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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 $P$?

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  • $\begingroup$ The notes say "but calculating $P$ directly is more difficult for this basis than it was with a basis of eigenvectors". Are you asking about how get the matrix using eigenvectors, or how to get the matrix directly? $\endgroup$ – Ben Grossmann Jan 20 '20 at 8:39
  • $\begingroup$ both if it is possible. I can't figure out why P is \begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}. $\endgroup$ – king Jan 20 '20 at 10:13
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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 $P=\frac{aa^{T}}{a^{T}a}$

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