# Introduction to Projective Geometry Solutions 5.4

## The Invariance of the Classification

A 3 minute read, posted on 22 Aug 2019
Last modified 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).

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