Question

Let B1={[1 1], [1 0], [1 0] and B2={[1 0], [0 1], [0 1]
[0 1] [0 1] [0 -1] } [0 0] [0 1] [0 -1] }
be 2 bases for span(B1) in M22, with the usual left to right ordering.
Let B3 be a basis for P1 and be the transition matrix from B2 to B3 giiven by [1 1 1]
[0 1 1] = PB2→B3
[0 0 1]
a) Find transition matrix PB1→B3
b) Use PB2→B3 to find B3

Answer 1

Answer on Question 44405 - Math - Linear Algebra

$$B _ {1} = \left\{\left[ \begin{array}{l l} 1 & 1 \\ 0 & 1 \end{array} \right]; \left[ \begin{array}{l l} 1 & 0 \\ 0 & 1 \end{array} \right]; \left[ \begin{array}{l l} 1 & 0 \\ 0 & - 1 \end{array} \right] \right\}$$

and

$$B _ {2} = \left\{\left[ \begin{array}{l l} 1 & 0 \\ 0 & 0 \end{array} \right]; \left[ \begin{array}{l l} 0 & 1 \\ 0 & 1 \end{array} \right]; \left[ \begin{array}{l l} 0 & 1 \\ 0 & - 1 \end{array} \right] \right\}$$

be 2 bases for $\operatorname{span} B_1 \subset M_{2x2}(\mathbb{R})$ with the usual left to right ordering. Let $B_3$ be a basis for $P_1$ and be a transition matrix from $B_2$ to $B_3$ given by

$$\left[ \begin{array}{c c c} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] = P B _ {2} \to B _ {3}$$

(a) Find transition matrix $PB_{1}\rightarrow B_{3}$ (b) Use $PB_{2}\to B_{3}$ to find $B_{3}$

Firstly, we notice that bases $B_{1}$ and $B_{2}$ are bases for linear subspace

$$B = \left\{\left[ \begin{array}{c c} x & y \\ 0 & z \end{array} \right], x, y, z \in \mathbb {R} \right\}$$

which is isomorphic to $\mathbb{R}^3$ . Let's find respective bases of $B$ in $\mathbb{R}^3$ for $B_1$ and $B_2$ . According to $B_1$

$$B = \left[ \begin{array}{c c} \alpha + \beta + \gamma & \alpha \\ 0 & \alpha + \beta - \gamma \end{array} \right], \alpha , \beta , \gamma \in \mathbb {R}$$

that's why it has a base

$$P B _ {1} = \left\{\left[ \begin{array}{l} 1 \\ 1 \\ 1 \end{array} \right], \left[ \begin{array}{l} 1 \\ 0 \\ 1 \end{array} \right], \left[ \begin{array}{l} 1 \\ 0 \\ - 1 \end{array} \right] \right\}$$

Analogically,

$$P B _ {2} = \left\{\left[ \begin{array}{l} 1 \\ 0 \\ 0 \end{array} \right], \left[ \begin{array}{l} 0 \\ 1 \\ 1 \end{array} \right], \left[ \begin{array}{l} 0 \\ 1 \\ - 1 \end{array} \right] \right\}$$

It's known that

$$B _ {3} \to P B _ {2} = (P B _ {2} \to B _ {3}) ^ {- 1} = \left[ \begin{array}{l l l} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right] ^ {- 1} = \left[ \begin{array}{l l l} 1 & - 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{l l l} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \end{array} \right]$$

Hence, we can find $B_{3}$ as a matrix $3x3$

$$B_{3} = B_{3} \rightarrow P B_{2} \cdot P B_{2} \cdot P B_{2} \rightarrow B_{3} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} =$$ $$= \begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 2 \\ 0 & 1 & 0 \end{bmatrix}$$

and $B_{3}$ as a span of $2x2$ matrices

$$B_{3} = \left\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}; \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}; \begin{bmatrix} -1 & 2 \\ 0 & 0 \end{bmatrix} \right\}$$

Due to standard theory of transition matrix

$$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = s_{11} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s_{21} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + s_{31} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$ $$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = s_{12} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s_{22} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + s_{32} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$ $$\begin{bmatrix} -1 \\ 2 \\ 0 \end{bmatrix} = s_{13} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} + s_{23} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + s_{33} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$

It's easy to show that

$$s_{11} = 0; \quad s_{21} = \frac{1}{2}; \quad s_{31} = \frac{1}{2}$$ $$s_{12} = 0; \quad s_{22} = \frac{1}{2}; \quad s_{32} = \frac{-1}{2}$$ $$s_{13} = 2; \quad s_{23} = \frac{-5}{2}; \quad s_{33} = \frac{-1}{2}$$

So, $S = P B_{1} \rightarrow B_{3} =$

$$\begin{bmatrix} 0 & 0 & 2 \\ \frac{1}{2} & \frac{1}{2} & \frac{-5}{2} \\ \frac{1}{2} & \frac{-1}{2} & \frac{-1}{2} \end{bmatrix}$$

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