Answer to Question #271357 in Linear Algebra for Gestavo

Question #271357

It is known that the vectors in the vector space R3:

⃗v1 = (1, 1, 1), v2 = (2, -1, 1), v3 = (0, 2, 1)

and

⃗w1 = (2, -1, 3), w2 = (3, -1, 7), w3 = (-1, 1, 1)

The vectors v1, v2, v3 are basis in R3. Transformation T : R3 → R3 is a linear transformation defined by:

 T( ⃗vi) = ⃗wi

Define:

1. Matrix transformation of T

2. Basis of Ker(T) and Range(T)



1
Expert's answer
2021-11-26T11:47:38-0500

1.


"\\vec e_1=\\begin{bmatrix}\n 1 \\\\\n 0 \\\\\n0\n\\end{bmatrix}=a\\begin{bmatrix}\n 1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}+b\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n1\n\\end{bmatrix}+c\\begin{bmatrix}\n 0 \\\\\n 2 \\\\\n1\n\\end{bmatrix}"

"\\begin{matrix}\n a+2b=1 \\\\\n a-b+2c=0 \\\\\na+b+c=0\n\\end{matrix}=>\\begin{matrix}\n a=3 \\\\\n b=-1 \\\\\n c=-2\n\\end{matrix}"

"\\vec e_1=3\\vec v_1-\\vec v_2-2\\vec v_3"


"T\\begin{pmatrix}\n \\begin{bmatrix}\n 1 \\\\\n 0 \\\\\n0\n\\end{bmatrix}\n\\end{pmatrix}= 3\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n3\n\\end{bmatrix}-\\begin{bmatrix}\n 3 \\\\\n -1 \\\\\n7\n\\end{bmatrix}-2\\begin{bmatrix}\n -1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}"

"=\\begin{bmatrix}\n 5 \\\\\n -4 \\\\\n0\n\\end{bmatrix}"


"\\vec e_2=\\begin{bmatrix}\n 0 \\\\\n 1 \\\\\n0\n\\end{bmatrix}=d\\begin{bmatrix}\n 1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}+f\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n1\n\\end{bmatrix}+g\\begin{bmatrix}\n 0 \\\\\n 2 \\\\\n1\n\\end{bmatrix}"

"\\begin{matrix}\n d+2f=0 \\\\\n d-f+2g=1 \\\\\nd+f+g=0\n\\end{matrix}=>\\begin{matrix}\n d=2 \\\\\n f=-1 \\\\\n g=-1\n\\end{matrix}"

"\\vec e_2=2\\vec v_1-\\vec v_2-\\vec v_3"


"T\\begin{pmatrix}\n \\begin{bmatrix}\n 0 \\\\\n 1 \\\\\n 0\n\\end{bmatrix}\n\\end{pmatrix} = 2\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n3\n\\end{bmatrix}-\\begin{bmatrix}\n 3 \\\\\n -1 \\\\\n7\n\\end{bmatrix}-\\begin{bmatrix}\n -1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}"

"=\\begin{bmatrix}\n 2 \\\\\n -2 \\\\\n -2\n\\end{bmatrix}"



"\\vec e_3=\\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n1\n\\end{bmatrix}=k\\begin{bmatrix}\n 1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}+m\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n1\n\\end{bmatrix}+n\\begin{bmatrix}\n 0 \\\\\n 2 \\\\\n1\n\\end{bmatrix}"

"\\begin{matrix}\n k+2m=0 \\\\\n k-m+2n=0 \\\\\nk+m+n=1\n\\end{matrix}=>\\begin{matrix}\n k=-4 \\\\\n m=2 \\\\\n n=3\n\\end{matrix}"

"\\vec e_3=-4\\vec v_1+2\\vec v_2+3\\vec v_3""T\\begin{pmatrix}\n \\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 1\n\\end{bmatrix}\n\\end{pmatrix} =-4\\begin{bmatrix}\n 2 \\\\\n -1 \\\\\n3\n\\end{bmatrix}+2\\begin{bmatrix}\n 3 \\\\\n -1 \\\\\n7\n\\end{bmatrix}+3\\begin{bmatrix}\n -1 \\\\\n 1 \\\\\n1\n\\end{bmatrix}"

"=\\begin{bmatrix}\n -5\\\\\n 5 \\\\\n 5\n\\end{bmatrix}"




"T(\\vec x)=A\\vec x"

"A=\\begin{bmatrix}\n 5 & 2 & -5 \\\\\n -4 & -2 & 5 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

2.


"A=\\begin{bmatrix}\n 5 & 2 & -5 \\\\\n -4 & -2 & 5 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

"R_1=R_1\/5"

"\\begin{bmatrix}\n 1 & 2\/5 & -1 \\\\\n -4 & -2 & 5 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

"R_2=R_2+4R_1"


"\\begin{bmatrix}\n 1 & 2\/5 & -1 \\\\\n 0 & -2\/5 & 1 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

"R_2=-(5R_2)\/2"


"\\begin{bmatrix}\n 1 & 2\/5 & -1 \\\\\n 0 & 1 & -5\/2 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

"R_1=R_1-(2R_2)\/5"


"\\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & -5\/2 \\\\\n 0 & -2 & 5\n\\end{bmatrix}"

"R_3=R_3+2R_2"


"\\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & -5\/2 \\\\\n 0 & 0 & 0\n\\end{bmatrix}"

"\\begin{bmatrix}\n 1 & 0 & 0 \\\\\n 0 & 1 & -5\/2 \\\\\n 0 & 0 & 0\n\\end{bmatrix}\\begin{bmatrix}\n x_1\\\\\n x_2 \\\\\n x_3\n\\end{bmatrix}=\\begin{bmatrix}\n 0\\\\\n 0 \\\\\n 0\n\\end{bmatrix}"

If we take "x_3=t," then "x_1=0, x_2=(5\/2)t"

Therefore the kernel of "T" has a basis formed by the set


"\\begin{bmatrix}\n 0\\\\\n 5\/2 \\\\\n 1\n\\end{bmatrix}"

"\\vec u_3=0\\vec u_1+(-5\/2)\\vec u_2"

The set "\\bigg\\{\\begin{bmatrix}\n 5\\\\\n -4 \\\\\n 0\n\\end{bmatrix}, \\begin{bmatrix}\n 2\\\\\n -2\\\\\n -2\n\\end{bmatrix}\\bigg\\}" forms a basis for the range of "T."



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