Question 2.16: Using N = {e1^→,e2^→}, the Cartesian coordinate system, let ...
Using N=\left\{\overrightarrow{e_{1} },\overrightarrow{e_{2} } \right\}, the Cartesian coordinate system, let the linear operator f: R²→ R² be defined as
f\left(x_{1},x_{2} \right)=\left(2x_{1}-3x_{2},-4x_{1}+6x_{2}\right)
= \overrightarrow{x} A, where \overrightarrow{x}=\left(x_{1},x_{2}\right) and A=\begin{bmatrix} 2 & -4 \\ -3 & 6 \end{bmatrix}.
Try to investigate its geometric mapping properties.
f is obviously linear. Now, the kernel of f is
Ker(f) = \left\{\left(x_{1},x_{2} \right)\mid 2x_{1}-3x_{2}=0 \right\} =\ll \left(3,2\right) \gg ,
which is a straight line passing through the origin. Hence f is not one-to-one and thus is not onto.
What is the range of f? Let \overrightarrow{y}=\left(y_{1},y_{2}\right)=\overrightarrow{x}A, then
\left \{ \begin{matrix} 2x_{1}-3x_{2}=y_{1} \\ -4x_{1}+6x_{2}=y_{2} \end{matrix} \right.
\Rightarrow 2y_{1}+y_{2}=0.
Therefore,
Im(f) = \left\{\overrightarrow{y}=\left(y_{1},y_{2}\right)∈ \mathbb{R}² \mid 2y_{1}+y_{2}=0\right\} = \ll \left(1,-2\right)\gg .
Deposit the x_{1}x_{2} -plane and y_{1}y_{2}-plane on the same plane.
The characteristic polynomial of A is
det \left(A-tI_{2} \right)=\left|\begin{matrix} 2-t & -4 \\ -3 & 6-t \end{matrix} \right| = t^{2} -8t.
Hence, f or A has two eigenvalues \lambda_{1 } =8 and \lambda_{2 } =0. Solve \overrightarrow{x}A=8\overrightarrow{x} and get the corresponding eigenvectors \overrightarrow{x_{1}}=t \left(1,-2 \right) for t ≠0. Similarly, solve \overrightarrow{x}A=0\overrightarrow{x}= \overrightarrow{0}, and the corresponding eigenvectors \overrightarrow{x_{2}}=t \left(3,2 \right) for t ≠ 0.
Therefore, \overrightarrow{v_{1}}=\left(1,-2 \right) and \overrightarrow{v_{2}}=\left(3,2 \right) are linearly independent and the eigenspaces are
\ll \overrightarrow{v_{1}} \gg = \left\{\overrightarrow{x}\in R^{2} \mid \overrightarrow{x}A=8\overrightarrow{x}\right\}= Im(f),
\ll \overrightarrow{v_{2}} \gg = \left\{\overrightarrow{x}\in R^{2} \mid \overrightarrow{x}A=\overrightarrow{0}\right\}= Ker(f),
which are invariant subspaces of f. Also,
R^{2} =Im(f) ⊕ Ker(f).
What follows can be easily handled by using classical high school algebra, but we prefer to adopt the vector method, because we are studying how to get into the realm of linear algebra.
How to find the images of lines parallel to the kernel \ll \overrightarrow{v_{2}} \gg ? Let \overrightarrow{x}=\overrightarrow{x_{0}}+t \overrightarrow{v_{2}} be such a line and suppose it intersects the line \ll \overrightarrow{v_{1}} \gg at the point t_{0}\overrightarrow{v_{1}}. Then
f\left(\overrightarrow{x} \right) =f\left(\overrightarrow{x_{0} }+t\overrightarrow{v_{2} } \right)=f\left(\overrightarrow{x_{0} }\right) +tf\left(\overrightarrow{v_{2} }\right) =f\left(\overrightarrow{x_{0} }\right)
= f\left( t_{0}\overrightarrow{v_{1}} \right)=t_{0}f\left( \overrightarrow{v_{1}} \right)=8 t_{0} \overrightarrow{v_{1}}.
This means that f maps the whole line \overrightarrow{x}=\overrightarrow{x_{0} }+t\overrightarrow{v_{2} } into a single point, 7 units away from the point of intersection of it with the line \ll \overrightarrow{v_{1}} \gg, along the line \ll \overrightarrow{v_{1}} \gg, and in the direction of the point of intersection. See Fig. 2.74.
What is the image of a line not parallel to the kernel \ll \overrightarrow{v_{2}} \gg? Let \overrightarrow{x}=\overrightarrow{x_{0} }+t \overrightarrow{u } be such a line where \overrightarrow{u } and \overrightarrow{v_{2}} are linearly independent.
Suppose \overrightarrow{u }=\overrightarrow{u_{1} }+\overrightarrow{u_{2} } with \overrightarrow{u_{1} } ∈Im(f) and \overrightarrow{u_{2} } ∈Ker(f). Then
f\left(\overrightarrow{x} \right)=f\left(\overrightarrow{x_{0} } \right)+tf\left(\overrightarrow{u} \right)
=f\left(\overrightarrow{x_{0} } \right) + tf\left(\overrightarrow{u_{1} }+\overrightarrow{u_{2} } \right) = f\left(\overrightarrow{x_{0} } \right)+ t \left(f\left(\overrightarrow{u_{1}} \right)+f\left(\overrightarrow{u_{2}} \right) \right)
=f\left(\overrightarrow{x_{0} }\right)+tf\left(\overrightarrow{u_{1}} \right)=f\left(\overrightarrow{x_{0} } \right)+8t\overrightarrow{u_{1} },
where f\left(\overrightarrow{x_{0} } \right)=8 t_{0}\overrightarrow{v_{1}} for some t_{0}. Since \overrightarrow{u_{1} }≠0, the image is a line and coincides with the range line \ll \overrightarrow{v_{1}} \gg . Also, f maps the line \overrightarrow{x}= \overrightarrow{x_{0}}+t\overrightarrow{u} one-to-one and onto the line \ll \overrightarrow{v_{1}} \gg and preserves ratios of signed lengths of segments along the line. See Fig. 2.75.
Now, we have a clearer picture about the mapping properties of f. For any \overrightarrow{x}= \left( x_{1},x_{2}\right), then
\overrightarrow{x}=\left( \frac{1}{4}x_{1}-\frac{3}{8}x_{2}\right) \overrightarrow{v_{1}} + \left( \frac{1}{4}x_{1}+\frac{1}{8}x_{2}\right) \overrightarrow{v_{2}}
\Rightarrow f\left(\overrightarrow{x} \right) = \left( \frac{1}{4}x_{1}-\frac{3}{8}x_{2}\right) f\left(\overrightarrow{v_{1}}\right) =\left( \frac{1}{4}x_{1}-\frac{3}{8}x_{2}\right) \cdot 8 \overrightarrow{v_{1}}= \left( 2x_{1}-3x_{2}\right)\overrightarrow{v_{1}}.
Define the mapping p: R² → R² by
p\left(\overrightarrow{x} \right)=\left(\frac{1}{4}x_{1}-\frac{3}{8}x_{2}\right)\overrightarrow{v_{1} }.
p is linear and has the same kernel and range as f does. But p keeps every \overrightarrow{x} in \ll \overrightarrow{v_{1}} \gg fixed, i.e. p² = p\circ p = p holds. This p is called the projection of R² onto \ll \overrightarrow{v_{1}} \gg along the kernel \ll \overrightarrow{v_{2}} \gg. Therefore,
\overrightarrow{x} → p\left(\overrightarrow{x} \right) → 8p\left(\overrightarrow{x} \right) = f\left(\overrightarrow{x} \right).
All one needs to do is to project \overrightarrow{x} onto \ll \overrightarrow{v_{1}} \gg along \ll \overrightarrow{v_{2}} \gg, and then to scalarly multiply the projected vector by 8.Finally, the resulting vector is f\left(\overrightarrow{x} \right). See Fig. 2.76 which is a special case of Fig. 2.73.
We can use (2.7.72) to reinterpret the content of the above paragraph.To set up the canonical decomposition of f, let B=\left\{\overrightarrow{v_{1} },\overrightarrow{v_{2} } \right\} which is a basis for R². Then
f\left(\overrightarrow{v_{1}} \right) = 8 \overrightarrow{v_{1}}=8\overrightarrow{v_{1}}+0\cdot \overrightarrow{v_{2}}
f\left(\overrightarrow{v_{1}} \right) = \overrightarrow{0}=0\cdot\overrightarrow{v_{1}}+0\cdot \overrightarrow{v_{2}}
\Rightarrow \left[f\right] _{B}=PAP^{-1}=\begin{bmatrix} 8 & 0 \\ 0 & 0 \end{bmatrix}, where P=\left[\begin{matrix} \overrightarrow{v_{1} } \\ \overrightarrow{v_{2} }\end{matrix} \right] =\begin{bmatrix} 1 & -2 \\ 3 & 2 \end{bmatrix}.
Let
A_{1}= P^{-1}=\begin{bmatrix} 1 & 0 \\0 & 0 \end{bmatrix}P =\frac{1}{8}\begin{bmatrix} 2 & 2 \\ -3 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ 3 & 2 \end{bmatrix}
=\frac{1}{8}\begin{bmatrix} 2 & -4 \\ -3 & 6 \end{bmatrix}=\begin{bmatrix} \frac{1}{4} & \frac{-1}{2} \\ \\ \frac{-3}{8} & \frac{3}{4} \end{bmatrix}=\frac{1}{8}A
as we might expect beforehand, i.e. A=8A_{1}. Note that
p\left(\overrightarrow{x} \right)=\overrightarrow{x}A_{1},
\left[P\right] _{N}= A_{1} and \left[P\right] _{B}=\begin{bmatrix} 1 & 0 \\0 & 0 \end{bmatrix}.
For the sake of later reference, we generalize (2.7.72) to linear operators on F^{n} or n × n matrix over a field F as follows.
