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Given a linear function $T$ such that $T(1,0) = 1$ and $T(0,1) = 0$ then determine whether is $T(1,1) = 1$ .

The given conditions are forming a basis matrix \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix} and if we form a linear equation Ax=b in two variables we can get the constants value as $1$ and $0$.

Based on it $ax+by$ linear equation can be calculated on values of $x$ and $y$ as $1$ and $1$ respectively so am getting the condition as true.

Is $T(1,1) = 1$ and is this a correct approach to solve it?

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    $\begingroup$ $$T(1,1)=T[(1,0)+(0,1)]=T(1,0)+T(0,1)=1+0=1$$ $\endgroup$ – Chinnapparaj R Apr 20 at 17:18
  • $\begingroup$ @ChinnapparajR can we find coefficients a and b of linear function ax+by by plugging in x and y values respectively and then find T(1,1) based on a and b? $\endgroup$ – Ten Doeschate Apr 20 at 17:21
  • $\begingroup$ @TenDoeschate: Yes, but that is a detour. (You should get $T(x,y)=x$). $\endgroup$ – Henning Makholm Apr 20 at 17:23
  • $\begingroup$ @HenningMakholm Got it thanks $\endgroup$ – Ten Doeschate Apr 20 at 17:25
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Since $T$ is a linear transform, $T(x+y) = T(x)+T(y)$ for all vectors $x$ and $y$.

Also, $(1,0) + (0,1) = (1,1)$.

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  • $\begingroup$ Thanks for the detailed explanation $\endgroup$ – Ten Doeschate Apr 20 at 19:15
  • $\begingroup$ You're welcome. $\endgroup$ – Vizag Apr 20 at 19:15
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Assuming $T:\mathbb{R}^2 \rightarrow \mathbb{R}$. Then $T$ can be represented with a $(1\times2)$-dimensional matrix.

Let this associated matrix be $\begin{pmatrix} a & b \end{pmatrix}$.

Then $T(1,0) \equiv \begin{pmatrix} a & b \end{pmatrix}*\begin{pmatrix} 1 & 0 \end{pmatrix}^T = a$.

Similarly, $T(0,1) \equiv \begin{pmatrix} a & b \end{pmatrix}*\begin{pmatrix} 0 & 1 \end{pmatrix}^T = b$.

Therefore it must be the case that a=1, b=0.

So $T(1,1) \equiv \begin{pmatrix} 1 & 0 \end{pmatrix}*\begin{pmatrix} 1 & 1 \end{pmatrix}^T = 1$.

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  • $\begingroup$ @H Park Thanks for the detailed explanation $\endgroup$ – Ten Doeschate Apr 20 at 19:14

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