Introduction to Projective Geometry Solutions 5.4

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

1 2 3 4 5

1. If $X_1$ and $X_2$ are linearly independent vectors, is it necessarily true that $B{X}_1$, $B{X}_2$ 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 $c_{1}B{X}_1+c_{2}B{X}_2=0$ 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 $DA{D}^{-1}$. But from Theorem 1, the matrices $A$ and $DA{D}^{-1}$ must have the same characteristic values. Hence $A$ and $B^{-1}AB$ have the same characteristic values.

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

$$BA{B}^{-1}=\left(\begin{array}{ccc}3& 1& 0\\ -4& -1& 0\\ -2& -1& 1\end{array}\right)$$

Both $A$ and $BA{B}^{-1}$ have a characteristic value of 1 repeated 3 times. Rank of both $A-I$ and $BA{B}^{-1}-I$ is 1. The characteristic vectors of $BA{B}^{-1}$ 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 $A=\left(\begin{array}{ccc}-3& -2& 2\\ 14& 7& -5\\ 10& 4& -2\end{array}\right)$ and $B=\left(\begin{array}{ccc}1& 2& 0\\ -1& 1& 1\\ -2& 1& 2\end{array}\right)$

$$BA{B}^{-1}=\left(\begin{array}{ccc}17& -36& 14\\ 18& -37& 14\\ 27& -57& 22\end{array}\right)$$

Both $A$ and $BA{B}^{-1}$ have a characteristic values 2, -1 and 1. Rank of both $A-kI$ and $BA{B}^{-1}-kI$ is 2 for each characteristic values. The characteristic vectors of $BA{B}^{-1}$ 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 $A=\left(\begin{array}{ccc}2& 1& -2\\ -2& 5& -7\\ -2& 2& -3\end{array}\right)$ and $B=\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& -1\\ 1& 1& 0\end{array}\right)$

$$BA{B}^{-1}=\left(\begin{array}{ccc}-7& -2& 7\\ -12& -1& 10\\ -12& -3& 12\end{array}\right)$$

Both $A$ and $BA{B}^{-1}$ have a characteristic values 3, -1 and 2. Rank of both $A-kI$ and $BA{B}^{-1}-kI$ is 2 for each characteristic values. The characteristic vectors of $BA{B}^{-1}$ 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).