Answer to Question #258889 in Linear Algebra for bruce

Question #258889

"W = {(x, y, z) \u2208R3|3x \u22122y + 4z = 0}."

1
Expert's answer
2021-11-01T18:33:59-0400
"3x\u22122y+4z=0"

Any vector "\\vec w=\\begin{bmatrix}\n x \\\\\n y \\\\\n z\n\\end{bmatrix}" in "W" satisfies "3x-2y+4z=0" or equivalently "x=\\dfrac{2}{3}y-\\dfrac{4}{3}z."

Thus we have

"\\vec w=\\begin{bmatrix}\n x \\\\\n y \\\\\n z\n\\end{bmatrix}=\\begin{bmatrix}\n \\dfrac{2}{3}y-\\dfrac{4}{3}z \\\\\n y \\\\\n z\n\\end{bmatrix}=y\\begin{bmatrix}\n 2\/3 \\\\\n 1 \\\\\n 0\n\\end{bmatrix}+z\\begin{bmatrix}\n -4\/3 \\\\\n 0 \\\\\n 1\n\\end{bmatrix}"

Let

"\\vec u_1=\\begin{bmatrix}\n 2\/3 \\\\\n 1 \\\\\n 0\n\\end{bmatrix}, \\vec u_2=\\begin{bmatrix}\n -4\/3 \\\\\n 0\\\\\n 1\n\\end{bmatrix}"

The above computation shows that any vector "\\vec w"  in "W" can be written as a linear combination of the vectors "\\vec u_1, \\vec u_2."

Hence the set "\\{\\vec u_1, \\vec u_2\\}" is a spanning set for the subspace "W."

We claim that "\\{\\vec u_1, \\vec u_2\\}" is a linearly independent set.

Consider


"a_1\\vec u_1+a_2\\vec u_2=\\vec 0"

The equation can be written as


"\\begin{bmatrix}\n (2\/3)a_1-(4\/3)a_2 \\\\\n a_1 \\\\\n a_2\n\\end{bmatrix}=\\begin{bmatrix}\n 0 \\\\\n 0 \\\\\n 0\n\\end{bmatrix}"

Comparing entries, we obtain "a_1=a_2=0."

Thus the equation "a_1\\vec u_1+a_2\\vec u_2=\\vec 0" has only the zero solution and hence the vectors "\\vec u_1, \\vec u_2"

are linearly independent.

We found a basis


"\\bigg\\{\\vec u_1=\\begin{bmatrix}\n 2\/3 \\\\\n 1 \\\\\n 0\n\\end{bmatrix}, \\vec u_2=\\begin{bmatrix}\n -4\/3 \\\\\n 0\\\\\n 1\n\\end{bmatrix}\\bigg\\}."



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