Question 1161581: Let u and v be vector in Rn that are expressed in column form and A be an invertible n×n matrix. Show that if u · v is the Euclidean inner product (or the dot product) on Rn, then
⟨u,v⟩ = vTATAu
also defines an inner product.
Answer by CPhill(2285)   (Show Source):
You can put this solution on YOUR website! To prove that $\langle u, v \rangle = v^T A^T A u$ defines a valid inner product on $\mathbb{R}^n$, we must verify that it satisfies the four fundamental axioms of a real inner product for all vectors $u, v, w \in \mathbb{R}^n$ and any scalar $c \in \mathbb{R}$:
1. **Symmetry (Commutativity):** $\langle u, v \rangle = \langle v, u \rangle$
2. **Additivity (Distributivity):** $\langle u + w, v \rangle = \langle u, v \rangle + \langle w, v \rangle$
3. **Homogeneity (Scalar Multiplication):** $\langle c u, v \rangle = c \langle u, v \rangle$
4. **Positive Definiteness:** $\langle u, u \rangle \ge 0$, and $\langle u, u \rangle = 0$ if and only if $u = \mathbf{0}$.
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### 1. Symmetry
Recall that the transpose of a $1 \times 1$ scalar is itself, and $(AB)^T = B^T A^T$.
$$\langle u, v \rangle = v^T A^T A u$$
Taking the transpose of the $1 \times 1$ real matrix (scalar) $v^T A^T A u$:
$$\langle u, v \rangle = (v^T A^T A u)^T = u^T A^T (A^T)^T (v^T)^T = u^T A^T A v = \langle v, u \rangle$$
Thus, **symmetry holds**.
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### 2. Additivity
Using matrix distribution properties:
$$\langle u + w, v \rangle = v^T A^T A (u + w)$$
$$\langle u + w, v \rangle = v^T A^T (A u + A w)$$
$$\langle u + w, v \rangle = v^T A^T A u + v^T A^T A w$$
$$\langle u + w, v \rangle = \langle u, v \rangle + \langle w, v \rangle$$
Thus, **additivity holds**.
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### 3. Homogeneity
Using scalar multiplication properties of matrices:
$$\langle c u, v \rangle = v^T A^T A (c u) = c (v^T A^T A u) = c \langle u, v \rangle$$
Thus, **homogeneity holds**.
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### 4. Positive Definiteness
Evaluate $\langle u, u \rangle$:
$$\langle u, u \rangle = u^T A^T A u = (A u)^T (A u) = (A u) \cdot (A u) = \Vert{}A u\Vert{}^2$$
* Since the standard Euclidean norm $\Vert{}A u\Vert{}^2$ is always non-negative:
$$\langle u, u \rangle = \Vert{}A u\Vert{}^2 \ge 0$$
* **Zero condition:**
$$\langle u, u \rangle = 0 \iff \Vert{}A u\Vert{}^2 = 0 \iff A u = \mathbf{0}$$
Because $A$ is an **invertible** $n \times n$ matrix, its kernel (null space) contains only the zero vector:
$$A u = \mathbf{0} \iff u = A^{-1}\mathbf{0} = \mathbf{0}$$
Thus, **positive definiteness holds**.
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### Conclusion
Since all four inner product axioms are satisfied, $\langle u, v \rangle = v^T A^T A u$ defines a valid inner product on $\mathbb{R}^n$.
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