Let $T$ be an $n\times n$ matrix. Define a form on $\mathbb{R^n}$ by $\langle x,y\rangle_T=\langle Tx,Ty\rangle$ for $x,y \in \mathbb{R^n}$. Show that this is an inner product iff T is invertable.

My solution,

=>: Suppose $\langle x,y\rangle_T$ is an inner product, $\langle x,x\rangle_T=0$ imply that Tx=0. I know that if T is invertable Tx=0 iff x=o but im not sure how to make the connextion to the previous statement.

<=: Suppose T is invertable, I can easily show bilearity and symmetry but im not sure how to show positive definiteness.

Any help would be appreciated,


  • $\begingroup$ All the properties of an inner-product carry on to $\langle x,y\rangle_T$ for any matrix $T$. Except for definiteness, which is equivalent to injectivity of $T$, hence to invertibility of $T$. $\endgroup$ – Julien Apr 21 '13 at 19:01

For the $\Rightarrow$ part, you are on the right track. The statement you need to conclude is that $T$ is invertible if and only if its kernel is trivial, along with some direct computation I'll mention in a second.

For the $\Leftarrow$ part, positive-definiteness and its strict version (that $\langle x,x \rangle_T = 0 \Leftrightarrow x = 0$) follow from direct computation: the entries of $Tx$ have some sign, and when you take their dot product you sum the entries squared, making all of the signs positive. The only way you could possible get zero is then if all the entries are zero. You can use invertibility of $T$ here to conclude.


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