linear_algebra.md

Linear Algebra Summary

מטריצות דומות (Similar Matrices)

A and B are similar matrices if there exists an invertible matrix P such that $A = PBP^{-1}$.

$$\boxed{\begin{aligned} A &\sim B \\ \text{Rank}(A) &= \text{Rank}(B) \\ \text{trace}(A) &= \text{trace}(B) \\ \det(A) &= \det(B) \\ \text{Same}&\text{ eigenvalues} \end{aligned}}$$

• מטריצות מייצגות דומות

הרחבה ליניארית (Linear Extension)

Let $B = {b_1, b_2, \dots, b_n}$ be a basis for $\mathbb{F}^n$. And $W = {w_1, w_2, \dots, w_n}$ where $w_j \in \mathbb{F}^n$ There exists a unique linear transformation $T$ such that:

$$\boxed{\begin{aligned} T(b_1) &= w_1\\ T(b_2) &= w_2\\ &\vdots\\ T(b_n) &= w_n \end{aligned}}$$

For any vector $v \in V$, let $(a_1, a_2, \dots, a_n)$ be the coordinates on basis $B$, i.e., $[v]_B = (a_1, a_2, \dots, a_n)$.

$$\boxed{\begin{aligned} v &= a_1 b_1 + a_2 b_2 + \dots + a_n b_n\\ T(v) &= T(a_1 b_1 + a_2 b_2 + \dots + a_n b_n)\\ T(v) &= a_1 T(b_1) + a_2 T(b_2) + \dots + a_n T(b_n) \end{aligned}}$$

If $T(b_1), \dots, T(b_n)$ are given, then we can find a general formula for $T$.

Adjugate Matrix & Cramer's Rule

• Why was the adjugate (adjoint) matrix created? Gives a closed-form, explicit formula for finding the inverse matrix. $$\det(A)A^{-1} = \text{adj}(A) = \text{cofactor}(A)^T$$

• Why was Cramer's rule created? Gives a closed-form, explicit formula for the entries of the unique solution.

בדיקה לסכום ישר V ⊕ W (Check for Direct Sum)

$S = V+W$ $V' = {v_1, v_2, \dots, v_k}$ $W' = {w_1, w_2, \dots, w_m}$ $V = \text{sp } V'$ $W = \text{sp } W'$

• $V' \cup W'$ vectors are linearly independent
• $\det(V' \cup W') \neq 0$
• $\text{Rank}(V' \cup W') = n$
• In general $V' \cup W'$ is invertible
• $\dim(V \cap W) = 0$
• $S = V \oplus W$

Grassmann's Formula (משפט המימד)

$\dim(V+W) = \dim(V) + \dim(W) - \dim(V \cap W)$

טרנספורמציית ההטלה (Projection Transformation)

Given two vector sub-spaces $V_1, V_2$.

• Check if $V = V_1 \oplus V_2$
• Let $B = V_1 \cup V_2$ basis
• Solve for a general vector $X$ composed of $x_1, x_2, x_3, \dots, x_n$ as a linear combination of vectors from $B$.
• Each scalar should be in terms of $x_1$ to $x_n$.
• For example: $X = (x_1+x_2)b_1 + x_3 b_2 + \dots$
• כדי לקבל הטלה על $V_1$, "נניח" את החלק שקשור ל-$V_2$ במשוואה שיצאה לנו ונחשב את הוקטור הכללי בעזרת $x_1, \dots, x_n$. (To get the projection on $V_1$, we "leave" the part related to $V_2$ in our equation and calculate the general vector using $x_1, \dots, x_n$.)
• For example, if $b_1$ is a vector from $V_1$ and the others are from $V_2$ then we calculate $T(x_1, x_2, \dots, x_n) = (x_1+x_2)b_1$.
• זה הכיוון ולא מספיק להכנה (This is the direction but not enough for preparation).

Linear Transformations

$T: V \to W$ $\text{Ker } T = {v \in V : T(v) = 0} \subseteq V$ $\dim(\text{Ker } T) \le \dim V$

$\text{Im } T = {T(v) : v \in V} \subseteq W$ $\dim(\text{Im } T) \le \dim W$

If ${v_1, v_2, \dots, v_n}$ spans $V$, then $\text{Im } T = \text{sp}{T(v_1), T(v_2), \dots, T(v_n)}$.

The Rank-Nullity Theorem (משפט המימד)

$\dim(V) = \dim(\text{Ker}(T)) + \dim(\text{Im}(T))$

• תמונות בת"ל אז מקורות בת"ל IF (If images are linearly independent then sources are linearly independent)
• If $T(v_1), T(v_2), \dots, T(v_k)$ בת"ל (are LI) Then ${v_1, v_2, \dots, v_k}$ בת"ל (are LI).
• הפרש תמונות שוות של טרנספורמציה נמצא בגרעין (The difference of equal images of a transformation is in the kernel) $T(v_1) = T(v_2) \implies T(v_1) - T(v_2) = 0 \implies T(v_1 - v_2) = 0 \implies v_1 - v_2 \in \text{Ker } T$
• הפרש פתרון של אי הומוגנית הוא פתרון של הומוגנית מאותה משוואה (The difference of solutions of a non-homogeneous system is a solution of the corresponding homogeneous system).

Isomorphism

Let $T$ be a linear transformation from subspaces $V$ to $W$. $T: V \to W$

• Injective (one-to-one) - חד-חד ערכית

• Surjective - על

• Bijective - חד-חד ערכית ועל

• T Injective

  • $\text{Ker } T = {0}$

• T Surjective

  • $\text{Im } T = W$
  • $\dim(\text{Im } T) = \dim(W)$

• $T$ is an isomorphism iff it is bijective (חח"ע ועל).

• If $\dim(V) = \dim(W)$ then $T$ is Injective IFF $T$ is Surjective.

The Rank-Nullity Theorem (משפט המימד)

$\dim(V) = \dim(\text{Ker}(T)) + \dim(\text{Im}(T))$

• If $\dim(V) > \dim(W)$ then $T$ is not Injective.
• If $\dim(V) < \dim(W)$ then $T$ is not Surjective.
• $TS$ is an Isomorphism IFF $T$ is Surjective and $S$ is Injective.

תנאים להפיכות (Invertibility Conditions)

$A$ is Invertible IFF:

• There exists $B$ s.t. $AB=BA=I$ ($B=A^{-1}$)
• $\det(A) \neq 0$
• $\text{Rank}(A) = n$
• $AX=0$ has only the trivial solution
• $AX=b$ has a unique solution
• Rows/Columns of $A$ span $\mathbb{F}^n$
• Rows/Columns of $A$ are linearly independent (בלתי תלויות ליניארית)
• $\det(A) = \gamma_1 \cdot \gamma_2 \cdots \gamma_n$ (for $\gamma_i$ eigenvalues of A)
• If $A$ is invertible -> $A^k$ is invertible
• If $A, B$ are invertible -> $AB$ is invertible
• $A$ is invertible IFF $\text{transpose}(A)$ is invertible
• Elementary Operations (Gauss) keep matrices invertible
• Transition Matrix from Basis to Basis is invertible
• If $A$ is Triangular or Diagonal and all diagonal elements are $\neq 0$ then $A$ is invertible.