Linear test 2

0.0(0)
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/41

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

42 Terms

1
New cards

Invertible matrix (DETREMINANT ad - bc DOES NOT EQUAL ZERO)

An n x n matrix A is invertible if there is an n x n matrix C such that CA = I and AC = I, where I is the n x n identity matrix. C, in this case, is the inverse of A.

* a non-invertible matrix is sometimes called a singular matrix, while and invertible matrix is called a nonsingular matrix.

*An n x n matrix is invertible if and only if A is row equivalent to the identity matrix

<p>An n x n matrix A is invertible if there is an n x n matrix C such that CA = I and AC = I, where I is the n x n identity matrix. C, in this case, is the inverse of A.</p><p>* a non-invertible matrix is sometimes called a singular matrix, while and invertible matrix is called a nonsingular matrix.</p><p>*An n x n matrix is invertible if and only if A is row equivalent to the identity matrix</p>
2
New cards

singular matrix AKA the name for a matrix that isn't invertible

A singular matrix is a square matrix with no inverse. It's determinant is zero.

<p>A singular matrix is a square matrix with no inverse. It's determinant is zero.</p>
3
New cards

Non-singular matrix AKA the name for a matrix that IS INVERTIBLE

A square matrix with a non-zero determinant.

<p>A square matrix with a non-zero determinant.</p>
4
New cards

A matrix is invertible if ad - bc DOES NOT EQUAL ZERO

cross multiply (ad - bc) to find the determinant if the determinant is NOT Zero keep going, you can now find the inverse by putting the determinant as the denominator (1/ determinant) and multiplying this fraction by the OG matrix now written as [ d -b]

[-c a]

<p>cross multiply (ad - bc) to find the determinant if the determinant is NOT Zero keep going, you can now find the inverse by putting the determinant as the denominator (1/ determinant) and multiplying this fraction by the OG matrix now written as [ d -b]</p><p>[-c a]</p>
5
New cards

Basis of a Matrix

The basis of a matrix refers to a set of linearly independent vectors that span the entire vector space of the matrix. It is the smallest set of vectors that can be used to express all other vectors in the matrix

6
New cards

Reduced Row Echelon Form (RREF)

1. In REF

2. All leading entries are 1

3. In a column with a leading 1, all other elements are 0.

<p>1. In REF</p><p>2. All leading entries are 1</p><p>3. In a column with a leading 1, all other elements are 0.</p>
7
New cards

nullity of a matrix

the dimension of the null space of a matrix

<p>the dimension of the null space of a matrix</p>
8
New cards

null space

Set of all solution to Ax = 0 could also be seen as the columns without the leading one

9
New cards

a basis for a subspace of R^n is set of vectors in S that

1) Span S (ITC, OUR S is R^2)

2) Linearly independent

VECTORS MUST MEET BOTH CONDITIONS TO BE CONSIDERED A BASIS OF R^n

10
New cards

Span of a matrix

the spn of a matrix i.e. R^2 is telling us that all vectors in that matrix have 2 #'s in their COLUMNS & AT LEAST6 have two columns!!!

For a matrix to span r^2 it should have BOTH conditions.. see pg in 3.5

<p>the spn of a matrix i.e. R^2 is telling us that all vectors in that matrix have 2 #'s in their COLUMNS & AT LEAST6 have two columns!!!</p><p>For a matrix to span r^2 it should have BOTH conditions.. see pg in 3.5</p>
11
New cards

AN INVERTIBLE MATRIX IS ALWAYS LINEARLY INDEPENDENT

AN INVERTIBLE MATRIX IS ALWAYS LINEARLY INDEPENDENT

12
New cards

when given matrix B has operations (either plus or minus) det(B) is ..

det(B) = (det A)

13
New cards

when given matrix B has swapped rows det(B) is ..

det(B)= -det(A)

(A row interchange changes the sign of the determinant)

14
New cards

when given matrix B has been multiplied by scalar(s), then det(B) is ..

det(B) = det(A)*scalar(s)

(MULTIPLIED BY EACH SCALAR IF THERE IS MULTIPLE!!!)

(A row scaling also scales the determinant by the same scalar factor.)

15
New cards

Know that Ax = b can NOT be consistent when there are more ROWS THAN COLUMNS

AKA Ax=b has a unique solution for every b

16
New cards

kernel is the same thing as a null space (interchange the word kernel for null)

REMEMBER THAT THE NULL SPACE IS THE SET OF VECTORS! THEREFORE THE KERNEL IS ALSO THE SET OF VECTORS, THAT MAKE A MATRIX EQUAL ZERO... only write the kernel as a number if asked for dim(kernel)

17
New cards

rank(T)= dim(range (T))

remember that these are the NUMBER VALUES NOT SETS OF VECTORS

18
New cards

range is the same thing as col (A)

remember that col(A) is the set of vectors that contains the leading ones...

19
New cards

rank + nullity = # of columns AND dim(range) + dim(kernel) = # of columns

these are the same as one another

20
New cards

what does it mean for something to be onto?

the rank of the matrix in RREF is equal to the codomain (AKA THE # OF ROWS)

21
New cards

domain or R^n

# of columns (m x n is row x columns)

22
New cards

trivial

has no free variables

23
New cards

nontrivial solution

free variable vector?? (a nonzero vector x that satisfies Ax=0)

24
New cards

the dimcol(A) and the rank are equal

TRUE by the Rank Theorem. Also since dimension of row space = number of nonzero rows in echelon form = number leading one columns = dimension of column space. (ur pretty much just counting the # of leading ones horizontally and then vertically)

25
New cards

the transpose of a matrix has the same value as the OG RREF rank

transpose? SAME RANK

26
New cards

if a matrix is invertible, then its is linearly independent right. then if its linearly independent then its RREF must be the Identity matrix!!!! (or at least you know that each row has a leading one)

invertible=linearly independent=identity matrix

<p>invertible=linearly independent=identity matrix</p>
27
New cards

codomian or R^m

# of rows (m x n is rows by columns)

28
New cards

is Ax=b consistent for R?

-if there's a leading 1 in each row, A IS CONSISTENT

- if there's NOT a leading 1 in each row then A IS NOT CONSITENT

29
New cards

is T one-to-one?

T is only ONE - TO - ONE WHEN THE COLUMNS ARE LINEARLY INDEPENDENT

30
New cards

Know that if a matrix is invertible (ad-bc = not zero), then it is also onto and also 1-1.. this is because the matrix as a whole is linearly independent (u can check linear independence by seeing if each row has leading one in RREF)

when a matrix is linearly independent...

31
New cards

Ax=b & linear dependence

-Ax=b has a unique solution means only one solution & when A is linearly independent/ invertible

-when rows > columns Ax = b isn't consistent (not linearly independent)

-when a matrix is linearly dependent, Ax=b has infinitely many solutions

32
New cards

Algebraic multiplicity

The number of times an Eigenvalue repeats.

33
New cards

Geometric multiplicity

dimension of eigenspace (number of free variables)

34
New cards

ONTO (is the matrix onto? AKA IS THERE A LEADING 1 IN EVERY ROW OF ITS RREF)

if there is a LEADING ONE in every row then YES the matrix is ONTO

35
New cards

ONE- TO -ONE (AKA is there a LEADING ONE IN EVERY COLUMN OF THE RREF)

if there is a leading 1 in every column of a matrix’s RREF, the matrix is one - to one

36
New cards

Find a basis is NOT THE SAME AS Find a basis for col(A)!!!!

Find a basis for col(A) means find the basis for the LEADING 1 COLUMNS !!

1) find RREF of matrix if not given and circle leading 1’s

2) write the vector columns with the LEADING 1’s out AS GIVEN DO NOT WRITE THE NUMBERS FROM THE RREF U FOUND LIKE U WOULD FOR A NORMAL BASIS

3) put basis brackets around leading one vectors that were written with the numbers from the non-RREF matrix

37
New cards

the rank of the coefficient matrix and the augmented matrix are the SAME NUMBER!!!

got wrong on test 2 page 3, rank of the coefficient matrix and the augmented matrix are the same number

38
New cards
39
New cards
40
New cards

41
New cards

The basis for a span must

1) HAVE EXACTLY THE # of VECTOR AS THE SPANS MINIMUM, a basis of span R² can ONLY have TWO vectors

  1. 2) must be linearly independent

42
New cards