Math 235

General Proofs Help Table

What you are trying to prove	What to use when specific theorems fail
$A\le B$	Prove A is a subset of B

7.1 Fundamental Subspaces

Concepts	Representations
$Col(A)$	Let $\vec{x} \in Col(A)$ and $\vec{r_1}, ... \vec{r_m}=rows \space in \space A$. $\vec{x}=c_1\vec{r_1}+...+c_m\vec{r_m}$
$Null(A)$	Let $\vec{x} \in Null(A)$ and $\vec{r_1}, ... \vec{r_m}=rows \space in \space A$. $\vec{0}=\begin{bmatrix}\vec{r_1}\vec{x}\\\vdots\\\vec{r_m}\vec{x}\end{bmatrix}$

Extensions from Theorems

Theorem 7.1.5: $rank(A)=rank(A^T)$

$ightarrow Col(A)=Row(A^T), Row(A)=Col(A^T)$

8.2 Linear Mappings

Inventing/Finding linear mappings

Common $V$ and $W$ vector spaces to use:

Vector Space	Dimension
$P_2(\R)$	# of $x^n$ terms e.g. $dim(a+bx+cx^2)=3$
$M_{2x2}(\R)$	4
$^n$	n

Practice: pg. 224 q3,4

Proofs Help

Definitions	Assumptions/Information	Interpretations/Applicable Theorems
$L: V\rightarrow W$ is linear mapping	$\{L(\vec{v_1}), ...L(\vec{v_k})\}$ spans $W$	$Range(L)=W$, $dim(Range(L))=k=rank(L)$
$L: V\rightarrow W$ is linear mapping		$dim(V)=n$ $rank(L)=dim(Range(L))$
$L: V\rightarrow W$ is linear mapping	$Ker(L)=\{\vec{0}\}$	$dim(Ker(L))=0=nullity(L)$
$L: U\rightarrow V$ is linear mapping $M: V\rightarrow W$ is linear mapping	$Range(M \circ L)$	Let $\vec{x}\in Range(M\circ L)$. Then there exists $\vec{v}\in V$ where $\vec{x}=(M\circ L)(\vec{v})=M(L(\vec{v}))\in Range(M)$. Therefore $Range(M \circ L)$ is a subset of $Range(M)$

What you are trying to prove	What to use
$dim V \le dim W$	1. Rank nullity theorem 2. Prove that V is a subset of the W
$rank(M \circ L) \le rank (M)$	Prove that the $Range(M \circ L)$ is a subset of $Range(M)$ - Can only use this if both ranges are in the same subspace
$rank(M \circ L) \le rank (L)$	Since they are not in the same subspace, we cannot use the above strategy Instead, analyze kernels and use rank-nullity theorem. $ightarrow$ Prove that $ker(M \circ L) \ge ker(L)$, then use rank-nullity theorem to do the rest

When you are given information about rank, try to turn it into information about nullity instead since you can generally do more with kernals then range.

8.3 Matrix of a Linear Mapping

Solving for $[\vec{x}_B]$:

• $\vec{x}=c_1\vec{B_1}+c_2\vec{B_2}...$ Solve for $c_1, c_2, ...$

  • If $\vec{x}$ is a polynomial, collect like terms on the RHS if $\vec{x}$ is in the form $a+b\vec{x}+...$ This way we can do coefficient equality in order to solve for the coefficients (pg. 226 example)

Solving for $[L(\vec{v})]_C$

• $L(\vec{v})=c_1C_1+...$ Solve for $c_1, ...$

8.4 Isomorphisms

Assumptions/Information	Interpretations/Applicable Theorems
$V$ is isomorphic to $W$	1. There exists $L: V \rightarrow W$ that is linear mapping. 2. L is 1-1 and onto $ightarrow$ $Range(L)=W$ and $Ker(L)=\{\vec{0}\}$

What you are trying to prove	What to use
$V$ is isomorphic to $W$	1. Define a basis for $V$ and a basis for $W$ 2. Define mapping $L: V \rightarrow W$ that maps the basis for $V$ to the basis for $W$ e.g. $V=\{ \vec{v_1},...\vec{v_2}\}$, $W=\{ \vec{w_1},...\vec{w_2}\}$ $L(t_1\vec{v_1}+...+t_n\vec{v_n})=t_1\vec{w_1}+...+t_n\vec{w_n}$ 3. Prove that the mapping is linear, 1-1, and onto.
$L$ is injective	Prove that $ker(L)=\{\vec{0}\}$