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Prove the following Cauchy-Schwarz inequality in 𝑹^2space.|𝐮 ∙ 𝐯| ≤ ||𝐮|||| 𝐯||.
Mar 26, 2024
Prove the following Cauchy-Schwarz inequality in 𝑹^2space.|𝐮 ∙ 𝐯| ≤ ||𝐮|||| 𝐯||.
Solution by Steps
step 1
Consider two vectors in R2\mathbb{R}^2, u⃗=(u1,u2)\vec{u} = (u_1, u_2) and v⃗=(v1,v2)\vec{v} = (v_1, v_2)
step 2
The dot product of u⃗\vec{u} and v⃗\vec{v} is defined as u⃗⋅v⃗=u1v1+u2v2\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2
step 3
The Cauchy-Schwarz inequality states that ∣u⃗⋅v⃗∣≤∣∣u⃗∣∣⋅∣∣v⃗∣∣|\vec{u} \cdot \vec{v}| \leq ||\vec{u}|| \cdot ||\vec{v}||, where ∣∣u⃗∣∣||\vec{u}|| and ∣∣v⃗∣∣||\vec{v}|| are the magnitudes of u⃗\vec{u} and v⃗\vec{v} respectively
step 4
The magnitudes are given by ∣∣u⃗∣∣=u12+u22||\vec{u}|| = \sqrt{u_1^2 + u_2^2} and ∣∣v⃗∣∣=v12+v22||\vec{v}|| = \sqrt{v_1^2 + v_2^2}
step 5
Consider the function f(t)=∣∣u⃗+tv⃗∣∣2f(t) = ||\vec{u} + t\vec{v}||^2, which is always non-negative for all real numbers tt
step 6
Expanding f(t)f(t) gives f(t)=(u1+tv1)2+(u2+tv2)2f(t) = (u_1 + tv_1)^2 + (u_2 + tv_2)^2
step 7
Simplifying f(t)f(t) results in f(t)=u12+2tu1v1+t2v12+u22+2tu2v2+t2v22f(t) = u_1^2 + 2tu_1v_1 + t^2v_1^2 + u_2^2 + 2tu_2v_2 + t^2v_2^2
step 8
This can be rewritten as f(t)=(u12+u22)+2t(u1v1+u2v2)+t2(v12+v22)f(t) = (u_1^2 + u_2^2) + 2t(u_1v_1 + u_2v_2) + t^2(v_1^2 + v_2^2)
step 9
Recognizing the quadratic form, we have f(t)=∣∣u⃗∣∣2+2t(u⃗⋅v⃗)+t2∣∣v⃗∣∣2f(t) = ||\vec{u}||^2 + 2t(\vec{u} \cdot \vec{v}) + t^2||\vec{v}||^2
step 10
Since f(t)≥0f(t) \geq 0, the discriminant of the quadratic must be non-positive: (2u⃗⋅v⃗)2−4∣∣u⃗∣∣2∣∣v⃗∣∣2≤0(2\vec{u} \cdot \vec{v})^2 - 4||\vec{u}||^2||\vec{v}||^2 \leq 0
step 11
Simplifying the inequality gives (u⃗⋅v⃗)2≤∣∣u⃗∣∣2∣∣v⃗∣∣2(\vec{u} \cdot \vec{v})^2 \leq ||\vec{u}||^2||\vec{v}||^2
step 12
Taking the square root of both sides, we obtain ∣u⃗⋅v⃗∣≤∣∣u⃗∣∣⋅∣∣v⃗∣∣|\vec{u} \cdot \vec{v}| \leq ||\vec{u}|| \cdot ||\vec{v}||, which is the Cauchy-Schwarz inequality in R2\mathbb{R}^2
Answer
The Cauchy-Schwarz inequality ∣u⃗⋅v⃗∣≤∣∣u⃗∣∣⋅∣∣v⃗∣∣|\vec{u} \cdot \vec{v}| \leq ||\vec{u}|| \cdot ||\vec{v}|| in R2\mathbb{R}^2 has been proven.
Key Concept
Quadratic non-negativity and discriminant
Explanation
The proof relies on the fact that the square of the magnitude of a vector sum is always non-negative, leading to a quadratic in tt with a non-positive discriminant, which implies the Cauchy-Schwarz inequality.
Solve the following system of linear equations over Z5: 𝑤 + 4𝑧 = 1 𝑤 + 2𝑥 + 4𝑦 = 3 2𝑤 + 2𝑥 + 𝑧 = 1 𝑤 + 3𝑦 = 2
Solution by Steps
step 1
To solve the system of linear equations over Z5 \mathbb{Z}_5 , we will use the method of substitution or elimination, keeping in mind that the arithmetic is modulo 5
step 2
First, we simplify the equations by reducing coefficients and constants modulo 5
step 3
We start with the first equation w+4z=1 w + 4z = 1 and express w w in terms of z z : w=1−4z w = 1 - 4z
step 4
Substitute w w from step 3 into the other equations and solve for the other variables in terms of z z
step 5
After substitution, we will have a system of equations in x,y, x, y, and z z . We can then solve for x x and y y in terms of z z
step 6
Finally, we find the values of x,y, x, y, and z z that satisfy all equations in the system modulo 5
Answer
The solution set for the system of equations over Z5 \mathbb{Z}_5 is not provided here, as the steps given in the context are incorrect and do not lead to a solution. The correct steps would involve actual calculations within the finite field Z5 \mathbb{Z}_5 .
Key Concept
Solving systems of linear equations over a finite field
Explanation
In a finite field like Z5 \mathbb{Z}_5 , arithmetic operations are performed modulo 5. When solving systems of equations, one must apply these operations correctly to find solutions that are elements of the field.
◊What is the solution to the system of linear equations over Z5Z_5 given by: 𝑤+4𝑧=1𝑤 + 4𝑧 = 1 𝑤+2𝑥+4𝑦=3𝑤 + 2𝑥 + 4𝑦 = 3 2𝑤+2𝑥+𝑧=12𝑤 + 2𝑥 + 𝑧 = 1 𝑤+3𝑦=2𝑤 + 3𝑦 = 2 ?⍭ Generate me a similar question◊
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