Just a simple example:

Let $V=\mathbb R^n$, Define $\langle x,y\rangle = x^Ty = x_1y_1+...+x_ny_n$. Verify that the function $\langle x,y\rangle $ satisfies the four conditions of being an inner product on $\mathbb R^n$.

My attempt:

  1. positivity: $\langle x,x\rangle = (x_1^2 +\dots+x_n^2)\geq0$

  2. symmetry: $\langle x,y\rangle = x_1y_1+\dots+x_ny_n$ and $\langle y,x\rangle =y_1x_1+\dots+y_nx_n$ therefore, $\langle x,y\rangle=\langle y,x\rangle$.

  3. linearity: $\langle x+y,z\rangle=(x_1+y_1)z_1+\dots+(x_1+y_1)z_n$ and $\langle x,z\rangle+ \langle y,z\rangle =(x_1z_1+\dots+x_nz_n)+(y_1z_1+...+y_nz_n)=(x_1z_1+y_1z_1)+\dots+(x_nz_n+y_nz_n)=(x_1+y_1)z_1+\dots+(x_n+y_n)z_n$

  4. linearity: $\langle cx,y\rangle=cx_1y_1+\dots+cx_ny_n=c(x_1y_1+\dots+x_ny_n)=c\langle x,y\rangle$

These trivial examples always confuse me, am I on the right track?

  • 2
    $\begingroup$ yes, you got it right $\endgroup$
    – gt6989b
    Apr 14, 2016 at 16:19
  • $\begingroup$ hey hey thanks! $\endgroup$
    – Bruce
    Apr 14, 2016 at 16:26
  • $\begingroup$ In (2), the second inner product must be $\;\langle y,x\rangle\;$ . Other than this it looks just fine to me. $\endgroup$
    – DonAntonio
    Apr 14, 2016 at 16:41

1 Answer 1


Wrong for no. 2 -- you need to show conjugate-symmetry.

But here your ground field is real, so you shouldn't have much trouble showing it. (Just a little more work.)


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