Find the Null Space [[2t],[t],[3t]]

Problem

Null Space*[[2*t],[t],[3*t]]

Solution

1. Identify the given vector as a linear transformation or a matrix representation. Since the vector is expressed in terms of a single variable t we can rewrite it as a product of a constant vector and the scalar t

[[2*t],[t],[3*t]]=t*[[2],[1],[3]]

2. Define the matrix A representing this system. If we interpret the task as finding the null space of the matrix A=[[2,1,3]] (treating the components as coefficients of a linear equation), we solve A*x=0

2*(x_1)+(x_2)+3*(x_3)=0

3. Solve for the free variables. We have one equation and three variables, meaning there are two free variables. Let (x_2)=s() and (x_3)=r

2*(x_1)=−s−3*r

(x_1)=−1/2*s−3/2*r

4. Express the solution in vector form to find the basis for the null space.

[[(x_1)],[(x_2)],[(x_3)]]=[[−1/2*s−3/2*r],[s()],[r]]

[[(x_1)],[(x_2)],[(x_3)]]=s*[[−1/2],[1],[0]]+r*[[−3/2],[0],[1]]

5. Simplify the basis vectors by multiplying by constants to remove fractions.

(v_1)=[[−1],[2],[0]]

(v_2)=[[−3],[0],[2]]

Final Answer

Null Space*[[2,1,3]]=span{[[−1],[2],[0]],[[−3],[0],[2]]}

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