# Introduction to Projective Geometry Solutions 5.4

## The Invariance of the Classification

A 3 minute read, posted on 22 Aug 2019

Tags computer vision, projective geometry, problem solution

Please read this introduction first before looking through the solutions. Here’s a quick index to all the problems in this section.

#### 1. If $X_1$ and $X_2$ are linearly independent vectors, is it necessarily true that $BX_1$, $BX_2$ are linearly independent?

This one is easy to contradict. If either or both $X_1$ and $X_2$ are in the nullspace of $B$, then for a non-zero value of at least one $c_i$ the equation $c_1BX_1 + c_2BX_2 = 0$ will hold. Hence the given statement is not true.

#### 2. Do $A$ and $B^{-1}AB$ always have the same characteristic values?

As $(B^{-1})^{-1} = B$, we can rewrite the second matrix as $B^{-1}A(B^{-1})^{-1}$. Writing $B^{-1}$ as $D$, the second matrix becomes $DAD^{-1}$. But from Theorem 1, the matrices $A$ and $DAD^{-1}$ must have the same characteristic values. Hence $A$ and $B^{-1}AB$ have the same characteristic values.

#### 3. Given $A = \begin{pmatrix}1 & 0 & 0 \\ 2 & 2 & -1 \\ 2 & 1 & 0\end{pmatrix}$ and $B = \begin{pmatrix}-1 & 0 & 1 \\ 4 & 1 & -3 \\ 2 & 1 & -2\end{pmatrix}$. Verify that $A$ and $BAB^{-1}$ have the same characteristic values. Verify that $A - kI$ and $BAB^{-1} -kI$ have the same rank for each characteristic value. Find the characteristic vectors of $BAB^{-1}$, and verify that multiplying each on the left by $B^{-1}$ yields a characteristic vector for $A$.

$$BAB^{-1} = \begin{pmatrix} 3 & 1 & 0 \\ -4 & -1 & 0 \\ -2 & -1 & 1 \end{pmatrix}$$

Both $A$ and $BAB^{-1}$ have a characteristic value of 1 repeated 3 times. Rank of both $A - I$ and $BAB^{-1} -I$ is 1. The characteristic vectors of $BAB^{-1}$ are $(1, -2, 0)$ and $(0, 0, 1)$. Multiplying them on the left by $B^{-1}$ we get, $(1, -2, 0)$, $(-1, 1, -1)$ which are characteristic vectors of $A$ (multiplying them by $A$ just scales the vector).

#### 4. Work Exercise $3$, given $A = \begin{pmatrix}-3 & -2 & 2 \\ 14 & 7 & -5 \\ 10 & 4 & -2\end{pmatrix}$ and $B = \begin{pmatrix}1 & 2 & 0 \\ -1 & 1 & 1 \\ -2 & 1 & 2\end{pmatrix}$

$$BAB^{-1} = \begin{pmatrix} 17 & -36 & 14 \\ 18 & -37 & 14 \\ 27 & -57 & 22 \end{pmatrix}$$

Both $A$ and $BAB^{-1}$ have a characteristic values 2, -1 and 1. Rank of both $A - kI$ and $BAB^{-1} -kI$ is 2 for each characteristic values. The characteristic vectors of $BAB^{-1}$ are $(2, 2, 3)$, $(7, 7, 10)$ and $(5, 6, 9)$. Multiplying them on the left by $B^{-1}$ we get, $(0, 1, 1)$, $(-1, 3, 2)$ and $(-1, 4, 2)$ which are characteristic vectors of $A$ (multiplying them by $A$ just scales the vector).

#### 5. Work Exercise $3$, given $A = \begin{pmatrix}2 & 1 & -2 \\ -2 & 5 & -7 \\ -2 & 2 & -3\end{pmatrix}$ and $B = \begin{pmatrix}1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 1 & 0\end{pmatrix}$

$$BAB^{-1} = \begin{pmatrix} -7 & -2 & 7 \\ -12 & -1 & 10 \\ -12 & -3 & 12 \end{pmatrix}$$

Both $A$ and $BAB^{-1}$ have a characteristic values 3, -1 and 2. Rank of both $A - kI$ and $BAB^{-1} -kI$ is 2 for each characteristic values. The characteristic vectors of $BAB^{-1}$ are $(1, 2, 2)$, $(5, 6, 6)$ and $(1, 6, 3)$. Multiplying them on the left by $B^{-1}$ we get, $(3, 3, 0)$, $(-1, -5, -4)$ and $(-2, 8, 4)$ which are characteristic vectors of $A$ (multiplying them by $A$ just scales the vector).