Linear Transformation
For 1 โ 4, we will conder the linear transformation P that projects vectors in ๐ 2 onto the line ๐ฆ = ๐ฅ. The
way to think about this projection is that your holding a flashlight a long ways away, and youโre shining it
perpendicularly to the line ๐ฆ = ๐ฅ. The โshadowโ that any vector casts on that line is the projection of
that vector. Here are some examples:
To find the matrix of this linear transformation (with respect to the standard basis) we just need to
figure out what this linear transformation does to the two standard basis vectors [ 1 0 ] and [0 1 ].
ย Letโs focus on the former:
Looking at that right triangle formed by the origin, the vector e1 and the projection, we can deduce that
the angles are as labeled. The hypothenuse is length 1, and the other two sides are equal, so squaring
the sides, adding them and setting them equal to 1 tells us that each side is length โ2
2 . So the projection
vector must be the vector [
โ2
2 cos
๐
4
โ2
2 sin
๐
4
] = [
1
2 1
2
].
A similar calculation (or staring at the picture convinced you that itโs symmetric) tells us that the
projection of e2 is also [12 12].
So the matrix of this linear transformation (with respect to the standard basis) is ๐ด = [ 1 2โ
1 2โ
1 2โ
1 2โ ]. Thatโs
the main thing we need.
