Answer to Question #115329 in Linear Algebra for Sourav mondal

Question

Check if (1, 3, 0) lies in the range of a linear
operator T on R3 given by
T (x1, x2 , x3) = (x1 , x2+ x3, x1– x2).
Is T one-one and onto ? Give reasons.

Answer 1

To check if b=(1,3,0) lies in the range of T:

The corresponding matrix A of the given linear map T is:

$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix}$

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the row reduced echelon form of this matrix is:

$M=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$ ...(1)

Now, let A(x₁,x₂,x₃)=b = (1,3,0)

Then, x₁-x₂=0 which implies x₁=x₂ ,

x₂+x₃= 3 which implies x₃= 2

x₁= 1 which implies x₂=1

Thus, there exists c=(1,1,2) in R³ such that T(1,1,2)=(1,3,0)

So,(1,3,0) lies in the range of T.

The linear map T is one to one if and only if null space of T has only a trivial solution.

null space of T is given by T(x₁,x₂,x₃) = (0,0,0)

now T(x₁,x₂,x₃) = (0,0,0)

that is A(x₁,x₂,x₃) = (0,0,0)

From (1),

x₁=x₂=x₃ = 0

Since this is a trivial solution, T is one to one.

T is onto:

T is onto if and only if columns of A span R³.

Since the row reduced echelon form of corresponding matrix A of the linear map T in (1) has a pivot in each row,

thus, columns of A are linearly independent, hence, columns of A span R³.

T is onto map.