Linear Algebra Vocab

Front 1

Rank

Back 1

the rank of a matrix is the dimension of the vector space spanned by either the matrix's columns or rows.

Rank(A)=dim(row(A))=dim(col(A))

Front 2

Nullity of (A)

Back 2

the dimension of the null space of the matrix

dim(null(A))

Front 3

null space

Back 3

given a matrix A, all solutions for v where

A*v=0

Front 4

kernel

Back 4

given a linear transform T, the kernel is the set of inputs that after you apply the linear transform, you end at the 0 vector

Front 5

Properties of null spaces

Back 5

if vector B is the RREF of vector A, then

row(A)=row(B)

null(A)=null(B)

col(A) =! col(B) most of the time

Front 6

if T is a linear transform, what condition must be met in order for T to be one to one?

Back 6

the kernel of T must be 0

kern(T)={0}=0

Front 7

Properties of bases (pronounced like baysees)

Back 7

We are allowed to

multiply an entry by a non zero scalar

add one entry of a basis to another

perform EROs on the vectors

Front 8

True/False: All basis of a given span are the same dimension

Back 8

True.They may not be identical, but they must have the same number of vectors, each with the same dimension

Front 9

Rank Nullity Theorem for a matrix

Back 9

Given a matrix A:

rank(A)+nullity(A)=# of columns in the matrix

Front 10

Range(A)

Back 10

the range of a given matrix is equal to the span of its columns

Range(A)=col(A)

Front 11

define hypervolume

Back 11

the "volume" of a matrix spanning more than R3

it is calculated using the determinant of the matrix

Front 12

cofactor matrix

Back 12

In a given 3*3 matrix: | 1 2 3 |

| 0 4 5 |

| 1 0 6 |

the 3*3 cofactor matrix looks like this

| (4*6-5*0) -(0*6-5*1) (0*0-4*1) |

the next entry is the determinant created by crossing out the 1st column and 2nd row, continuing to alternate between positive and negative

| -(2*6-3*0) (1*6-3*1) -(1*0-2*1) |

| (2*5-3*4) -(1*5-3*0) (1*4-2*0) |

Front 13

determinant of a diagonal matrix

| 3 0 0 |

| 0 5 0 |

|0 0 2 |

Back 13

determinants of diagonal matrices are equal to the product of their leading entries in each row.

3*5*2= 30

THIS IS ALSO TRUE FOR ANY TRIANGULAR MATRIX

Front 14

Properties of determinants, given Matrices A and B

Back 14

a.) A is invertable if and only if det(A)=0

b.) det(A) = det(A)^T (linear transforms have no impact on the determinant of a matrix)

c.) det(AB) = det(A) * det(B)

d.) If A is triangular, then det(A) is the product of its diagonal entries

Front 15

adjugate matrix

Back 15

the transpose of a cofactor matrix

Front 16

define transpose of a matrix

Back 16

reflect matrix A over its diagonal ( \ )

write the rows of A as the columns of A

Front 17

If matrix (A) has a rank of 0, what do we know about its possible solutions?

Back 17

If a rank(A)=0, the corresponding linear system will have only the trivial solution, therefor its columns are linearly independent

Front 18

when are linear transformations invertable?

Back 18

linear transformations are only invertable when the domain and codomain have the same dimension

Front 19

Steps to compute the basis of the null space of matrix A

Back 19

1.) row reduce A

2.) express the general solution to Ax=0 in vector form

3.) those vectors are a basis for A

Front 20

Properties for bases

Back 20

(S is a subspace, U is the set of vectors in S)

a.) if U is a basis for S, every vector in S can be written as a unique linear combination of U

b) if U is linearly independent, and not a basis for S, then we can add some vectors from S to U in order to find a basis. All subspaces have a basis, just keep adding vectors in the subspace until you find it.

c.) if U is linearly dependent, and spans S, we can remove some excess vectors to get a basis for S

d.) if U has dim(S) vectors, and is either linearly independent or spans S, then U is already a basis for S

e.) if U has fewer than dim(S) vectors, U cannot span S