint ($A^* = A$)? Some authors assume this and others do not. $\endgroup$
    – user169852
    Jun 23, 2020 at 1:59
  • 1
    $\begingroup$ Yes. Positive semidefinite and positive definite operators are self-adjoint by definition, accordingly to Halmos. $\endgroup$ Jun 23, 2020 at 2:02
  • $\begingroup$ Every positive self-adjoint operator has a square root, no matter if the underlying Hilbert space is finite-dimensional or not (if the space is not complete, it might map to the completion, but that does not matter for your proof). But I guess you don't know that yet? $\endgroup$
    – MaoWao
    Jun 23, 2020 at 5:51
  • $\begingroup$ The question doesn't say that the underlying inner product space is a Hilbert space (i.e., a complete inner product space). The book introduces the concept of Hilbert spaces much later. So, like elsewhere in the book, the space is to be understood as a not-necessarily-finite-dimensional space. $\endgroup$ Jun 23, 2020 at 6:37
  • $\begingroup$ @Omnomnomnom Why? It certainly has a square root in the completion, but why should it map the inner product space into itself? $\endgroup$
    – MaoWao
    Jun 23, 2020 at 7:52

2 Answers 2


The map $(y,z)\mapsto \langle Ay,z\rangle$ is a semi-inner product (i.e. it satisfies the same conditions as an inner product except for positive definiteness, which is replaced by positive semi-definiteness). In particular, the Cauchy-Schwarz inequality applied to $y=x$ and $z=Ax$ gives: $$ \lvert \langle Ax,Ax\rangle\rvert\leq \langle Ax,x\rangle^{1/2}\langle A(Ax),Ax\rangle^{1/2}=0. $$ Thus $Ax=0$.


Suppose that $(Ax,x) = 0$, but $Ax \neq 0$. Consider the vector $v = Ax + tx$, with $t \in \Bbb R$. We have $$ (Av,v) = (A^2x+tAx,Ax+tx) = (x,Ax)t^2 + [(A^2x,x) + (Ax,Ax)]t + (A^2x,Ax)\\ = 2\|Ax\|^2 t + (A^2x,Ax). $$ We see that for a "sufficiently negative" $t$ ($t < -\frac{(A^2x,Ax)}{2\|Ax\|^2}$), $(Av,v)$ must be negative. So, $A$ cannot be positive semidefinite.


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