Ax=0 Unique Solution

First if Ax b has a unique solution call it x. Any multiple of this vector is in the nullspace.

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For any square linear system Avec xvec b over some field there exists a unique solution iff det Aneq 0 as then we can use the inverse matrix.

Ax=0 unique solution. I think this calculation says x 0 must be a solution but dose not guarantee the uniqueness. So one solution is x 1 because the second column is just twice the 0 0 first column. The trap is that Ax b may not have any solutions and the problem cleverly omitted the assumption Ax b is consistent.

From Theorem 44 we know that Ax0 implies that x0necessarily if and only if all the columns ajof Aare linearly independent. If the augmented matrix does not tell us there is no solution and if there is no free variable ie. For an invertible A the inverse A 1 is well-defined thus we can left-multiply A 1 to each side of the equation to get A 1 A x A 1 0 which leads to I x 0 and thus x 0.

Ker A x Rn Ax 0. In this case the span of Ais the whole space SARnand. But in general x0may not be the unique solution.

Let mathzmath be the unique solution of mathAxbmath and let mathymath be a nonzero solution for mathAx0math. That is x0is the unique solution to Ax0if and only if rankAn. If the matrix A is square then this cannot be done and indeed from uniqueness of solution of A x 0 we can deduce that every system A x b has a unique solution.

A n v n 0 has only the trivial solution a 1. If Ax 0 has a unique solution x must be the zero vector implying that As columns are linearly independent. 2 0 x 2 1.

In general suppose you have two solutions x1 and x2 to Ax 0. Let A A1A2An. As a consequence if n mie if the number of unknowns is larger than the number of equations then the system will have infinitely many solutions.

Definition The solution set of a homogeneous equation Ax 0 is called the kernel of A. Ax0 Unique solution trivial solution x0 No free variables Every column of Ais pivot column rank A of columns of A Infinitely many nontrivial solutions Some free variables Some columns of A are not pivot columns rank A. Ax b has the unique solution x A1b if A 6 0.

The homogeneous system Ax 0 has a non-trivial solution if and only if the equation has at least one free variable or equivalently if and only if A has a column with no pivots. Let A A 1A 2A n. Prove or give a counterexample.

Import numpy as py from scipylinalg import solve A 1 1 1 0 1 -3 2 1 5 b 2 1 0 x solve Ab x. Then Ax b has a unique solution if and only if the only solution of Ax 0 is x 0. A rephrasing of this is in the square case Ax b has a unique solution exactly when A1A2An is a linearly independent set.

Lets use python and see what answer we get. A rephrasing of this is in the square case Ax b has a unique solution exactly when fA 1A 2A ngis a linearly independent set. Ax 0 has non-trivial solutions A 0.

Consequently Ax b if it has a solution that solution is also unique. For instance if A 2 4 1 0 0 1 0 0 3 5and b 2 4 0 0 1 3 5 then the equation Ax 0. But is this the only solution.

Which impossible 0 cannot equal -3. Let A A1A2An. Then Ax b has a unique solution if and only if the only solution of Ax 0 is x 0.

So if x is a solution to Ax 0 any other solution can be written as the sum of x and a vector in the nulispace. Letting a different free variable equal 1 and setting the other free variables equal to zero gives us other vectors in the nullspace. Ax b and Ax 0 Theorem 1.

Then by linearity mathAzy. If A 0 then Ax b usually has no solutions but does have solutions for some b. The case of a unique solution.

In all other cases it will have infinitely many solutions. If A has column vectors v 1 v n then those vectors are linearly independent if and only if the equation a 1 v 1. ProblemGiven Ax0 whereAin Rmtimes nxin Rn What is the solution.

The vector x2 x1 will be in the nullspace of A as Ax2xizAx9Ax1bbz0. Every column other than the right-most column is a pivot column then the system has a unique solution. Ax 0 will have a unique solution the trivial solution x 0 if and only if rankA n.

To see this write the linear independence expression in matrix notation. In particular the matrix has a right inverse and so it is invertible. It is not a proof.

When will x0 be the unique solution to Ax0. We can write x2 x1 x2 xi. If Ax 0 has only the trivial solution x 0 then Ax b always has a unique solution.

The solution x 0 is called the trivial solution. The statements 20 and 21 are proved since we have a formula for the solution. Avec xvec bLongleftrightarrow A-1Avec xA-1vec bLongleftrightarrow A-1vec bvec x.

A n 0 This is equivalent to saying A x 0 has only the trivial solution. Therefore this system of linear equations has no solution. Thus the solution set to Ax 0 is Spanuvw or parametrically x ru sv tw where rst R are parameters.

Let A be a square n n matrix.

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