# Assume that T is a linear transformation. Find the standard matrix of T.T=RR^2rarrRR^4 such that T(e_1)=(7,1,7,1),and T(e_2)=(-8,5,0,0),where e_1=(1,0), and e_2=(0,1).

Assume that T is a linear transformation. Find the standard matrix of T.
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$\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{e}_{2}=\left(0,1\right)$

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Let $T:V\to W$ be a linear transformation. Suppose dim $V=n,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}S=\left\{{v}_{1},\dots ,{v}_{n}\right\}$ is an ordered basis for V and suppose dim $W=m\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\left\{{w}_{1},\dots ,{w}_{m}\right\}$ is an ordered basis for W
1.Calculate $T\left({v}_{1}\right),T\left({v}_{2}\right),\dots ,T\left({v}_{n}\right)$
2.Find the coordinate vectors ${\left(T\left({v}_{1}\right)\right)}_{B}\mid {\left(T\left({v}_{2}\right)\right)}_{B},\dots ,{\left(T\left({v}_{m}\right)\right)}_{B}$
3.Write the matrix with columns as the column vectors calculated in Step 2:
$M=\left[{\left(T\left({v}_{1}\right)\right)}_{B}|{\left(T\left({v}_{2}\right)\right)}_{B}|\dots \mid {\left(T\left({v}_{m}\right)\right)}_{B}\right]$
Clearly, here $V={\mathbb{R}}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}W={\mathbb{R}}^{4}.HereS=\left\{{e}_{1}=\left(1,0\right),{e}_{2}=\left(0,1\right)\right\}$is the standard ordered basis for $V\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}B=\left\{{ϵ}_{1}=\left(1,0,0,0\right),{ϵ}_{2}=\left(0,1,0,0\right),{ϵ}_{3}=\left(0,0,0,1\right),{ϵ}_{4}=\left(0,0,0,1\right)\right\}$ is the standard ordered basis for W.
Now we calculate  it follows that
$T\left({e}_{1}\right)=7{ϵ}_{1}+1{ϵ}_{2}+7{ϵ}_{3}+1{ϵ}_{4}⇒{\left[T\left({e}_{1}\right)\right]}_{B}=\left(7,1,7,1\right)$
$T\left({e}_{2}\right)=8{ϵ}_{1}+5{ϵ}_{2}+0{ϵ}_{3}+0{ϵ}_{4}⇒{\left[T\left({e}_{2}\right)\right]}_{B}=\left(-8,5,0,0\right)$
Now we create the matrix M with columns as the column as the column vectors of ${\left[T\left({e}_{1}\right)\right]}_{B},{\left[T\left({e}_{2}\right)\right]}_{B}$.
Therefore