Philadelphia university Department of basic Sciences Final exam(linear algebra 250241)
... ) T(x,y,z)=(3,y+z,x-y+z) is linear transformation from R3 to R3 )If A and B are a square matrix of order n, then ...
... ) T(x,y,z)=(3,y+z,x-y+z) is linear transformation from R3 to R3 )If A and B are a square matrix of order n, then ...
Lecture 30 - Math TAMU
... operator L : Rn → Rn given by L(x) = Ax. Let v1 , v2 , . . . , vn be a nonstandard basis for Rn and B be the matrix of the operator L with respect to this basis. Theorem The matrix B is diagonal if and only if vectors v1 , v2 , . . . , vn are eigenvectors of A. If this is the case, then the diagonal ...
... operator L : Rn → Rn given by L(x) = Ax. Let v1 , v2 , . . . , vn be a nonstandard basis for Rn and B be the matrix of the operator L with respect to this basis. Theorem The matrix B is diagonal if and only if vectors v1 , v2 , . . . , vn are eigenvectors of A. If this is the case, then the diagonal ...
A.1 Summary of Matrices
... This produces a third order polynomial in λ whose three roots are the eigenvalues λ i. There are several characteristics of the operator A that determine the character of the eigenvalue. Briefly summarized they are (a) if A is hermitian, then the eigenvalues are real and the eigenvectors are orthogo ...
... This produces a third order polynomial in λ whose three roots are the eigenvalues λ i. There are several characteristics of the operator A that determine the character of the eigenvalue. Briefly summarized they are (a) if A is hermitian, then the eigenvalues are real and the eigenvectors are orthogo ...
Lab # 7 - public.asu.edu
... by typing C:=matrix([[1,2],[3,4]]); or C:=matrix(2,2,[1,2,3,4]); You can enter the vector (1,2,3) by typing b := vector( [1,2,3]); etc. (To solve system of linear equations Ax=b, type linsolve(A,b);) Also, matrix multiplication is different than ordinary multiplication, so Maple uses &* for matrix p ...
... by typing C:=matrix([[1,2],[3,4]]); or C:=matrix(2,2,[1,2,3,4]); You can enter the vector (1,2,3) by typing b := vector( [1,2,3]); etc. (To solve system of linear equations Ax=b, type linsolve(A,b);) Also, matrix multiplication is different than ordinary multiplication, so Maple uses &* for matrix p ...
(pdf)
... R is the reduced echelon form of A. Mark as true or false the following statements: • row(A) = row(R). • col(A) = col(R). • null(A) = null(R). Find a basis for the vector space col(A). List the free variables for the system Ax = b and find a basis for the vector space null(A). Find the rank(A). 3. E ...
... R is the reduced echelon form of A. Mark as true or false the following statements: • row(A) = row(R). • col(A) = col(R). • null(A) = null(R). Find a basis for the vector space col(A). List the free variables for the system Ax = b and find a basis for the vector space null(A). Find the rank(A). 3. E ...
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Final Exam [pdf]
... (b) If a 6 × 10 matrix is row equivalent to an echelon matrix with 4 non-zero rows, then the dimension of the null space of A is 2. ...
... (b) If a 6 × 10 matrix is row equivalent to an echelon matrix with 4 non-zero rows, then the dimension of the null space of A is 2. ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.