If
is an inner product on a vector space V, and
is an invertible linear transformation, then
is also an inner product on V.

Question

If 
  is an inner product on a vector space V, and 
  is an invertible linear transformation, then 
  is also an inner product on V.

Quizplus · Accepted Answer

Let's denote the first inner product as $\langle \cdot, \cdot angle_1$ and the second inner product as $\langle \cdot, \cdot angle_2$. We want to show that $\langle \cdot, \cdot angle_2$ is also an inner product on $V$.

Since $\langle \cdot, \cdot angle_1$ is an inner product on $V$, it satisfies the following properties for all $u,v,w \in V$ and $a,b \in \mathbb{R}$:

1. Non-negativity: $\langle u, u angle_1 \geq 0$ and $\langle u, u angle_1 = 0$ if and only if $u = 0$.
2. Linearity: $\langle au + bv, w angle_1 = a\langle u,w angle_1 + b\langle v,w angle_1$ for all $a,b \in \mathbb{R}$.
3. Symmetry: $\langle u,v angle_1 = \langle v,u angle_1$.
4. Positive definiteness: $\langle u,v angle_1 = 0$ for all $v \in V$ if and only if $u = 0$.

Now, let $T$ denote the invertible linear transformation given by $T(\cdot) = A(\cdot)$, where $A$ is the matrix representing $T$ with respect to some (any) basis of $V$. Since $T$ is invertible, $A$ is invertible.

We can define $\langle u, v angle_2 = \langle T(u), T(v) angle_1$ for all $u,v \in V$. We want to show that this satisfies the four properties of an inner product.

1. Non-negativity: For all $u \in V$, 
$$\langle u, u angle_2 = \langle T(u), T(u) angle_1 \geq 0.$$
Also, if $\langle u,u angle_2 = 0$, then $\langle T(u), T(u) angle_1 = 0$. But since $\langle \cdot, \cdot angle_1$ is an inner product, this implies that $T(u) = 0$, which in turn implies that $u = 0$ (since $T$ is invertible). Hence, $\langle \cdot, \cdot angle_2$ satisfies non-negativity.

2. Linearity: For all $u,v,w \in V$ and $a,b \in \mathbb{R}$, we have
\begin{align*}
\langle au+bv, w angle_2 &= \langle T(au+bv), T(w) angle_1 \
&= \langle aT(u) + bT(v), T(w) angle_1 \
&= a\langle T(u), T(w) angle_1 + b\langle T(v), T(w) angle_1 \
&= a\langle u, w angle_2 + b\langle v, w angle_2.
\end{align*}
Hence, $\langle \cdot, \cdot angle_2$ satisfies linearity.

3. Symmetry: For all $u,v \in V$, we have
$$\langle u, v angle_2 = \langle T(u), T(v) angle_1 = \langle T(v), T(u) angle_1 = \langle v, u angle_2.$$
Hence, $\langle \cdot, \cdot angle_2$ satisfies symmetry.

4. Positive definiteness: For all $u \in V$, if $\langle u,v angle_2 = 0$ for all $v \in V$, then in particular, $\langle u,u angle_2 = 0$. But we showed earlier that this implies $u = 0$, so $\langle \cdot, \cdot angle_2$ satisfies positive definiteness.

Hence, $\langle \cdot, \cdot angle_2$ satisfies all four properties of an inner product, so it is also an inner product on $V$.

If Is an Inner Product on a Vector Space