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I have the following problem:

Let's have the scalar product in space $ℝ^2$ given by the expression:

$$ <x, y> = 2x_1 y_1 + x_2 y_2 + x_1 y_2 + x_2 y_1 $$

  1. For a defined dot product, formulate the precise definition of the Cauchy-Schwarz inequality.

  2. For a given dot product, define the projection matrix onto the span{a} of the line a defined by the vector $ v = (1, 0)^T$.


For task 1, I was not really sure how to express the norm inducted by the dot product; therefore, I am struggling with the RHS of the inequality:

$$ |<x, y>| ≤ ||x|| \ ||y|| $$

More precisely:

$$ 2x_1 y_1 + x_2 y_2 + x_1 y_2 + x_2 y_1 ≤ \lVert x \rVert \ \lVert y \rVert $$

Should I express the euclidian norm in the way of $ \sqrt{<x, x>} $, then get rid of the root and express it in some more suitable way?


For task 2, I was able to determine the projection onto the line generated by span(a):

$$ x_u = \frac{ax^T}{a^Ta} $$

from that, we get after substituting into the defined dot product the following equation:

$$ \frac{<x, v>}{\lVert v \rVert} v = \frac{(2x_1 + x_2)}{\sqrt{2}} . (1, 0)^T $$

I believe this should be it, but now I need to express it using the projection matrix. I believe it by multiplying given vectors; I got the following matrix:

$$ \begin{bmatrix} 2 & 0 \\ 1 & 0 \end{bmatrix} $$

But somehow, It does not seem right.

Could anyone help me out?

Thanks in advance.

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  • $\begingroup$ for (1), writing it in matrix form could be helpful: $A:=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$, which is symmetric PD. Then $\mathbf x^T A\mathbf y \leq \big \Vert A^\frac{1}{2}\mathbf x\big\Vert_2\big \Vert A^\frac{1}{2}\mathbf y\big\Vert_2$ by Cauchy-Schwarz, where $\big \Vert \mathbf z\big\Vert_2$ denotes the standard 2-norm $\endgroup$ Jan 14, 2023 at 17:36

2 Answers 2

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For part 1.

Any inner product $\langle ,\rangle$ will induce the norm

$$ \Vert \cdot \Vert : x \mapsto \langle x,x\rangle^{\frac{1}2}$$

If the inner product is the usual dot product $ \cdot :(x,y) \mapsto \sum_i x_iy_i$ then you find the usual euclidian norm. If the inner product is different then you usually find a different norm. In our case norm is given by

$$ \Vert(x_1,x_2)\Vert^2 = 2x_1x_1 + x_2x_2 + x_1x_2 + x_2x_1 = 2x_1^2 + x_2^2 + 2x_1x_2 = x_1^2 + (x_1+x_2)^2$$


For part 2.

To find the projection of a vector $x$ onto the span of $a$. You need to find $\lambda \neq 0$ such that

$$ \langle \lambda a, x- \lambda a\rangle = 0 $$

Solving for $\lambda$ gives

$$ \lambda\langle a, x\rangle - \lambda^2\langle a,a\rangle = 0 \iff \lambda = \frac{\langle a, x\rangle}{\langle a, a\rangle} $$

So our projection is the following map

$$P : x \mapsto \frac{\langle a, x\rangle}{\langle a, a\rangle} \times a $$

Finding the matrix from here should not be too difficult. Indeed we have $$ \langle a, a\rangle = 2$$ and $$ \langle a, x\rangle = 2x_1 + x_2$$ so our projection is the map

$$ P : (x_1,x_2) \mapsto \left(\frac{2x_1+x_2}{2},0\right) = \left(x_1 + \frac{1}{2}x_2,0 \right)$$

In matrix form we have

$$ \begin{bmatrix} 1 & 1/2 \\ 0 & 0\end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix} = \begin{bmatrix} x_1 + \frac{1}{2}x_2 \\ 0\end{bmatrix}.$$

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The projection in this case (we are projecting onto a line) is given by $$\overset{\longrightarrow}{{\rm Proj}}_{a}(x):=\frac{\langle x,a\rangle}{\langle a,a\rangle}\cdot a$$ for every inner product (or "dot product") in a vector space $V$ with inner product $\langle \cdot,\cdot \rangle$. In particular in the plane we can write the linear transformation as $$\overset{\longrightarrow}{{\rm Proj}}_{\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}\left(\begin{bmatrix}x\\y\end{bmatrix} \right)=\frac{\left\langle \begin{bmatrix}x\\y\end{bmatrix},\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}\right\rangle}{\left\langle \begin{bmatrix}a_{1}\\a_{2}\end{bmatrix},\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}\right\rangle} \begin{bmatrix} a_{1}\\a_{2}\end{bmatrix}$$

Using your definition of inner product, we have

$$\overset{\longrightarrow}{{\rm Proj}}_{\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}\left(\begin{bmatrix}x\\y\end{bmatrix} \right)=\frac{2a_{1}x+a_{2}y+a_{2}x+a_{1}y}{2a_{1}^{2}+a_{2}^{2}+2a_{1}a_{2}}\begin{bmatrix}a_{1}\\a_{2} \end{bmatrix}$$ $$\overset{\longrightarrow}{{\rm Proj}}_{\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}\left(\begin{bmatrix}x\\y\end{bmatrix} \right)=\frac{1}{2a_{1}^{2}+a_{2}^{2}+2a_{1}a_{2}}\begin{bmatrix}(2a_{1}^{2}+a_{1}a_{2})x+(a_{1}^{2}+a_{1}a_{2})y\\(a_{2}^{2}+a_{2}a_{1})x+(a_{2}^{2}+a_{1}a_{2})y \end{bmatrix}$$ $$\overset{\longrightarrow}{{\rm Proj}}_{\begin{bmatrix}a_{1}\\a_{2}\end{bmatrix}}\left(\begin{bmatrix}x\\y\end{bmatrix} \right)=\frac{1}{2a_{1}^{2}+a_{2}^{2}+2a_{1}a_{2}}\begin{bmatrix} 2a_{1}^{2}+a_{1}a_{2} &a_{1}^{2}+a_{1}a_{2}\\a_{2}^{2}+a_{2}a_{1}& a_{2}^{2}+a_{1}a_{2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$$ Therefore, the projection matrix onto the line through $a$ with your inner product is given by

$$P=\frac{1}{2a_{1}^{2}+a_{2}^{2}+2a_{1}a_{2}}\begin{bmatrix} 2a_{1}^{2}+a_{1}a_{2} &a_{1}^{2}+a_{1}a_{2}\\a_{2}^{2}+a_{2}a_{1}& a_{2}^{2}+a_{1}a_{2}\end{bmatrix}$$

Let's check it: using $a=\begin{bmatrix}1\\0\end{bmatrix}$, we have $\boxed{P=\frac{1}{2}\begin{bmatrix} 2&1\\0& 0\end{bmatrix}}$ which is of course the correct and complete answer given by @digitallis.

In 1) notice that Cauchy-Schwarz's inequality says $|\langle x,y\rangle|^{2}\leqslant \langle x,x\rangle\cdot \langle y,y\rangle$ for all $x,y$ in a vector space $V$.

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