07. 좌표변환Change of Coordinates

Solution. Here $\beta = \{(1,0),(0,1)\}$ is the standard ordered basis and $\beta' = \{(1,1),(1,2)\}$.

(1) The desired matrix has as its columns the $\beta'$-coordinates of the standard basis vectors. Writing $e_1=(1,0)$, $e_2=(0,1)$ as linear combinations of $\beta'$,

so $[e_1]_{\beta'} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$ and $[e_2]_{\beta'} = \begin{pmatrix} -1 \\ 1 \end{pmatrix}$. Hence the matrix changing $\beta$-coordinates to $\beta'$-coordinates is

(Check: $Q = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$ changes $\beta'$-coordinates to $\beta$-coordinates and $\det Q = 1$, so inverting directly gives the same $Q^{-1}$.)

(2) Since $T(a,b) = (2a+b, a-3b)$, $T(1,0) = (2,1)$ and $T(0,1) = (1,-3)$, so

By the change-of-coordinates theorem, $[T]_{\beta'} = Q^{-1}[T]_\beta Q$. First,

and then

(Check: $T(1,1) = (3,-2) = 8(1,1) - 5(1,2)$ and $T(1,2) = (4,-5) = 13(1,1) - 9(1,2)$, matching the first and second columns of $[T]_{\beta'}$.)

Solution. The desired matrix has as its columns the $\beta$-coordinates of the vectors of $\beta'$.

(1) Write $(0,10) = a(-1,3) + b(2,-1)$, so $-a+2b=0$ and $3a-b=10$. The first equation gives $a=2b$; substituting, $6b-b=10$, so $b=2$, $a=4$. Hence $[(0,10)]_\beta = (4,2)$.

Write $(5,0) = c(-1,3)+d(2,-1)$, so $-c+2d=5$ and $3c-d=0$. The second equation gives $d=3c$; substituting, $-c+6c=5$, so $c=1$, $d=3$. Hence $[(5,0)]_\beta = (1,3)$.

Therefore the desired matrix is

(Check: $4(-1,3)+2(2,-1) = (-4,12)+(4,-2) = (0,10)$ and $(-1,3)+3(2,-1) = (-1,3)+(6,-3) = (5,0)$, both correct.)

(2) For $\beta = \{x^2,x,1\}$, the $\beta$-coordinates of a polynomial are exactly its coefficient vector: $a_2x^2+a_1x+a_0$ has $\beta$-coordinates $(a_2,a_1,a_0)$, and likewise the other two vectors have $\beta$-coordinates $(b_2,b_1,b_0)$ and $(c_2,c_1,c_0)$. Therefore the desired matrix is

Solution. By the change-of-coordinates theorem, if $Q$ is the matrix whose columns are the vectors of $\beta$ (i.e. their standard coordinates), then $[L_A]_\beta = Q^{-1}AQ$. Hence

is the desired matrix.

Check. $\det Q = 1\cdot2 - 1\cdot1 = 1 \neq 0$, so $Q$ is invertible with $Q^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$. First,

and then

On the other hand, $L_A(1,1) = A\begin{pmatrix}1\\1\end{pmatrix} = \begin{pmatrix}4\\2\end{pmatrix} = 6(1,1) - 2(1,2)$ and $L_A(1,2) = A\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}7\\3\end{pmatrix} = 11(1,1) - 4(1,2)$, so

which agrees exactly with $Q^{-1}AQ$ computed above.

Solution. Let $\beta' = \{(1,m), (-m,1)\}$. Since $(1,m)\cdot(-m,1) = -m+m = 0$, the two vectors are perpendicular, and since $(1,m) \neq (0,0)$ they are linearly independent; so $\beta'$ is an ordered basis for $\mathbb{R}^2$, with $(1,m)$ a direction vector for $L$ and $(-m,1)$ a direction vector perpendicular to $L$.

The reflection $T$ fixes vectors along $L$ and negates vectors perpendicular to $L$. So $T(1,m) = (1,m)$ and $T(-m,1) = -(-m,1) = (m,-1)$, giving

Let $\beta$ be the standard ordered basis for $\mathbb{R}^2$. The matrix changing $\beta'$-coordinates to $\beta$-coordinates is the matrix with the vectors of $\beta'$ as columns,

and by the change-of-coordinates theorem $[T]_\beta = Q[T]_{\beta'}Q^{-1}$. Since $\det Q = 1+m^2$,

First,

and then

That is, for every $(a,b) \in \mathbb{R}^2$,

(Check: for $m=1$, the reflection about $y=x$, this gives $\frac{1}{2}\begin{pmatrix} 0 & 2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, i.e. $(a,b) \mapsto (b,a)$, which is indeed the reflection about $y=x$.)

Proof. Since $A, B$ are similar, there is an invertible matrix $Q$ with $B = Q^{-1}AQ$.

(1) Suppose $A, B$ are invertible. As a product of invertible matrices, $Q^{-1}AQ$ is invertible, so $B$ is invertible, and

Hence $A^{-1}$ and $B^{-1}$ are similar, via the same $Q$.

(2) Transposing $B = Q^{-1}AQ$ and using $(XY)^t = Y^tX^t$,

Transposing $QQ^{-1} = I$ gives $(Q^{-1})^tQ^t = I$, so $(Q^{-1})^t = (Q^t)^{-1}$. Thus

Set $P := (Q^t)^{-1}$; since $Q^t$ is invertible, so is $P$, and $Q^t = P^{-1}$. Then

so $A^t$ and $B^t$ are similar, via the invertible matrix $P = (Q^t)^{-1} = (Q^{-1})^t$.

Proof. Since $A, B$ are similar, there is an invertible matrix $Q$ with $B = Q^{-1}AQ$.

(1) First we show $\operatorname{tr}(XY) = \operatorname{tr}(YX)$ for any $n \times n$ matrices $X, Y$. Since $(XY)_{ii} = \sum_{k=1}^n X_{ik}Y_{ki}$,

since the order of summation in a finite double sum may be interchanged. Now taking $X = Q^{-1}$, $Y = AQ$,

(2) Let $V$ be a finite-dimensional vector space and $T$ a linear operator on $V$. Fix an ordered basis $\beta$ for $V$ and define

For this to be well-defined, the value must not depend on the choice of $\beta$.

Let $\beta'$ be another ordered basis for $V$. By the change-of-coordinates theorem, there is an invertible matrix $Q$ with $[T]_{\beta'} = Q^{-1}[T]_\beta Q$, i.e. $[T]_\beta$ and $[T]_{\beta'}$ are similar matrices. By part (1),

so the value of $\operatorname{tr}(T)$ does not depend on the choice of basis. Hence the definition above is well-defined.