Linear Algebra Exam Notes

Definition of Linear Operator L: The linear operator $L: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ is defined by $L(x, y, z) = (2y + z, x - 4y, 3x)$ .
Bases: We have the standard basis $S$ for $\mathbb{R}^3$ and another basis $S' = {(1, 1, 1), (1, 1, 0), (1, 0, 0)}$ .

The transition matrix $A$ from $S$ to $S'$ expresses the standard basis vectors in terms of the $S'$ basis.
We need to find scalars $a, b, c$ such that:
- $(1, 0, 0) = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0)$
- $(0, 1, 0) = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0)$
- $(0, 0, 1) = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0)$
Solving these systems:
- For $(1, 0, 0)$ : $a + b + c = 1, a + b = 0, a = 0$ . Thus, $a = 0, b = 0, c = 1$ .
- For $(0, 1, 0)$ : $a + b + c = 0, a + b = 1, a = 0$ . Thus, $a = 0, b = 1, c = -1$ .
- For $(0, 0, 1)$ : $a + b + c = 0, a + b = 0, a = 1$ . Thus, $a = 1, b = -1, c = 0$ .
The transition matrix $A$ is therefore: $A = \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & -1 \ 1 & -1 & 0 \end{bmatrix}$ .

The transition matrix $B$ from $S'$ to $S$ expresses the $S'$ basis vectors in terms of the standard basis $S$ .
$B$ is simply formed by taking the vectors of $S'$ as columns:
$B = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}$ .
Note: $B = A^{-1}$ . Calculating the inverse of $A$ :
$A^{-1} = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix}$

[L]_S: This is the matrix representation of $L$ with respect to the standard basis.
- $L(1, 0, 0) = (0, 1, 3)$
- $L(0, 1, 0) = (2, -4, 0)$
- $L(0, 0, 1) = (1, 0, 0)$
  $[L]_S = \begin{bmatrix} 0 & 2 & 1 \ 1 & -4 & 0 \ 3 & 0 & 0 \end{bmatrix}$ .
[L]{S'}: This is the matrix representation of $L$ with respect to the basis $S'$ .
$[L]{S'} = A^{-1} [L]S A = B [L]S A$
$[L]{S'} = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 0 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 2 & 1 \ 1 & -4 & 0 \ 3 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & -1 \ 1 & -1 & 0 \end{bmatrix}$ $[L]{S'} = \begin{bmatrix} 4 & -2 & 1 \ 1 & -2 & 1 \ 0 & 2 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \ 0 & 1 & -1 \ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -3 & 6 \ 1 & -3 & 3 \ 1 & 1 & -2 \end{bmatrix}$ .

First, find $L(v)$ : $L(1, 0, 1) = (2(0) + 1, 1 - 4(0), 3(1)) = (1, 1, 3)$ .
Now, express $L(v)$ in terms of the $S'$ basis. We want to find $a, b, c$ such that $(1, 1, 3) = a(1, 1, 1) + b(1, 1, 0) + c(1, 0, 0)$ .
$a + b + c = 1, a + b = 1, a = 3$ .
Solving gives $a = 3, b = -2, c = 0$ .
Therefore, $[L(v)]_{S'} = \begin{bmatrix} 3 \ -2 \ 0 \end{bmatrix}$ .

The Linear System:
- $x + 2y + 2z - s + 2t = 0$
- $x + 2y + 3z + s + t = 0$
- $3x + 6y + 8z + s + 5t = 0$
Solving the System: We can represent this system as an augmented matrix and row reduce it.
$\begin{bmatrix} 1 & 2 & 2 & -1 & 2 & 0 \ 1 & 2 & 3 & 1 & 1 & 0 \ 3 & 6 & 8 & 1 & 5 & 0 \end{bmatrix}$
Perform $R2 \rightarrow R2 - R1$ and $R3 \rightarrow R3 - 3R1$ :
$\begin{bmatrix} 1 & 2 & 2 & -1 & 2 & 0 \ 0 & 0 & 1 & 2 & -1 & 0 \ 0 & 0 & 2 & 4 & -1 & 0 \end{bmatrix}$
Perform $R3 \rightarrow R3 - 2R2$ : $\begin{bmatrix} 1 & 2 & 2 & -1 & 2 & 0 \ 0 & 0 & 1 & 2 & -1 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$ Perform $R1 \rightarrow R1 - 2R2$ :
$\begin{bmatrix} 1 & 2 & 0 & -5 & 4 & 0 \ 0 & 0 & 1 & 2 & -1 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$
Perform $R1 \rightarrow R1 - 4R3$ and $R2 \rightarrow R2 + R3$ :
$\begin{bmatrix} 1 & 2 & 0 & -5 & 0 & 0 \ 0 & 0 & 1 & 2 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 \end{bmatrix}$
From the row-reduced echelon form, we have:
- $x + 2y - 5s = 0$
- $z + 2s = 0$
- $t = 0$
  Thus, $x = -2y + 5s, z = -2s, t = 0$ . The variables $y$ and $s$ are free variables.
Basis: The solution can be written as:
$\begin{bmatrix} x \ y \ z \ s \ t \end{bmatrix} = y \begin{bmatrix} -2 \ 1 \ 0 \ 0 \ 0 \end{bmatrix} + s \begin{bmatrix} 5 \ 0 \ -2 \ 1 \ 0 \end{bmatrix}$
A basis for the solution space is ${(-2, 1, 0, 0, 0), (5, 0, -2, 1, 0)}$ .
Dimension: Since there are two free variables, the dimension of the solution space is 2.

The coefficient matrix is:
$\begin{bmatrix} 1 & 2 & 2 & -1 & 2 \ 1 & 2 & 3 & 1 & 1 \ 3 & 6 & 8 & 1 & 5 \end{bmatrix}$
From the row reduction in part (a), we found that there are three non-zero rows in the row-reduced echelon form. Thus, the rank of the coefficient matrix is 3.