Practice: Linear Transformations

(a) $F$ is linear. For $u=(x_1,y_1,z_1)$ and $v=(x_2,y_2,z_2)$, the first component of $F(u+v)$ is $(x_1+x_2)-(z_1+z_2)=(x_1-z_1)+(x_2-z_2)$, the sum of the first components of $F(u)$ and $F(v)$; the second behaves the same way. For scaling, each component of $F(av)$ carries out one factor of $a$. Both conditions hold for arbitrary arguments. Matrix. $F(1,0,0)=(1,2)$, $F(0,1,0)=(0,1)$, $F(0,0,1)=(-1,-2)$, so

Kernel. $F(x,y,z)=(0,0)$ requires $x-z=0$ and $2x+y-2z=0$. The first gives $x=z$; substituting, $2z+y-2z=y=0$. So the kernel is $\{(z,0,z):z\in \mathbb{R}\}$, the line spanned by $(1,0,1)$, of dimension 1. Checking: $F(1,0,1)=(1-1,2+0-2)=(0,0)$ ✓. Image. Columns 1 and 2 are $(1,2)$ and $(0,1)$, which are independent, so the image is all of ${\mathbb{R}}^2$, dimension 2. (Column 3 is $-1$ times column 1, contributing nothing new.) Rank–nullity. $1+2=3=\dim {\mathbb{R}}^3$ ✓. (b) $G$ is not linear. It fails homogeneity, and additivity as well. Take $v=(1,1)$ and $a=2$: $G(2v)=G(2,2)=(4,2)$, while $2G(v)=2(1,1)=(2,2)$. These differ, so $G(av)\ne aG(v)$. Additivity also fails: $G(1,0)+G(0,1)=(0,0)+(0,1)=(0,1)$, while $G(1,1)=(1,1)$. The origin test does not detect it. $G(0,0)=(0\cdot 0,0)=(0,0)$, so $G$ fixes the origin. That test can only rule a map out; passing it establishes nothing, and here the failure is visible only under scaling or addition. (c) $H$ is linear. $H(p)=p-p(0)$ subtracts the constant term. For $p,q\in P_2$,

using that evaluation at 0 is itself linear, and $H(ap)=ap-ap(0)=aH(p)$. Matrix with respect to $1,x,x^2$. $H(1)=1-1=0$, so column 1 is zero. $H(x)=x-0=x$, giving coordinates $(0,1,0)$. $H(x^2)=x^2-0=x^2$, giving $(0,0,1)$. So

Kernel. $H(p)=0$ means $p=p(0)$, a constant. The kernel is the constants, dimension 1. Image. $H$ leaves the $x$ and $x^2$ terms alone and kills the constant, so the image is $\{bx+c{x}^2\}$, the polynomials with zero constant term, dimension 2. Rank–nullity. $1+2=3=\dim P_2$ ✓. Note $H\circ H=H$: subtracting the constant term twice changes nothing after the first time. $H$ is a projection onto the polynomials vanishing at 0, along the constants. (d) What is lost, and why injectivity fails. $F$ has kernel the line spanned by $(1,0,1)$: it collapses that direction to the origin. Concretely $(1,0,1)$ and $(0,0,0)$ have the same image, as do any $u$ and $u+t(1,0,1)$. $H$ has kernel the constants: it forgets a polynomial's value at 0, so $x^2+5$ and $x^2$ have the same image. In general, if $T(u)=0$ for some $u\ne 0$, then for any $v$,

so $v$ and $v+u$ are distinct inputs with equal outputs and $T$ is not injective. Conversely, if the kernel is trivial and $T(v_1)=T(v_2)$, then $T(v_1-v_2)=0$ forces $v_1=v_2$. So injectivity is exactly the condition $\ker T=\{0\}$, which is why the kernel, rather than any count of inputs, is what one computes to decide it.