# Find the linear transformation of a matrix knowing 4 linear transformations

I'm stuck with an exercise where they give me 4 linear transformations $$T:M_{2\times2}\to\Bbb R$$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix.

$$T \begin{pmatrix} 1 &0\\ 0 & 0\\ \end{pmatrix} =3, T\begin{pmatrix} 0 &1\\ 1 & 0\\ \end{pmatrix} =-1, T \begin{pmatrix} 1 &0\\ 1 & 0\\ \end{pmatrix} =0, T \begin{pmatrix} 0 &0\\ 0 & 1\\ \end{pmatrix} =0, T \begin{pmatrix} a &b\\ c & d\\ \end{pmatrix} =?$$

and I have worked with this situation for $$T:\Bbb R^n \to\Bbb R^m$$ but never with matrices so I have the doubt that how is supposed to find 4 constants that will multiply de 4 matrices I know to compose the fifth matrix to finally can find the linear transformation of this last one. (I have tried adding up the 4 matrices as follows)

$$\begin{pmatrix} 1 &0\\ 0 & 0\\ \end{pmatrix} + \begin{pmatrix} 0 &1\\ 1 & 0\\ \end{pmatrix} + \begin{pmatrix} 1 &0\\ 1 & 0\\ \end{pmatrix} + \begin{pmatrix} 0 &0\\ 0 & 1\\ \end{pmatrix}= \begin{pmatrix} a &b\\ c & d\\ \end{pmatrix}$$

$$\begin{cases} 2=a & \\ 1=b & \\ 2=c\\ 1=d\\ \end{cases}$$

and then use the coefficients $$2\begin{pmatrix} 1 &0\\ 0 & 0\\ \end{pmatrix} + 1\begin{pmatrix} 0 &1\\ 1 & 0\\ \end{pmatrix} + 2\begin{pmatrix} 1 &0\\ 1 & 0\\ \end{pmatrix} + 1\begin{pmatrix} 0 &0\\ 0 & 1\\ \end{pmatrix}= \begin{pmatrix} a &b\\ c & d\\ \end{pmatrix}$$

and I just want to know if I continue in this way or I have committed any mistakes, thanks for your time <3

• You haven't stated anything incorrect, but it doesn't look like you're on the right track Dec 8, 2020 at 20:32

What is given can be translated as follows. Denote, in a standard way, $$E_1=\begin{pmatrix}1&0\\0&0 \end{pmatrix},\quad E_2=\begin{pmatrix}0&1\\0&0 \end{pmatrix}, \quad E_3=\begin{pmatrix}0&0\\1&0 \end{pmatrix},\quad E_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ We have $$\:M=\begin{pmatrix}a&b\\c&d \end{pmatrix}=aE_1+bE_2+cE_3+dE_4$$, so $$T(M)=aT(E_1)+bT(E_2)+cT(E_3)+dT(E_4).$$
Now, we are given that $$T(E_1)=3,\enspace T(E_2)+T(E_3)=-1,\enspace T(E_1)+T(E_3)=0, \enspace T(E_4)=0$$. Can you proceed?
One standard approach is to find numbers $$x_ij$$ with $$1 \leq i,j \leq 4$$ for which $$x_{11} \pmatrix{1&0\\0&0} + x_{12} \pmatrix{0&1\\1&0} + x_{13} \pmatrix{1&0\\1&0} + x_{14} \pmatrix{0&0\\0&1} = \pmatrix{1&0\\0&0}\\ x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1} = \pmatrix{0&1\\0&0}\\ x_{31} \pmatrix{1&0\\0&0} + x_{32} \pmatrix{0&1\\1&0} + x_{33} \pmatrix{1&0\\1&0} + x_{34} \pmatrix{0&0\\0&1} = \pmatrix{0&0\\1&0}\\ x_{41} \pmatrix{1&0\\0&0} + x_{42} \pmatrix{0&1\\1&0} + x_{43} \pmatrix{1&0\\1&0} + x_{44} \pmatrix{0&0\\0&1} = \pmatrix{0&0\\0&1}.\\$$ Once you have done this, we find that $$T\pmatrix{a&b\\c&d} \\ = a\cdot T\left[ x_{11} \pmatrix{1&0\\0&0} + x_{12} \pmatrix{0&1\\1&0} + x_{13} \pmatrix{1&0\\1&0} + x_{14} \pmatrix{0&0\\0&1}\right]\\ + b\cdot T\left[x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1} \right]\\ + c \cdot T\left[x_{31} \pmatrix{1&0\\0&0} + x_{32} \pmatrix{0&1\\1&0} + x_{33} \pmatrix{1&0\\1&0} + x_{34} \pmatrix{0&0\\0&1}\right]\\ + d \cdot T \left[x_{41} \pmatrix{1&0\\0&0} + x_{42} \pmatrix{0&1\\1&0} + x_{43} \pmatrix{1&0\\1&0} + x_{44} \pmatrix{0&0\\0&1} \right] \\ = a\cdot \left[x_{11} T\pmatrix{1&0\\0&0} + x_{12} T\pmatrix{0&1\\1&0} + x_{13} T\pmatrix{1&0\\1&0} + x_{14} T\pmatrix{0&0\\0&1}\right]\\ + b\cdot \left[x_{21} \pmatrix{1&0\\0&0} + x_{22} \pmatrix{0&1\\1&0} + x_{23} \pmatrix{1&0\\1&0} + x_{24} \pmatrix{0&0\\0&1}\right] \\ + c \left[x_{31} T\pmatrix{1&0\\0&0} + x_{32} T\pmatrix{0&1\\1&0} + x_{33} T\pmatrix{1&0\\1&0} + x_{34} T\pmatrix{0&0\\0&1}\right]\\ + d \cdot \left[x_{41} T\pmatrix{1&0\\0&0} + x_{42} T\pmatrix{0&1\\1&0} + x_{43} T\pmatrix{1&0\\1&0} + x_{44} T\pmatrix{0&0\\0&1} \right].$$ In other words, if $$X$$ is the matrix whose entries are $$x_{ij}$$, then we have $$T \pmatrix{a&b\\c&d} = \pmatrix{3&-1&0&0} X \pmatrix{a\\b\\c\\d}.$$ So, we need to find the matrix $$X$$. If we write the full system of equations as a matrix, then we can see that the system of equations can be written as $$\pmatrix{1&0&1&0\\0&1&0&0\\0&1&1&0\\0&0&0&1} \pmatrix{x_{11}&x_{12}&x_{13}& x_{14}\\ x_{21} & x_{22} & x_{23} & x_{24}\\ x_{31} & x_{32} & x_{33} & x_{34}\\ x_{41} & x_{42} & x_{43} & x_{44}} = \pmatrix{1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1}.$$ In other words, $$X = \pmatrix{1&0&1&0\\0&1&0&0\\0&1&1&0\\0&0&0&1}^{-1}.$$