I am learning the matrix of linear transformation but have trouble understanding it when reading the notes. The part I am stuck at is as follows:
To see how important the choice of basis is, let's use the standard basis for the linear transformation that projects the plane onto a line at a $45^\circ$ angle. If we choose $\boldsymbol{\mathrm v_1} = \boldsymbol{\mathrm w_1} = \begin{bmatrix}1 \\ 0\end{bmatrix}$ and $\boldsymbol{\mathrm v_2} = \boldsymbol{\mathrm w_2} = \begin{bmatrix}0 \\ 1\end{bmatrix}$, we get the projection matrix $P = \dfrac{\boldsymbol{\mathrm{aa}}^T}{\boldsymbol{\mathrm a}^T\boldsymbol{\mathrm a}} = \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix}$. We can check by graphing that this is the correct matrix, but calculating $P$ directly is more difficult for this basis than it was with a basis of eigenvectors.
My question is:
How to get the projection matrix $P$?
Let $a$ and $b$ be two vectors in $R^m$.
Note that the projection of vector $b$ onto vector $a$ is $\frac{b^{T}a}{a^{T}a}a$
Rewrite the same as $a\frac{a^{T}b}{a^{T}a}=\frac{aa^{T}}{a^{T}a}b=Pb$ [Note that $\frac{b^{T}a}{a^{T}a}$ is a scaler and that $P=\frac{aa^{T}}{a^{T}a}$ is a rank 1 matrix which projects $b$ onto $a$ ]
Now as per your question, choose $a$ to be any vector on the line $y=x$, for example $a=(1,1)^{T}$ and get $P$ from $P=\frac{aa^{T}}{a^{T}a}$