Rank, Nullity, Dimension and Elimination
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1 Rank, Nullity, Dimension and Elimination Last Frame Test Last Frame Example Definition: Reduced Echelon System Definition: Rank and Nullity Definition: Dimension Theorem: Rank plus Nullity equals Variable Count Elimination Theorem: Elimination An Elimination Algorithm Example
2 Last Frame Test In a sequence of combo-swap-mult operations, a frame without a signal equation 0 = 1 is called the Last Frame provided: Each nonzero equation has a lead variable. This means the variable has leading coefficient 1 and it appears nowhere else in the system. Equations 0 = 0 appear at the end. Variables within an equation appear in variable list order. Nonzero equations are listed with lead variables in variable list order.
3 Last Frame Example Let s assume variable list order x, y, z. x + y = 2, z = 3, 0 = 0. The lead variables are x, z and the free variable is y.
4 Definition 1 (Reduced Echelon System) A linear system which passes the last frame test is called a reduced echelon system. Definition 2 (Rank and Nullity) Assume the last frame test has been passed. Then Rank = number of lead variables, Nullity = number of free variables (non-lead variables). Determining the rank and nullity of a system Display a frame sequence whose first frame is that system, and whose last frame passes the Last Frame Test. The last frame is a reduced echelon system. The rank of the system is the number of lead variables in the last frame. The nullity of the system is the number of variables minus the rank. It is exactly the free variable count.
5 Theorem 1 (Rank and Nullity) The following equations hold: lead variable count + free variable count = number of variables, or rank + nullity = number of variables.
6 Definition 3 (Dimension) The dimension of a system of equations is the number of free variables, which is the number of invented symbols assigned in the Last Frame Algorithm. The case of a homogeneous 3 3 system gives geometric meaning to the term dimension. In this case, dimension 0 is the unique solution case, geometrically a point, and dimensions 1, 2, 3 enumerate the possibilities for the general solution to describe a line, plane or all of space. Dimension is devoid of geometric meaning. It is a gross measure of the complexity of the solution set of the system of equations.
7 Elimination The elimination algorithm applies at each algebraic step one of the three toolkit rules swap, multiply and combination. The objective of each algebraic step is to increase the number of lead variables. The process stops when a signal equation (typically 0 = 1) is found. Otherwise, it stops when no more lead variables can be found, and then the last system of equations is a reduced echelon system. A detailed explanation of the process appears infra. Reversibility of the algebraic steps means that no solutions are created nor destroyed throughout the algebraic steps: the original system and all systems in the intermediate steps have exactly the same solutions. The final reduced echelon system has either a unique solution or infinitely many solutions. In both cases we report the general solution. In the infinitely many solution case, the last frame algorithm is used to write out a general solution.
8 Theorem 2 (Elimination) Every linear system has either no solution or else it has exactly the same solutions as an equivalent reduced echelon system, obtained by repeated application of the toolkit rules swap, multiply and combination.
9 An Elimination Algorithm An equation is said to be processed if it has a lead variable. Otherwise, the equation is said to be unprocessed. 1. If an equation 0 = 0 appears, then move it to the end. If a signal equation 0 = c appears (c 0 required), then the system is inconsistent. In this case, the algorithm halts and we report no solution. 2. Identify the first symbol x r, in variable list order x 1,..., x n, which appears in some unprocessed equation. Apply the multiply rule to insure x r has leading coefficient one. Apply the combination rule to eliminate variable x r from all other equations. Then x r is a lead variable: the number of lead variables has been increased by one. 3. Apply the swap rule repeatedly to move this equation past all processed equations, but before the unprocessed equations. Mark the equation as processed, e.g., replace x r by boxed symbol x r. 4. Repeat steps 1 3, until all equations have been processed once. Then lead variables x i1,..., x im have been defined and the last system is a reduced echelon system.
10 1 Example (Elimination) Solve the system. w + 2x y + z = 1, w + 3x y + 2z = 0, x + z = 1. Solution The answer using the natural variable list order w, x, y, z is the standard general solution Details appear in the next few slides. w = 3 + t 1 + t 2, x = 1 t 2, y = t 1, z = t 2, < t 1, t 2 <.
11 Elimination Details We will apply the three rules swap, multiply and combination for equivalent equations to obtain a frame which passes the Last Frame Test. Then we apply the Last Frame Algorithm to obtain the general solution of the system. Let s mark processed equations with a box enclosing the lead variable (w is marked w ). w + 2x y + z = 1 w + 3x y + 2z = 0 x + z = 1 1 w + 2x y + z = x z = 1 x + z = 1 w + 2x y + z = 1 x + z = 1 0 = 0 w + 0 y z = 3 x + z = 1 0 =
12 Documentation of the steps 1 Original system. Identify the variable order as w, x, y, z. 2 Choose w as a lead variable. Eliminate w from equation 2 by using combo(1,2,-1). 3 The w-equation is processed. Let x be the next lead variable. Eliminate x from equation 3 using combo(2,3,-1). 4 Eliminate x from equation 1 using combo(2,1,-2). Mark the x-equation as processed. Last Frame Test passed. The four frames make the frame sequence which takes the original system into the last frame. Basic exposition rules apply: 1. Variables in an equation appear in variable list order. 2. Equations inherit variable list order from the lead variables.
13 Last Frame Algorithm The last frame of the sequence, which must pass the Last Frame Test, is used to write out the general solution, as follows. w = 3 + y + z x = 1 z y = t 1 z = t 2 Solve for the lead variables w, x. Assign invented symbols t 1, t 2 to the free variables y, z. w = 3 + t 1 + t 2 x = 1 t 2 y = t 1 z = t 2 Back-substitute free variables into the lead variable equations to display the standard general solution. Requirement: variables must be listed in variable list order [w, x, y, z].
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