05. 선형변환Linear Transformation

Solution. We check additivity and homogeneity directly for each function.

1. $T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$ is linear. For $x=(a_1,a_2,a_3)$, $y=(b_1,b_2,b_3)$, $$T(x+y) = ((a_1+b_1)-(a_2+b_2),\, 2(a_3+b_3)) = (a_1-a_2,2a_3)+(b_1-b_2,2b_3) = T(x)+T(y),$$ $$T(cx) = (ca_1-ca_2,\, 2ca_3) = c(a_1-a_2,2a_3) = cT(x),$$ so both additivity and homogeneity hold.
2. $T(f(x)) = xf(x)+f'(x)$ is linear. For $f,g\in P_2(\mathbb{R})$, $c\in\mathbb{R}$, $$T(f+g) = x(f+g)(x)+(f+g)'(x) = \big(xf(x)+f'(x)\big)+\big(xg(x)+g'(x)\big) = T(f)+T(g),$$ $$T(cf) = x(cf)(x)+(cf)'(x) = c\big(xf(x)+f'(x)\big) = cT(f).$$
3. $T(A)=\operatorname{tr}(A)$ is linear, since the trace is the sum of the diagonal entries, so $\operatorname{tr}(A+B)=\operatorname{tr}(A)+\operatorname{tr}(B)$ and $\operatorname{tr}(cA)=c\operatorname{tr}(A)$.
4. $T(a_1,a_2)=(a_1+1,a_2)$ is not linear. Since $T(0,0)=(1,0)\ne(0,0)$, the necessary condition $T(0)=0$ fails. (Directly: $T(1,0)+T(1,0)=(2,0)+(2,0)=(4,0)$ while $T\big((1,0)+(1,0)\big)=T(2,0)=(3,0)$, so additivity also fails.)
5. $T(a_1,a_2)=(\sin a_1,0)$ is not linear. Taking $x=(\pi/2,0)$ and $c=2$, $$T(cx) = T(\pi,0) = (\sin\pi,0)=(0,0), \qquad cT(x) = 2(\sin(\pi/2),0)=(2,0),$$ so $T(cx)\ne cT(x)$ and homogeneity fails.

Solution. Write $(2,3)$ as a linear combination of $(1,0)$ and $(1,1)$. Setting $(2,3)=a(1,0)+b(1,1)=(a+b,\,b)$ gives $b=3$ and $a+b=2$, so $a=-1$, i.e., $$(2,3) = -(1,0) + 3(1,1).$$ Since $T$ is linear, $$T(2,3) = -T(1,0) + 3T(1,1) = -(1,4) + 3(2,5) = (-1,-4)+(6,15) = (5,11).$$ Hence $T(2,3) = (5,11)$.

Solution. The $xy$-plane consists of the points with $z$-coordinate $0$, so reflection about this plane keeps the $x$- and $y$-coordinates fixed and negates the $z$-coordinate. That is, $$T(a,b,c) = (a, b, -c).$$ This is linear, since $T(x+y)=T(x)+T(y)$ and $T(cx)=cT(x)$ hold componentwise trivially.

Solution. Only (a), (b), (c) of Exercise 1 are linear, so we compute for those; (d) and (e) are not linear and hence not applicable.

1. $T(a_1,a_2,a_3)=(a_1-a_2,2a_3)$. $\operatorname{Ker}(T) = \{(a,a,0): a\in\mathbb{R}\} = \operatorname{span}\{(1,1,0)\}$, so the nullity is $1$. By the dimension theorem the rank is $3-1=2$, which equals the dimension of the codomain $\mathbb{R}^2$, so $T$ is onto. But $\operatorname{Ker}(T)\ne 0$, so $T$ is not one-to-one.
2. $T(f(x))=xf(x)+f'(x)$. For $f(x)=a+bx+cx^2$, $T(f) = b + (a+2c)x + bx^2 + cx^3$. If $T(f)=0$ then $b=0$, $c=0$, and $a+2c=0$ gives $a=0$, so $\operatorname{Ker}(T)=0$ and the nullity is $0$. The rank is $3-0=3$, while the codomain $P_3(\mathbb{R})$ has dimension $4$, so $T$ is not onto. Since $\operatorname{Ker}(T)=0$, $T$ is one-to-one.
3. $T(A)=\operatorname{tr}(A)$. $\operatorname{Ker}(T)$ consists of the traceless matrices, of dimension $n^2-1$, so the nullity is $n^2-1$. The rank is $n^2-(n^2-1)=1$, which equals the dimension of the codomain $F$ (e.g., $\operatorname{tr}(E_{11})=1$ generates $F$), so $T$ is onto. For $n\ge 2$, $\operatorname{Ker}(T)\ne 0$, so $T$ is not one-to-one. (For $n=1$, $T$ is the trivial isomorphism.)
4. Not applicable (not linear).
5. Not applicable (not linear).

Proof.

1. Let $x=(a_1,a_2,\cdots)$, $y=(b_1,b_2,\cdots)\in V$ and $c\in\mathbb{R}$. $$T(x+y) = (a_2+b_2,a_3+b_3,\cdots) = (a_2,a_3,\cdots)+(b_2,b_3,\cdots) = T(x)+T(y), \quad T(cx) = (ca_2,ca_3,\cdots) = cT(x),$$ so $T$ is linear. Likewise, $$U(x+y) = (0,a_1+b_1,a_2+b_2,\cdots) = (0,a_1,a_2,\cdots)+(0,b_1,b_2,\cdots) = U(x)+U(y), \quad U(cx) = (0,ca_1,ca_2,\cdots) = cU(x),$$ so $U$ is linear as well.
2. We show $T$ is onto. For any $y=(b_1,b_2,\cdots)\in V$, let $x=(0,b_1,b_2,\cdots)\in V$; then $T(x) = (b_1,b_2,\cdots) = y$, so $T$ is onto. On the other hand, for $x_0=(1,0,0,\cdots)\ne 0$ we have $T(x_0) = (0,0,\cdots) = 0$, so $x_0\in\operatorname{Ker}(T)$ with $x_0\ne 0$. Hence $\operatorname{Ker}(T)\ne 0$ and $T$ is not one-to-one.
3. We show $U$ is one-to-one. Suppose $U(x)=0$ for $x=(a_1,a_2,\cdots)$; then $(0,a_1,a_2,\cdots)=(0,0,0,\cdots)$, so $a_i=0$ for all $i$, i.e., $x=0$. Thus $\operatorname{Ker}(U)=0$ and $U$ is one-to-one. On the other hand, consider $y_0=(1,0,0,\cdots)\in V$. Since the first coordinate of $U(x)$ is always $0$, no $x\in V$ satisfies $U(x)=y_0$. Hence $U$ is not onto.

Solution. Define $T,U:\mathbb{R}^2\to\mathbb{R}^2$ by $$T(a,b) = (a,0), \qquad U(a,b) = (2a,0).$$ $T$ is linear as a coordinate projection, and $U=2T$ is a scalar multiple of a linear transform, hence also linear.

$\operatorname{Ker}(T) = \{(0,b): b\in\mathbb{R}\}$ and $\operatorname{Ker}(U) = \{(a,b): 2a=0\} = \{(0,b): b\in\mathbb{R}\}$, so $\operatorname{Ker}(T)=\operatorname{Ker}(U)$.

$\operatorname{Im}(T) = \{(a,0): a\in\mathbb{R}\}$ and $\operatorname{Im}(U) = \{(2a,0): a\in\mathbb{R}\} = \{(a,0): a\in\mathbb{R}\}$, so $\operatorname{Im}(T)=\operatorname{Im}(U)$.

However, $T(1,0)=(1,0)\ne(2,0)=U(1,0)$, so $T\ne U$. Thus these $T,U$ are distinct linear transforms satisfying both conditions.

Proof.

1. We show $T^{-1}(0) = \operatorname{Ker}(T)$. If $v\in T^{-1}(0)$, then by definition of the preimage $T(v)=0$, so $v\in\operatorname{Ker}(T)$. Conversely, if $v\in\operatorname{Ker}(T)$, then $T(v)=0$, so $v\in T^{-1}(0)$. Both inclusions hold, so $T^{-1}(0)=\operatorname{Ker}(T)$.
2. Suppose $T(x)=y$; we show $T^{-1}(y) = x+\operatorname{Ker}(T)$. ($\subseteq$) Let $z\in T^{-1}(y)$. Then $T(z)=y=T(x)$, so by linearity of $T$, $T(z-x) = T(z)-T(x) = 0$. Hence $z-x\in\operatorname{Ker}(T)$ and $z = x+(z-x)\in x+\operatorname{Ker}(T)$. ($\supseteq$) Let $z\in x+\operatorname{Ker}(T)$, so $z=x+k$ for some $k\in\operatorname{Ker}(T)$. By linearity, $T(z) = T(x)+T(k) = y+0 = y$, so $z\in T^{-1}(y)$. Both inclusions give $T^{-1}(y) = x+\operatorname{Ker}(T)$.