Section C.2 Unit 2 applied, geometric, and computational interpretation
These solution sketches correspond to the additional applied and computational interpretation exercises at the end of Unit 2.
Subsection Projection and forgotten height in code
The matrix \(P\) keeps the first two coordinates and drops the third:
\begin{equation*}
P
\begin{bmatrix}
x\\y\\z
\end{bmatrix}
=
\begin{bmatrix}
x\\y
\end{bmatrix}.
\end{equation*}
The two inputs \(\mathbf{x}\) and \(\mathbf{x}+6\mathbf{z}\) have the same first two coordinates, so their outputs agree. Since \(P\mathbf{z}=\mathbf{0}\text{,}\) the height direction is forgotten. The rank is \(2\) and the nullity is \(1\text{,}\) so two input directions are transmitted and one input direction is forgotten.
Subsection Reachable outputs from augmented rref
For \(\mathbf{b}_1\text{,}\) the rref is
\begin{equation*}
\left[
\begin{array}{cc|c}
1\amp0\amp1\\
0\amp1\amp2\\
0\amp0\amp0
\end{array}
\right],
\end{equation*}
so the system is consistent and \(\mathbf{x}=\begin{bmatrix}1\\2\end{bmatrix}\) is a solution.
For \(\mathbf{b}_2\text{,}\) the rref is
\begin{equation*}
\left[
\begin{array}{cc|c}
1\amp0\amp0\\
0\amp1\amp0\\
0\amp0\amp1
\end{array}
\right].
\end{equation*}
The last row is \([0\ 0\mid 1]\text{,}\) which means \(0=1\text{.}\) Thus \(\mathbf{b}_2\) is not reachable. A pivot to the right of the vertical line means inconsistency.
Subsection Line of intersection from an rref output
The rref gives
\begin{equation*}
x+\frac{5}{3}z=1,
\qquad
y-\frac{2}{3}z=0.
\end{equation*}
Let \(z=t\text{.}\) Then
\begin{equation*}
x=1-\frac{5}{3}t,
\qquad
y=\frac{2}{3}t.
\end{equation*}
So the solution set is
\begin{equation*}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}
=
\begin{bmatrix}
1\\0\\0
\end{bmatrix}
+
t
\begin{bmatrix}
-5/3\\
2/3\\
1
\end{bmatrix}.
\end{equation*}
There is one free parameter, so the solution set is a line. Geometrically, it is the line of intersection of the two planes.
Subsection Plane through three points from a null-space computation
The rows of \(M\) are displacement vectors in the plane. A vector in \(\operatorname{null}(M)\) is perpendicular to both rows, so it is normal to the plane. The output gives
\begin{equation*}
\mathbf{n}
=
\begin{bmatrix}
2\\-1\\1
\end{bmatrix}.
\end{equation*}
Using \(P=(1,1,-2)\text{,}\) the plane equation is
\begin{equation*}
2(x-1)-(y-1)+(z+2)=0,
\end{equation*}
or
\begin{equation*}
2x-y+z=-1.
\end{equation*}
Subsection Reading a null-space basis in code
The vector
\begin{equation*}
\mathbf{z}
=
\begin{bmatrix}
-1\\-1\\1
\end{bmatrix}
\end{equation*}
is an input direction. Direct multiplication gives \(A\mathbf{z}=\mathbf{0}\text{.}\) If \(A\mathbf{x}=\mathbf{y}\text{,}\) then
\begin{equation*}
A(\mathbf{x}+t\mathbf{z})
=
A\mathbf{x}+tA\mathbf{z}
=
\mathbf{y}.
\end{equation*}
Thus different inputs can give the same output.
Subsection Redundant square-footage features in code
The equation
\begin{equation*}
X
\begin{bmatrix}
1\\1\\-1
\end{bmatrix}
=
\mathbf{0}
\end{equation*}
means
\begin{equation*}
\text{first-level area}+\text{second-level area}=\text{total area}
\end{equation*}
for every row. The rank is \(2\text{,}\) so the three feature columns contain only two independent directions.
The two coefficient vectors differ by the null-space direction:
\begin{equation*}
\mathbf{c}-\mathbf{c}_{\mathrm{alt}}
=
\begin{bmatrix}
1\\1\\-1
\end{bmatrix}.
\end{equation*}
Since \(X\mathbf{z}=\mathbf{0}\text{,}\) the two coefficient vectors make the same predictions. With redundant features, individual coefficient importance is not uniquely determined by the prediction vector.
Subsection Difference matrix: levels versus changes
The product \(D\mathbf{x}\) gives consecutive differences:
\begin{equation*}
5-2=3,
\qquad
9-5=4,
\qquad
10-9=1.
\end{equation*}
Adding \(10\) to every entry changes the level but not the differences, so the first two outputs agree. Since
\begin{equation*}
D\mathbf{1}=\mathbf{0},
\end{equation*}
the constant direction is forgotten. The rank is \(3\text{.}\) Since \(D\) has four columns, the nullity is \(1\text{.}\)
Subsection Debugging a column-space basis from code
The pivot indices
(0, 1) mean mathematical columns 1 and 2. Row reduction identifies pivot positions, but row operations do not preserve the original columns. A basis for \(\operatorname{col}(A)\) must use columns of the original matrix \(A\text{,}\) not columns of \(R\text{.}\)
Thus a basis is
\begin{equation*}
\left\{
\begin{bmatrix}
1\\2\\0
\end{bmatrix},
\begin{bmatrix}
2\\4\\1
\end{bmatrix}
\right\}.
\end{equation*}
Subsection Invertibility gallery in code
A \(2\times 2\) matrix is invertible exactly when its determinant is nonzero, equivalently when its rank is \(2\text{.}\) The invertible matrices are the second, third, and fourth matrices. The first and fifth matrices are singular because their determinants are \(0\) and their ranks are \(1\text{.}\)
The first matrix
\begin{equation*}
\begin{bmatrix}
0\amp0\\
0\amp1
\end{bmatrix}
\end{equation*}
forgets the \(x\)-direction:
\begin{equation*}
\begin{bmatrix}
0\amp0\\
0\amp1
\end{bmatrix}
\begin{bmatrix}
1\\0
\end{bmatrix}
=
\begin{bmatrix}
0\\0
\end{bmatrix}.
\end{equation*}
The fifth matrix
\begin{equation*}
\begin{bmatrix}
1\amp0\\
0\amp0
\end{bmatrix}
\end{equation*}
forgets the \(y\)-direction:
\begin{equation*}
\begin{bmatrix}
1\amp0\\
0\amp0
\end{bmatrix}
\begin{bmatrix}
0\\1
\end{bmatrix}
=
\begin{bmatrix}
0\\0
\end{bmatrix}.
\end{equation*}
Subsection Reading an inverse-computation output
The left side of the vertical line has reduced to \(I_3\text{,}\) so \(A\) is invertible. The right side is the inverse:
\begin{equation*}
A^{-1}
=
\begin{bmatrix}
3\amp -1\amp -10\\
-2\amp 1\amp 6\\
0\amp 0\amp 1
\end{bmatrix}.
\end{equation*}
Row-reducing \([A\mid I_3]\) applies the same row operations to \(A\) and \(I_3\text{.}\) If the left side becomes \(I_3\text{,}\) then the right side is the matrix that undoes \(A\text{,}\) namely \(A^{-1}\text{.}\)
