LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034 1
... 15. Prove that a basis of a subspace S, can always be selected out of a set of vectors which span S. 16. Show that the set of 3 vectors X1 = (1 0 0 ), X2 = (0 1 0) and X3 = (0 0 1) is linearly independent. 17. (a) State the properties of linear transformation. (b) Prove that a linear transformation ...
... 15. Prove that a basis of a subspace S, can always be selected out of a set of vectors which span S. 16. Show that the set of 3 vectors X1 = (1 0 0 ), X2 = (0 1 0) and X3 = (0 0 1) is linearly independent. 17. (a) State the properties of linear transformation. (b) Prove that a linear transformation ...
MTL101:: Tutorial 3 :: Linear Algebra
... (14) Use standard inner product on R2 over R to prove the following statement: “A parallelogram is a rhombus if and only if its diagonals are perpendicular to each other.” (15) Find with respect to the standard inner product of R3 , an orthonormal basis containing (1, 1, 1). (16) Find an orthonormal ...
... (14) Use standard inner product on R2 over R to prove the following statement: “A parallelogram is a rhombus if and only if its diagonals are perpendicular to each other.” (15) Find with respect to the standard inner product of R3 , an orthonormal basis containing (1, 1, 1). (16) Find an orthonormal ...
Math 314 Exam 1
... This part of the exam is take home open book. You may not discuss it with anyone. It is due at 10:30 a.m. on February 20. Clearly show your work and your answers. ...
... This part of the exam is take home open book. You may not discuss it with anyone. It is due at 10:30 a.m. on February 20. Clearly show your work and your answers. ...
Let u1,u2,... ,uk ∈ Rn, and let v1,v2,... ,vm ∈ span(u 1,u2,... ,uk).
... then we have determined that there are x1 , x2 , . . . , xm , not all 0, such that x1 v1 + · · · + xm vm = O and hence that { v1 , v2 , . . . , vm } is linearly dependent. But every homogeneous system of linear equations with more variables than equations has a nontrivial solution, and since k < m, ...
... then we have determined that there are x1 , x2 , . . . , xm , not all 0, such that x1 v1 + · · · + xm vm = O and hence that { v1 , v2 , . . . , vm } is linearly dependent. But every homogeneous system of linear equations with more variables than equations has a nontrivial solution, and since k < m, ...
Numbers and Vector spaces
... 7. Rational functions are ratios of polynomials. Like (x + 1)/(x2 + 1). Strictly speaking, they are not functions on the real line, because the denominator can be zero at some point. Nevertheless it is clear what is a sum or product of two rational functions. Verify that all rational functions with ...
... 7. Rational functions are ratios of polynomials. Like (x + 1)/(x2 + 1). Strictly speaking, they are not functions on the real line, because the denominator can be zero at some point. Nevertheless it is clear what is a sum or product of two rational functions. Verify that all rational functions with ...
Solutions for Midterm I - Stony Brook Math Department
... Now we see that the subspace V is spanned by two vectors, (1, 0, 0, 1) and (2, −1, 1, 0). They comprise a basis of V . Note that a basis is not unique and other answers are possible. One can get a basis of V in a short cut: V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0} = {(x1 , x2 ...
... Now we see that the subspace V is spanned by two vectors, (1, 0, 0, 1) and (2, −1, 1, 0). They comprise a basis of V . Note that a basis is not unique and other answers are possible. One can get a basis of V in a short cut: V = {(x1 , x2 , x3 , x4 ) ∈ R4 | x1 + 2x2 − x4 = 0, x2 + x3 = 0} = {(x1 , x2 ...
Solutions
... Recall that a set of vectors {v1 , . . . , vn } ⊂ V is linearly dependent if there are real numbers x1 , . . . , xn , not all of which are zero, such that 0 = x1 · v 1 + · · · + xn · v n Such a set is linearly independent if it is not linearly dependent; that is, if whenever we write 0 = x1 · v 1 + ...
... Recall that a set of vectors {v1 , . . . , vn } ⊂ V is linearly dependent if there are real numbers x1 , . . . , xn , not all of which are zero, such that 0 = x1 · v 1 + · · · + xn · v n Such a set is linearly independent if it is not linearly dependent; that is, if whenever we write 0 = x1 · v 1 + ...
Basis (linear algebra)
Basis vector redirects here. For basis vector in the context of crystals, see crystal structure. For a more general concept in physics, see frame of reference.A set of vectors in a vector space V is called a basis, or a set of basis vectors, if the vectors are linearly independent and every vector in the vector space is a linear combination of this set. In more general terms, a basis is a linearly independent spanning set.Given a basis of a vector space V, every element of V can be expressed uniquely as a linear combination of basis vectors, whose coefficients are referred to as vector coordinates or components. A vector space can have several distinct sets of basis vectors; however each such set has the same number of elements, with this number being the dimension of the vector space.