# Given a linear function T determine whether $T(1,1)=1$

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?

• $$T(1,1)=T[(1,0)+(0,1)]=T(1,0)+T(0,1)=1+0=1$$ – Chinnapparaj R Apr 20 at 17:18
• @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? – Ten Doeschate Apr 20 at 17:21
• @TenDoeschate: Yes, but that is a detour. (You should get $T(x,y)=x$). – Henning Makholm Apr 20 at 17:23
• @HenningMakholm Got it thanks – Ten Doeschate Apr 20 at 17:25

## 2 Answers

Since $$T$$ is a linear transform, $$T(x+y) = T(x)+T(y)$$ for all vectors $$x$$ and $$y$$.

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

• Thanks for the detailed explanation – Ten Doeschate Apr 20 at 19:15
• You're welcome. – Vizag Apr 20 at 19:15

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

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