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let $f$ be a function defined from $R3$ to $R2$, with matrix equal to $$ A = \begin{bmatrix}2&0&-1\\0&-1&1\end{bmatrix}$$ with respect to these two basis: $$ B = ((1, -1, 0), (0, 1, 0), (0, 0, 2))$$ and $$ B' = ((0, 2), (-1, 0)) $$

I need to find the linear transformation. for example, if I have $$ R3 -> R2 $$ defined as $$ f: (x, y, z) -> (2x+y, z-y) $$

I need to find this part $((2x+y, z-y))$, but in the other exercise.

I have no clue how to do it, I don't even know how to start. I know that if I have this part $((2x+y, z-y))$, it's pretty easy to find the matrix, just find the image of each vector of the basis of R3, and then use the knowledge of span to find the scalars (i.e the entry of the matrix).

but I don't know how to do the reverse, from matrix to linear transformation.

I hope this question is clear, if it's not, let me know and I'll edit it.

EDIT: the solution must be f((x, y, z)) = (x + y − z/2, 4x − z) but I got another solution,

$(x, y, z) = α*(1,-1,0) + β*(0,1,0)+γ*(0,0,2)$ with a linear system of three equations I found α, β, and γ.

$α = x$

$β = y+x$

$γ = (1/2)*z$

and by substituting α, β , and γ in this equation: $T(x,y,z)=x*T(1,−1,0)+(y+x)*T(0,1,0)+((1/2)*z)*T(0,0,2)$ I got: $(x,y)=(2x-(1/2)*z, -y-x, -(1/2)*z + (1/2)*z)$

$-(1/2)*z + (1/2)*z$ the first term and the second term, of course, cancel out.

But it's not the correct solution, and I don't know why.

my exercise

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1 Answer 1

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The idea is: given $(x,y,z)$, write it as a linear combination in its basis $B$, this is, $$ (x,y,z) = \alpha (1,-1,0)+ \beta (0,1,0) + \gamma (0,0,2). $$

Find $\alpha, \beta,\gamma$ as a function of $x,y$ and $z$.Let $T$ be the linear transformation you are looking for. When you already know $\alpha, \beta,\gamma$ , then applying $T$ in the equation above, you find $$ T(x,y,z) = \alpha T(1,-1,0)+ \beta T(0,1,0) + \gamma T(0,0,2). $$ Due to $A$ matrix definition, you already have to $T(1,-1,0)= (2,0)$, $T(0,1,0) = (0,-1)$ and $T(0,0,2)=(-1,1)$.

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  • $\begingroup$ after that, how can I obtain the linear transformation formula if I don't have it explicitly? I'm still confused $\endgroup$ Commented Dec 23, 2021 at 9:55
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    $\begingroup$ but you'll get it, if you follow the step-by-step I wrote above, as I said, essentially just find the scalars $\alpha, \beta, \gamma.$ $\endgroup$
    – Ilovemath
    Commented Dec 23, 2021 at 19:38
  • $\begingroup$ the correct solution of this exercise must be f((x, y, z)) = (x + y − z/2, 4x − z) and I got another solution (see photo in my question) $\endgroup$ Commented Dec 24, 2021 at 6:04
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    $\begingroup$ @GabrielBurzacchini . The method in this answer is correct and is perhaps the least laboursome . Otherwise you have to find the matrix wrt standard ordered basis of $\mathbb{R}^{2}$ and $\mathbb{R^{3}}$. and then the calculation of the transformation will be easier. But this method requires less effort. If you have not made any calculation errors then the answer you derive by this method is correct. And you cannot expect people to verify answers. They are here to "help" you . Not to provide full solutions to you. $\endgroup$ Commented Dec 24, 2021 at 7:34

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