Notation question: defining a matrix as $P=\sum_{i=1}^{k}v_{i}v_{i}^{T}$

Question

I've seen in a paper the following sentence:

Let $V$ be a $k$-dimensional subspace of $\mathbb R ^d$ and let $v_1 ,...,v_k$ be an orthonormal basis of $V$. Define $P=\sum_{i=1}^{k}v_{i}v_{i}^{T}$. The matrix P is called a projection matrix into the subspace....

Later on, the basis $\{v_1,...,v_k\}$ is completed to an orthonormal basis of $\mathbb R ^d$: $\{v_1,...,v_d\}$

Given a vector $x \in \mathbb R ^d$, it is written as $x=\sum_{j=1}^{d}\alpha_{j}v_{j}$.

Then this equation appears:

$Px=\left(\sum_{i=1}^{k}v_{i}v_{i}^{T}\right)\left(\sum_{j=1}^{d}\alpha_{j}v_{j}\right)=\left(\sum_{i=1}^{k}v_{i}v_{i}^{T}\right)\left(\sum_{j=1}^{d}\alpha_{j}v_{j}\right)=\sum_{i=1}^{k}\sum_{j=1}^{d}\alpha_{j}v_{i}v_{i}^{T}v_{j}=\sum_{j=1}^{k}\alpha_{j}v_{j}$

My question is this:

$P=\sum_{i=1}^{k}v_{i}v_{i}^{T}:= \left[\left.\begin{array}{c} \\ v_{1}\\ \\ \end{array}\right|\left.\begin{array}{c} \\ \cdots\\ \\ \end{array}\right|\begin{array}{c} \\ v_{k}\\ \\ \end{array}\right] \left[\left.\begin{array}{c} \\ v_{1}\\ \\ \end{array}\right|\left.\begin{array}{c} \\ \cdots\\ \\ \end{array}\right|\begin{array}{c} \\ v_{k}\\ \\ \end{array}\right]^{T}? $

If so, isn't this notation a bit confusing when written in the equation which follows it?

Answer 1

you confused the dimension of basis vectors. We can suppose that $v_i=(v_{i1},\dots,v_{id})^T$ where $i=1\dots k$ and $v_i$ are the basis vectors of V. then $$v_iv_i^T=\left(\begin{matrix}v_{i1}\\v_{i2}\\\vdots\\v_{id}\end{matrix}\right)(v_{i1},\dots,v_{id})=\begin{pmatrix}v_{i1}^2 & \dots & v_{i1}v_{in} &\dots & v_{i1}v_{id}\\\vdots&\ddots&\vdots&&\vdots\\v_{im}v_{i1}&\dots& v_{im}v_{in}&\dots&v_{im}v_{id}\\\vdots&&\vdots&\ddots&\vdots\\v_{id}v_{i1}&\dots&v_{id}v_{in}&\dots&v_{id}^2\end{pmatrix}$$ which is a $d\times d$ matrix.

Here we can pose an example. For example, $d=3$, and $V=R^2$ is a subspace of $R^3$. Then we can write $v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $v_2=\begin{pmatrix}0\\1\\0\end{pmatrix}$, then $$P=\sum_{i=1}^2v_iv_i^T=\begin{pmatrix}1 &0 & 0\\0 &1&0\\0&0&0\end{pmatrix}$$

Answer 2

If it helps with the notation you can think of the last part of the equation:$P=\sum_{i=1}^{k}v_{i}v_{i}^{T}:= \left[\left.\begin{array}{c} \\ v_{1}\\ \\ \end{array}\right|\left.\begin{array}{c} \\ \cdots\\ \\ \end{array}\right|\begin{array}{c} \\ v_{k}\\ \\ \end{array}\right] \left[\left.\begin{array}{c} \\ v_{1}\\ \\ \end{array}\right|\left.\begin{array}{c} \\ \cdots\\ \\ \end{array}\right|\begin{array}{c} \\ v_{k}\\ \\ \end{array}\right]^{T}$ as a set of elements $\mathbf{v}=\begin{pmatrix}v_1^T\\\vdots\\v_k^T\end{pmatrix}$ for which $P$ is an inner product $\mathbf{v}^T\mathbf{v}=v_1v_1^T+\ldots+v_kv_k^T$