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$1)$ Let $T:\mathbb{R}^2->\mathbb{R}^2$ be given by

$$T(x,y)^T=(2x+y,-x+2y)^T.$$ Find a matrix with respect to $[(-1,0)^T,(1,1)^T]$ and $[(0,1)^T,(-1,0)^T]$

I solved this using the following method:

$$T[(-1,0)^T]=(-2,1)^T=1 (0,1)^T+2(-1,0)^T$$ $$T(1,1)^T=(3,1)^T=1 (0,1)^T+(-3)(-1,0)^T$$ So, I got the matrix as $\begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix}$

But in this question

$2)$ Let $T$ be a linear map form $\mathbb{R}^3->\mathbb{R}^3$ defined by the formula: $$T[(x,y,z)^T]=(3x+y,2x-z,y)^T$$ Find the matrix of $T$ with respect to the standard basis of $\mathbb{R}^3.$

I tried to solve this by the same above method. So, I took the standard basis of $\mathbb{R}^3$ as $e_1=(1,0,0)^T,e_2=(0,1,0)^T,e_3=(0,0,1)^T$ and I proceeded as followed.. $$T(1,0,0)^T=(3,2,0)^T=?$$ $$T(0,1,0)^T=(1,0,1)^T$$ $$T(0,0,1)^T=(0,-1,0)^T$$

How can I proceed further? Can anyone please explain how to solve the $2^{nd}$ problem.

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It is the exact same procedure.

$$T((1,0,0)) = (3,2,0) = 3e_1 + 2e_2$$ $$T((0,1,0)) = (1,0,1) = e_1 + e_3$$ $$T((0,0,1)) = (0,-1,0) = -e_2$$

So, $$T = \begin{pmatrix} 3 & 1 & 0 \\ 2 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}$$

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  • $\begingroup$ Can you show me how to solve the first problem same as the one you showed for the second. $\endgroup$
    – tien lee
    Jun 24 '18 at 5:37
  • $\begingroup$ You already showed it :)! The second problem can be restated as this: Find the matrix of T with respect to $\{e_1, e_2, e_3\}$ and $\{e_1, e_2 , e_3\}$. In this case, we happen to use the same basis for the domain and codomain to represent T. $\endgroup$
    – James Yang
    Jun 24 '18 at 11:50

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