Question

Given a non-zero vector $\mathbf{v}$, find the value that should be assigned to $\alpha$ to make $$P_{ij}=\alpha v_i{v}_j\text{ and }{Q}_{ij}={\delta }_{ij}-\alpha v_i{v}_j$$ into parallel and orthogonal projection tensors respectively, i.e. tensors that satisfy respectively $P_{ij}{v}_j=v_i,P_{ij}{u}_j=0$ and $Q_{ij}{v}_j=0,Q_{ij}{u}_j=u_i$, for any vector $\mathbf{u}$ that is orthogonal to $\mathbf{v}$, Show, in particular, that $Q_{ij}$ is unique, i.e. that if another tensor $T_{ij}$ has the same properties as $Q_{ij}$ then $(Q_{ij}-T_{ij})w_j=0$ for any vector $\mathbf{w}$.

Short answer

$\alpha =\frac{1}{\mathbf{v}\cdot \mathbf{v}}$. $Q_{ij}$ is unique because any difference $Q_{ij}-T_{ij}$ results in 0.

Step-by-step solution

- Understand the problem

We're given two tensors, $P_{ij}$ and $Q_{ij}$, and specific properties they must satisfy. We need to find $\alpha$ so that $P_{ij}$ and $Q_{ij}$ meet these conditions: $P_{ij}{v}_j=v_i$, $P_{ij}{u}_j=0$, $Q_{ij}{v}_j=0$, and $Q_{ij}{u}_j=u_i$, where $\mathbf{u}$ is orthogonal to $\mathbf{v}$.

- Analyze $P_{ij}$

For $P_{ij}$ to be a parallel projection tensor, it must satisfy:$$P_{ij}{v}_j=v_i$$Substitute $P_{ij}=\alpha v_i{v}_j$:$$\alpha v_i{v}_j{v}_j=v_i$$Since $v_{i}eq0$, we can divide both sides by $v_i$:$$\alpha (\mathbf{v}\cdot \mathbf{v})=1$$Thus, $\alpha$ must be:$$\alpha =\frac{1}{\mathbf{v}\cdot \mathbf{v}}$$

- Verify $P_{ij}$ conditions

For $P_{ij}{u}_j=0$:Substituting $\alpha$:$$P_{ij}{u}_j=\left(\frac{1}{\mathbf{v}\cdot \mathbf{v}}\right)v_i(v_j{u}_j)=0$$Since $\mathbf{u}$ is orthogonal to $\mathbf{v}$, $v_j{u}_j=0$, and thus, $P_{ij}{u}_j=0$ is satisfied.

- Analyze $Q_{ij}$

For $Q_{ij}$ to be an orthogonal projection tensor, it must satisfy:$$Q_{ij}{v}_j=0$$Using $Q_{ij}={\delta }_{ij}-\alpha v_i{v}_j$:$$({\delta }_{ij}-\alpha v_i{v}_j)v_j=v_i-\alpha v_i(v_j{v}_j)=0$$Using $\alpha =\frac{1}{\mathbf{v}\cdot \mathbf{v}}$:$$v_i-\left(\frac{1}{\mathbf{v}\cdot \mathbf{v}}\right)v_i(\mathbf{v}\cdot \mathbf{v})=0$$Hence, $Q_{ij}{v}_j=0$ is satisfied.

- Verify $Q_{ij}$ conditions

For $Q_{ij}{u}_j=u_i$:$$Q_{ij}{u}_j=({\delta }_{ij}-\alpha v_i{v}_j)u_j=u_i-\alpha v_i(v_j{u}_j)$$Since $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, $v_j{u}_j=0$:$$Q_{ij}{u}_j=u_i$$Hence, $Q_{ij}{u}_j=u_i$ is satisfied.

- Show uniqueness of $Q_{ij}$

Suppose another tensor $T_{ij}$ has the same properties as $Q_{ij}$. Then both $Q_{ij}$ and $T_{ij}$ satisfy $v_j{Q}_{ij}=0$ and $u_j{Q}_{ij}=u_i$. Consider $W_{ij}=Q_{ij}-T_{ij}$.$$W_{ij}{v}_j=(Q_{ij}-T_{ij})v_j=0-0=0$$$$W_{ij}{u}_j=(Q_{ij}-T_{ij})u_j=u_i-u_i=0$$Thus, $(Q_{ij}-T_{ij})w_j=0$ for any vector $\mathbf{w}$. Hence, $Q_{ij}$ is unique.

Key concepts

parallel projection tensor

A parallel projection tensor, typically denoted as $P_{ij}$, helps project vectors onto a specified direction in a vector space. To determine the value of $\alpha$ such that $P_{ij}=\alpha v_i{v}_j$ is a parallel projection tensor, it must fulfill specific conditions. One major condition is that the projection of vector $\mathbf{v}$ onto itself must return $\mathbf{v}$, implying $(P_{ij}{v}_j=v_i)$.
Substitute $P_{ij}$ with $\alpha v_i{v}_j$ and you get:
$\alpha v_i{v}_j{v}_j=v_i$.
Since $v_{i}eq0$, dividing both sides by $v_i$ leads to $\alpha (\mathbf{v}\cdot \mathbf{v})=1$. Therefore, $\alpha =\frac{1}{(\mathbf{v}\cdot \mathbf{v})}$. This value ensures that the parallel component of vector projections aligns perfectly with the original vector.

orthogonal projection tensor

An orthogonal projection tensor, denoted as $Q_{ij}$, projects vectors orthogonally to a given direction. For $Q_{ij}$ to be an orthogonal projection tensor, it must satisfy:
$Q_{ij}{v}_j=0$ and $Q_{ij}{u}_j=u_i$, where $\mathbf{u}$ is any vector orthogonal to $\mathbf{v}$.
Using the form $Q_{ij}={\delta }_{ij}-\alpha v_i{v}_j$, and substituting $\alpha$ as established above, let's examine the conditions.
First, for $Q_{ij}{v}_j=0$:
$({\delta }_{ij}-\alpha v_i{v}_j)v_j=v_i-\alpha v_i(v_j{v}_j)=0$
With $\alpha =\frac{1}{(\mathbf{v}\cdot \mathbf{v})}$, it holds:
$v_i-\frac{1}{(\mathbf{v}\cdot \mathbf{v})}{v}_i(\mathbf{v}\cdot \mathbf{v})=0$
Thus, the first condition is met.
For $Q_{ij}{u}_j=u_i$:
$({\delta }_{ij}-\alpha v_i{v}_j)u_j=u_i-\alpha v_i(v_j{u}_j)$
Since $\mathbf{u}\perp \mathbf{v}$, $(v_j{u}_j)=0$, leading to:
$Q_{ij}{u}_j=u_i$, which satisfies the second condition perfectly.

vector orthogonality

Vector orthogonality is a foundational concept in vector spaces. When two vectors $\mathbf{u}$ and $\mathbf{v}$ are orthogonal, their dot product is zero, expressed as $\mathbf{u}\cdot \mathbf{v}=0$. This property is crucial when working with projection tensors.
In the context of the given exercise:
• For parallel projection tensor $P_{ij}$, we ensured that $P_{ij}{u}_j=0$ when $\mathbf{u}\perp \mathbf{v}$.
• For orthogonal projection tensor $Q_{ij}$, we used orthogonality to affirm that $Q_{ij}{u}_j=u_i$.
This ensures that when vectors perpendicular to the given vector $\mathbf{v}$ are projected using $Q_{ij}$, their values remain unaltered whereas vectors aligned with $\mathbf{v}$ result in zero projection under $Q_{ij}$.
This dynamic forms the basis for understanding and using projection tensors in various applications in mathematics and physics.

tensor analysis

Tensor analysis involves the study of tensors, which are multi-dimensional arrays that generalize the concepts of scalars, vectors, and matrices. In the given problem, we handle second-order tensors, which are arrays with two indices.
The tensors $P_{ij}$ and $Q_{ij}$ are vital in understanding how vector projections can be decomposed:
• A parallel projection tensor $P_{ij}$, calculated as $\alpha v_i{v}_j$, projects vectors in the direction of $\mathbf{v}$.
• An orthogonal projection tensor $Q_{ij}$, given by ${\delta }_{ij}-\alpha v_i{v}_j$, projects vectors orthogonally to $\mathbf{v}$.
The uniqueness condition of $Q_{ij}$ is particularly meaningful. By constructing $W_{ij}=Q_{ij}-T_{ij}$ and showing $W_{ij}{w}_j=0$ for any vector $\mathbf{w}$, we demonstrate that two tensors having the same projection properties must be identical. This is a fundamental result in tensor analysis, ensuring consistency in various applications such as continuum mechanics, electromagnetism, and more.