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In $\Bbb R^n,$ if vectors are written as column-vectors (i.e., $n \times 1$ column-matrices) like $u=\begin{bmatrix}{c}u_{1}\\\vdots\\ u_{n}\end{bmatrix}$ then the standard dot-product in $\mathbb{R}^{n}$ can be written as $$u\cdot v=v^Tu$$ by using the meaning of "transpose of a matrix' and multiplication of matrices.

(a) Let $A \in \mathcal{M}_{n \times n}$ be an $n \times n$ square-matrix. Using the properties of 'transpose of a matrix' and the definition given above, show that $$ Au\cdot v=u\cdot A^T v\forall u,v \in \Bbb R^n$$

(b) Now, let $A$ be the $2 \times 2$ symmetric-matrix given by $A=\begin{bmatrix}1 & 1 \\ 1 & 2\end{bmatrix}.$ Show that the real-valued function $\langle\cdot \cdot \cdot\rangle: \Bbb R^2\times \Bbb R^2\to\Bbb R$ defined by $$\langle u, v\rangle:=u\cdot A v$$ is an inner-product on $\Bbb R^2$

How to prove (b) is an inner product? I proved $\langle u,v\rangle=\langle v,u \rangle $. How to prove $ \langle u+v,w\rangle =\langle u,w\rangle +\langle v+w\rangle $, $\langle ku,v\rangle =k\langle u,v\rangle $ and $\langle v,v\rangle \geq 0$ ??

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  • $\begingroup$ Please include the problem text in your post instead of linking to a picture. Images are not searchable, so your post does no one else any good. You’re asking us to spend time answering your questions; it’s only fair that you spend some of your time on it as well. $\endgroup$
    – amd
    Commented Dec 1, 2016 at 19:37

2 Answers 2

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The properties $\langle u+v, w \rangle = \langle u, w \rangle + \langle v,w \rangle$ and $\langle ku,v \rangle = k \langle u,v \rangle$ follow trivially from linearity of the dot product. For example, $$\langle u+v, w \rangle = (u+v)\cdot Aw = u\cdot Aw + v\cdot Aw = \langle u, w \rangle + \langle v,w \rangle.$$ For positive definiteness, you need to use the specific matrix given. If $v = \binom{v_1}{v_2}$, then $$\langle v,v\rangle = v \cdot \left( \begin{matrix} 1 & 1 \\ 1 & 2\end{matrix} \right)v = \binom{v_1} {v_2} \cdot \binom{v_1 + v_2}{v_1 + 2v_2} = v_1^2 + 2v_2 + 2v_2^2 = (v_1 + v_2)^2 + v_2^2.$$

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Inner product satisfies that following

  1. Symmetry: $\newcommand{\inp}[1]{\left\langle #1 \right\rangle}$ $\inp{u, v} = \inp{v, u}$

  2. Linearity: $\inp{k u, v} = k \inp{u, v}$.

  3. Positive-definite: $\inp{u, u} \ge 0$ with equality if and only if $u = 0$.

So the question is prove $\inp{u, v} = v^TA^Tu$ is an inner product for the given $A$.

  1. Symmetry: Trivial (you proved it as well).
  2. Linearity: $\inp{k u, v} = v^TA^T k u = k.v^TA^Tu = k.\inp{u, v}$
  3. Positive-definite: $\inp{u, u} = u^TA^T u > 0$ since the determinanats of the leading principal minors of $A$ are positive definite ($1 > 0$ and $det(A) > 0$)
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