Each archetype below is a linear transformation. Compute the kernel for each.

Archetype M, Archetype N, Archetype O, Archetype P, Archetype Q, Archetype R, Archetype S, Archetype T, Archetype U, Archetype V, Archetype W, Archetype X

The linear transformation $\ltdefn{T}{\complex{4}}{\complex{3}}$ is not injective. Find two inputs $\vect{x},\,\vect{y}\in\complex{4}$ that yield the same output (that is $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$). \begin{equation*} \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4}}= \colvector{ 2x_1+x_2+x_3\\ -x_1+3x_2+x_3-x_4\\ 3x_1+x_2+2x_3-2x_4 } \end{equation*}

Define the linear transformation \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{2}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3} \end{equation*} Find a basis for the kernel of $T$, $\krn{T}$. Is $T$ injective?

Let $A = \begin{bmatrix} 1 & 2 & 3 & 1 & 0\\ 2 & -1 & 1 & 0 & 1\\ 1 & 2 & -1 & -2 & 1\\ 1 & 3 & 2 & 1 & 2 \end{bmatrix}$ and let $\ltdefn{T}{\complex{5}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Is $T$ injective? (Hint: No calculation is required.)

Let $\ltdefn{T}{\complex{3}}{\complex{3}}$ be given by $\lt{T}{\colvector{x\\y\\z}} = \colvector{2x + y + z\\ x - y + 2z\\ x + 2y - z}$. Find $\krn{T}$. Is $T$ injective?

Let $A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & -1 & 1 & 0 \\ 1 & 2 & -1 & -2 \\ 1 & 3 & 2 & 1 \end{bmatrix}$ and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Find $\krn{T}$. Is $T$ injective?

Let $A = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 2 & 1 & 2 \\ 1 & 2 & 1 & 1 \end{bmatrix}$ and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Find $\krn{T}$. Is $T$ injective?

Let $T : M_{22} \rightarrow P_2$ be given by $T\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = (a + b) + (a + c)x + (a + d)x^2$. Is $T$ injective? Find $\krn{T}$.

Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{3}}$, $\lt{T}{\colvector{x\\y\\z}} = \colvector{2x + y\\2y + z\\x + 2z}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2},\, \lt{T}{\vect{e}_3} }$ is a linearly independent set.

Given that the linear transformation $\ltdefn{T}{\complex{2}}{\complex{3}}$, $\lt{T}{\colvector{x\\y}} = \colvector{x+y\\2x + y\\x + 2y}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2} }$ is a linearly independent set.

Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{5}}$, $\lt{T}{\colvector{x\\y\\z}} = \begin{bmatrix} 1 & 3 & 2\\ 0 & 1 & 1\\ 1 & 2 & 1\\ 1 & 0 & 1\\ 3 & 1 & 2 \end{bmatrix} \colvector{x\\y\\z}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2},\, \lt{T}{\vect{e}_3} }$ is a linearly independent set.

Show that the linear transformation $R$ is not injective by finding two different elements of the domain, $\vect{x}$ and $\vect{y}$, such that $\lt{R}{\vect{x}}=\lt{R}{\vect{y}}$. ($S_{22}$ is the vector space of symmetric $2\times 2$ matrices.) \begin{equation*} \ltdefn{R}{S_{22}}{P_1}\quad \lt{R}{\begin{bmatrix}a&b\\b&c\end{bmatrix}}=(2a-b+c)+(a+b+2c)x \end{equation*}

Suppose $U$ and $V$ are vector spaces. Define the function $\ltdefn{Z}{U}{V}$ by $\lt{Z}{\vect{u}}=\zerovector_{V}$ for every $\vect{u}\in U$. Then by Exercise LT.M60, $Z$ is a linear transformation. Formulate a condition on $U$ that is equivalent to $Z$ being an injective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): $Z$ is injective if and only if $U$ is $\Rule{1in}{1pt}{0pt}$. (See Exercise SLT.M60, Exercise IVLT.M60.)

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. For which vectors $\vect{v}\in V$ is $\preimage{T}{\vect{v}}$ a subspace of $U$?

Suppose that that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are linear transformations. Prove the following relationship between kernels. \begin{equation*} \krn{T}\subseteq\krn{\compose{S}{T}} \end{equation*}

Suppose that $A$ is an $m\times n$ matrix. Define the linear transformation $T$ by \begin{equation*} \ltdefn{T}{\complex{n}}{\complex{m}},\quad \lt{T}{\vect{x}}=A\vect{x} \end{equation*} Prove that the kernel of $T$ equals the null space of $A$, $\krn{T}=\nsp{A}$.