I have a linear map $L:\Bbb R^3\to \Bbb R^4$ such that $$L(1,1,2)=(1,0,1,0),$$ $$L(1,2,2)=(2,0,2,0),$$ $$L(2,1,5)=(1,2,2,1).$$ How can i find the formula that linear map is defined by?
Formula linear map defined by
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linear-algebra
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1Do the vectors $(1,1,2), (1,2,2), (2,1,5)$ constitute a basis for $\mathbb{R}^3$? Note that, for example, $(1,2,2)-(1,1,2) = (0,1,0)$. – 2017-02-04
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2Hint: $(1,2,2)-(1,1,2)=(0,1,0)$, $(1,2,2)+(2,1,5)-3(1,1,2)=(0,0,1)$ – 2017-02-04
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0Ok, i`ve got, that (x1,x2,x3)=a(1,1,2)+b(1,2,2)+c(2,1,5) and received that a=(8x-y-3z), b=(-3x+y+z), c=(z-2x). But should i do next? – 2017-02-04
1 Answers
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Form the matrix $$ \left(\begin{array}{c|c} b_1 & c_1 \\ b_2 & c_2 \\ b_3 & c_3 \end{array} \right) $$ and transform it to reduced row-echelon form. You'll get $$ \left( \begin{array}{ccc|ccc} 1 & 1 & 2 & 1 & 0 & 1 & 0\\ 1 & 2 & 2 & 2 & 0 & 2 & 0\\ 2 & 1 & 5 & 1 & 2 & 2 & 1 \end{array} \right) \to \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & -4 & -2 & -2\\ 0 & 1 & 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 2 & 1 & 1 \end{array} \right). $$ It then follows \begin{align*} L(1,0,0) &=(0 , -4 , -2 , -2),\\ L(0,1,0) &=(1 , 0 , 1 , 0),\\ L(0,0,1) &=(0 , 2 , 1 , 1) \end{align*} (why?) and hence $$ L(x_1,x_2,x_3)=(x_2,-4x_1+2x_3, -2x_1+x_2+x_3, -2x_1+x_3) $$ for all $x_1,x_2,x_3 \in \mathbf R.$