# Penn State: Math 220 Cheat Sheet by luckystarr

Cheat sheet for MATH 220 Penn State Students, Matrices Course

### 1.1

 A Matrix row, columns Coeffi­cients Matrix Just Left Hand Side Augmented Matrix Left and Right Hand Side Solving Linear Systems (1) Augmented Matrix (2) Row Operations (3) Solution to Linear System The RHS is the solution One Solution Upper triangle with Augmented Matrix No Solution Last row is all zeros = RHS number Infinitely Many Solutions Last row (including RHS) is all zeros Incons­istent Has No Solution

### 1.2

 Echelon Matrix (1) Zero Rows at the bottom (2) Leading Entries are down and to the right (3) Zeros are below each leading entry Reduced Echelon Matrix (1) The leading entry of each nonzero row is 1 (2) Zeros are below AND above each 1 Pivot Position Location of Matrix that Corres­ponds to a leading 1 in REF Pivot Column Column in Matrix that contains a pivot To get to EF down and right To get to REF up and left Free Variables Variables that don't correspond to pivot columns Consistent System Pivot in every Column

### 1.3

 RR2 Set of all vectors with 2 rows

### 1.4

 Vector Equation x1a1­+­x2­a2­+­x3­a3 =b Matrix Equation Ax­=b If A is an m x n matrix the following are all true or all false Ax = b has a solution for every b in RRm Every b in RRm is a lin. combo of columns in A Columns of A span RRm Matrix A has a pivot in every row (i.e. no row of zeros)
Anything in Bold means it is a vector.

### 1.5

 Homoge­neous Ax = 0 Trivial Solution Ax = 0 if at lease one column is missing a pivot Determine if homogenous Linear System has a non trivial solution (1) Write as Augmented Matrix (2) Reduce to EF (3) Determine if there are any free variab­les­(column w/o pivot) (4) If any free variables, than a non-tr­ivial solution exists (5) Non-Tr­ivial Solution can be found by further reducing to REF and solving for x If Ax = 0 has one free variable Than x is a line that passes through the origin If Ax = 0 has two free variables Than x has a plane that passes through the origin

### 1.7

 Linear Indepe­ndence No free Variables, none of the vectors are multiples of each other To check ind/dep reduce augmented matrix to EF and see if there are free variab­les(ie. every column must have a pivot to be linearly indepe­ndent) To check if multiples u = c * v find value of c, then it is a multiple therefore linearly dependent Linearly Dependent If there are more columns than rows

### 1.8

 Every Matrix Transf­orm­ation is a: Linear Transf­orm­ation T(x) = A(x) If A is m x n Matrix, then the properties are (1) T(u + v) = T(u) + T(v) (2) T(cu) = cT(u) (3) T(0) = 0 (4) T(cu + dv) = cT(u) + dT(v)

### 1.9

 RRn --> RRm is said to be 'onto' Equation T(x) =Ax­=b has a unique solution or more than one solution each row has a pivot RRn --> RRm is said to be one-to-one Equation T(x) =Ax­=b has a unique solution or no solution each row has a pivot

### 2.1

 Addition of Matrices Can Add matrices if they have same # of rows and columns (ie A(3x4) and B(3x4) so you can add them) Multiply by Scalar Multiply each entry by scalar Matrix Multip­lic­ation (A x B) Must each row of A by each column of B Powers of a Matrix Can compute powers by if the matrix has the same number of columns as rows Transpose of Matrix row 1 of A becomes column 1 of A row 2 of A becomes column 2 of A Properties of Transpose (1) if A is m x n, then AT is n x m (2) (A­T)T = A (3) (A + B)T = A T + B T (4) (tA­)T = tAT (5) (A B)T = BT AT

### 2.2

 Singular matrix A matrix that is NOT investable Determ­inate of A (2 x 2) Matrix det A = ad - bc If A is invertable & (nxn) There will never be no solution or infinitely many solutions to Ax = b Properties of Invertable Matricies (A­-1­)-1 = A (assuming A & B are invest­able) (AB)-1 = B-1 A-1 (AT)-1 = (A-1)T Finding Inverse Matrix [A | I ] --> [ I | A-1] Use row operat­ions STOP when you get a row of Zeros, it cannot be reduced

### 2.8

 A subspace S of RRn is a subspace is S satisfies: (1) S contains zero vector (2) If u & v are in S, then u + v is also in S (3) If r is a real # & u is in S, then ru is also in S Subspace RR3 Any Plane that Passes through the origin forms a subspace RR3 Any set that contains nonlinear terms will NOT form a subspace RR3 Null Space (Nul A) To determine in u is in the Nul(A), check if: Au = 0 If yes --> then u is in the Nullspace

### 2.9

 Dimension of a non-zero Subspace # of vectors in any basis; it is the # of linearly indepe­ndent vectors Dimension of a zero Subspace is Zero Dimension of a Column Space # of pivot columns Dimension of a Null Space # of free variables in the solution Ax­=0 Rank of a Matrix # of pivot columns The Rank Theorem Matrix A has n columns: rank A (# pivots) + dim Nul A (# free var.) = n
dim = dimension; var. = variable

### 3.1

 Calcul­ating Determ­inant of Matrix A is another way to tell if a linear system of equations has a solution (1) Det(A) not =0, then Ax­=b has a unique solution (2) Det(A) =0, then Ax­=b has no solutions or inf many If Ax not= 0 A-1 exist If Ax = 0 A-1 Does NOT exist Cofactor Expansion Use row/column w/ most zeros If Matrix A has an upper or lower triangle of zeros The det(A) is the multip­lic­ation down the diagonals

### 3.2

 Determ­inate Property 1 If a multiple of 1 row of A is added to another row to produce Matrix B, then det(B)­=det(A) Determ­inate Property 2 If 2 rows of A are interc­hanged to produce B, then det(B)­=-d­et(A) Determ­inate Property 3 If one row of A is multiplied to produce B, then det(B)­=k*­det(A) Assuming both A & B are n x n Matrices (1) det(AT) = det(A) (2) det(AB) = det(A)­*det(B) (3) det(A-1) = 1/det(A) (4) det(cA) = cn det(A) (5) det(Ar) = (detA)r

### 3.3 AKA Cramer's Rule

 Cramer's Rule Can be used to find the solution to a linear system of equations Ax­=b when A is an investable square matrix Def. of Cramer's Rule Let A be an n x n invertible matrix. For any b in RRn, the unique solution x of Ax­=b has entries given by xi = detAi(­b­)­/det(A) i = 1,2,...n Ai(b) is the matrix A w/ column i replaced w/ vector b

### 5.1

 Au­=λu A is an nxn matrix. A nonzero vector u is an eigenv­ector of A if there exists such a scalar λ To determine if λ is an eigenvalue reduce [(A-λI)|0] to echelon form and see if it has any free variables. yes -> λ is Eigenvalue no -> λ is not eigenvalue To determine if given vector is an eigenv­ector Ax=λx Eigenspace of A = Nullspace of (A-λI) Eigenv­alues of triangular Matrix entries along diagonal *you CANNOT row reduce a matrix to find its eigenv­alues

### 5.2

 If λ is an eigenvalue of a Matrix A then (A-λI)­x­=­0 will have a nontrivial solution A nontrivial solution will exist if det(A-­λI)=0 (Chara­cte­ristic Equation) A is nxn Matrix. A is invertible if and only if (1) The # 0 is NOT an λ of A (2) The det(A) is not zero Similar Matrices If nxn Matrices A and B are similar, then they have the same charac­ter­istic polynomial (same λ) with same multip­lic­ities

### 5.3

 A matrix A written in diagonal form A=PDP-1 Power of Matrix Ak = Diagonal matrix and #'s on diagonal get raised to the k Determ­ining if Matrix is Diagon­ali­zable λ of a nxn matrix n distinct (or real) λ then matrix is diagon­ali­zable less than n λ, it may or may not be diagon­ali­zable; it will be if # of linearly dependent eigenv­ectors = n eigenv­ectors of nxn matrix n linearly indepe­ndent eigenv­ectors, then diagon­ali­zable less than n linearly indepe­ndent eigenv­ectors, then matrix is NOT diagon­lizable D matrix w/ λ down diagonal P columns of P have linearly n linearly indepe­ndent eigenv­ectors Finding P solve A-λI and plug in the λ values. Reduce to EF, solve for x, & find eigenv­ector

### 6.1

 Length of vector x ||x|| = sqrt(x­12­+x­22) Length fo vector x in RR2 ||x|| = sqrt(x • x) The Unit Vector u = v/||v|| Two vectors u & v in RRn, the distance between u & v ||u - v|| Two vectors u & v are orthogonal if and only if ||u+v|­|2= ||u||2 +||v||2 u • v = 0

### 6.2

 The distance from y to the line through u & the origin ||z|| = ||y - y-ha­t||

### 6.4

 Gram- Schmidt Process Overview take a given set of vectors & transform them into a set of orthogonal or orthon­ormal vectors Given x1 & x2, produce v1 & v2 where the v's are perp. to each other (1) Let v1­=x1 (2) Find v2; v2­=x2 - x2hat x2 hat (x2•v1­)/(­v1•v1) * v1 Orthogonal Basis {v1­,­v2­,...,­­vn} Orthon­ormal Basis {v1­/|­|­v1||, v2/ ||v2­|­|,..., vn­/||­v­n||}

### 7.1

 Symmetric Matrix A square matrix where AT=A If A is a symmetric Matrix then eigenv­ectors associated w/ distinct eigenv­alues are orthogonal If a matrix is symmet­rical, it has an orthogonal & orthon­ormal basis of vectors Orthogonal matrix is a square matrix w/ orthon­ormal columns (1) Matrix is square (2) Columns are orthogonal (3) Columns are unit vectors If Matrix P has orthon­ormal columns PTP=I If P is a nxn orthogonal matrix PT=P-1 A=PDPT A must be symmetric, P must be normalized

### 7.1 Example (2.2)

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