Busy. Please wait.

Forgot Password?

Don't have an account?  Sign up 

show password


Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.

By signing up, I agree to StudyStack's Terms of Service and Privacy Policy.

Already a StudyStack user? Log In

Reset Password
Enter the email address associated with your account, and we'll email you a link to reset your password.

Remove ads
Don't know (0)
Know (0)
remaining cards (0)
To flip the current card, click it or press the Spacebar key.  To move the current card to one of the three colored boxes, click on the box.  You may also press the UP ARROW key to move the card to the "Know" box, the DOWN ARROW key to move the card to the "Don't know" box, or the RIGHT ARROW key to move the card to the Remaining box.  You may also click on the card displayed in any of the three boxes to bring that card back to the center.

Pass complete!

"Know" box contains:
Time elapsed:
restart all cards

Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.

  Normal Size     Small Size show me how

Definitions Math250B


Linear Independence Let V be a vector space and let v1, ... , vn be vectors in V. We say that v1, ... ,vn are linearly independent if: c1v1 + c2v2 + ... +cn vn = 0
Define the set of all possible linear combinations of the Vi's. Let V = vector space & v1, v2, ... ,vn = vectors in V. The span of v1, ... ,vn "span {v1, ...vn}" is the subspace defined by span {v1, ..., vn} = {c1v1, c2v2, ..., cnvn : c1, c2, ... cn E R}.
When does vector v1, ... ,vn span the vector space V? Vectors v1, ..., vn span the vector space V if any vector v E V can be written as a linear combinations of the vector v1, ... vn.
Define spanning sets. Let V ba a vector space and v1,... vn be vectors in V. By a linear combination of the vectors v1, ... vn, we ge the form c1v1 + c2v2 + ... + cnvn E V with c1, ...cn E R.
What is the relationship between zero vector and subspace? Every subspace must contain the zero vector of V. If 0 not= W then W not a subsapce of V
Let V be a vector space & let W c V be a non-empty subset. We say W is a subspace of V if: (2 conditions) 1-Closure under Addition w1, w2 then w1 + w2 E W 2-Closure under Scalar Multiplication a E R, w E W then w E W
Define subspace. Let V be a vector space vector. A subset W c V is said to be a subspace of V if W is closed under additiona dn scalar multiplication.
Define Vector Space Meets all 10 axioms (1-closure under Addition, 2- Closure under Scalar Multiplication, ...)
Non-empty set of V is called a vector space if: any 2 elements in V can be: 1- added together, i.e. x E V , y E V , then x + y E V (closure under addition) 2-Multiplied by scalar, i.e. x E V , scalar a E R or a E C then, ax E V (closure under scalar multiplication)
Real vector vs complect vectors The elements of V are called vectors. If the scalar are reals, then V is a real vector space. If scalar is complex, then V is a complex vector space.
The adjoint matrix of an n x n matrix A is also an n x n matrix defined by: adj(A) = [Cij]T of n x n = [C11 C12 ... C1n Cn1 Cn2 ... Cnn]transpose
Describe Determint process Let A be n x n matrix. The minor Mij of A is the determint of the (n-1) x (n-1) matrix obtained by deleting row i & column j of A
Relationship between Inverse matrix and determint A matrix is invertible if and only if its determint is NON-ZERO.
Properties of the DETERMINT: (4) 1- If [A] has a row or column of 0, det (A) = 0 2-If [A] has 2 identifcal rows or columns, det(A) = 0. 3- If we switch 2 rows or columns in [A], det (A) is multiplied by "-". 4-If a row of [A] is a multiple of k, k can be pulled outside the determint.
What effect does row-elemetry operation have on a determint. The determint remained unchanged under the row-elemtry operation Aij (k). (add rows)
Determints ca be computed along ... any row or column.
We say that an n x n matrix A is invertible if: (formula) A(-1) * A = In = A * A (-1)
A system will have infintely many solutions if: Rank (A) < # of columns of A (# variable) # of free variable = Rank(A) - # of variables
Homogeneous systems always: have at least 1 solution (0,0,0).
If Rank(A) = # of columns of A then, the system has a unique solution.
If Rank(A)< Rank(A#) then, the system has no solution.
Definition of Rank of A. Let A be an m x n matrix. The # of non-zero rows of its row-echlen form is called the rank of A.
If A is an m x n matrix, the nullspace of A is defined by: nullspace(A) = {x E R: Ax=0}
Created by: DrMolina