Linear Transformations and Operators

Let \(\displaystyle u_1=\left[\begin{array}{c}x_1\\y_1\end{array}\right]\) and \(\displaystyle u_2=\left[\begin{array}{c}x_1\\y_1\end{array}\right]\) be two vectors, and \(\alpha_1\) and \(\alpha_2\) two real numbers. Then we have: \begin{eqnarray*} L\left(\alpha_1 u_1+\alpha_2u_2\right)&=&L\left(\left[\begin{array}{c}\alpha_1x_1+\alpha_2x_2\\ \alpha_1y_1+\alpha_2y_2\end{array}\right]\right) = \left[\begin{array}{c}\alpha_1x_1+\alpha_2x_2+\alpha_1y_1+\alpha_2y_2\\\alpha_1x_1+\alpha_2x_2-\alpha_1y_1-\alpha_2y_2\end{array}\right]\\ &=& \left[\begin{array}{c}\alpha_1x_1+ \alpha_1y_1 \\\alpha_1x_1 -\alpha_1y_1 \end{array}\right]+\left[\begin{array}{c} \alpha_2x_2 +\alpha_2y_2\\ \alpha_2x_2-\alpha_2y_2\end{array}\right]=\alpha_1L(u_1)+\alpha_2L(u_2). \end{eqnarray*}

Let \(\displaystyle u=\left[\begin{array}{c}1\\0\end{array}\right]\) and \(\displaystyle v=\left[\begin{array}{c}0\\1\end{array}\right]\). Notice that \(\displaystyle \left[\begin{array}{c}3\\7\end{array}\right]=3u+7v \), hence \begin{eqnarray*}L\left[\begin{array}{c}3\\7\end{array}\right]&=&L(3u+7v)=3L(u)+7L(v)=3\left[\begin{array}{c}3\\-2\end{array}\right]+7\left[\begin{array}{c}2\\5\end{array}\right]=\left[\begin{array}{c}9\\-6\end{array}\right]+ \left[\begin{array}{c}14\\35\end{array}\right]=\left[\begin{array}{c}23\\29\end{array}\right]. \end{eqnarray*}

Let \(\displaystyle u=\left[\begin{array}{c}2\\-3\end{array}\right]\), \(\displaystyle v=\left[\begin{array}{c}3\\-1\end{array}\right]\), and \(\displaystyle w=\left[\begin{array}{c}1\\2\end{array}\right]\). We will first find scalars \(\alpha\) and \(\beta\) such that \(w=\alpha u+\beta v\). Then we will use \(L(w)=L(\alpha u+\beta v)=\alpha L(u)+\beta L(v)\).

The scalars \(\alpha\) and \(\beta\) must satisfy: \begin{eqnarray*} 2\alpha+3\beta&=&1\\ -3\alpha-\beta&=&2. \end{eqnarray*} We n multiply the second equation by \(3\) and add it to the first to obtain: \(-7\alpha=7\), hence \(\alpha=-1\). From the first equation we get \(\beta=\frac13(1-2\alpha)=1\). Thus \[L(w)=-L(u)+L(v)=\left[\begin{array}{c}2\\1\end{array}\right] - \left[\begin{array}{c}1\\-2\end{array}\right]=\left[\begin{array}{c}1\\3\end{array}\right].\]

We will first define \(L(w)\) for each \(w\in\mathbb R^m\). Assume that \(w\in\mathbb R^m\) is given. Since \(\{e_1, \dots, e_m\}\) is a basis of \(\mathbb R^m\) there are scalars \(\alpha_1\), \(\dots\), \(\alpha_m\) such that \(w=\alpha_1e_1+\cdots+\alpha_me_m\). Then we define \(L(w)=\alpha_1f_1+\cdots+\alpha_mf_m\). It is very easy to prove that \(L\) is linear.

Assume now that \(L\) and \(K\) are two linear transformations such that \(L(e_1)=K(e_1)\), \(\dots\), \(L(e_m)=K(e_m)\). Let \(w\in\mathbb R^m\). We will prove that \(L(w)=K(w)\). There are scalars \(\alpha_1\), \(\dots\), \(\alpha_m\) such that \(w=\alpha_1e_1+\cdots+\alpha_me_m\). Since \(L\) and \(K\) are linear we must have: \[L(w)=L(\alpha_1e_1+\cdots +\alpha_me_m)=\alpha_1L(e_1)+\cdots+\alpha_mL(e_m)=\alpha_1K(e_1)+\cdots+\alpha_mK(e_m)=K(\alpha_1e_1+\cdots \alpha_me_m)=K(w).\]

Let us denote by \(e_1\), \(e_2\), and \(e_3\) the vectors of the standard basis of \(\mathbb R^3\). In other words, we have: \(\displaystyle e_1=\left[\begin{array}{c}1\\0\\0\end{array}\right]\), \(\displaystyle e_1=\left[\begin{array}{c}0\\1\\0\end{array}\right]\), and \(\displaystyle e_3=\left[\begin{array}{c}0\\0\\1\end{array}\right]\). In order to determine the matrix of \(L\) we need to find \(L(e_1)\), \(L(e_2)\), and \(L(e_3)\). Let \(\displaystyle f_2=\left[\begin{array}{c}0\\1\\1\end{array}\right]\) and \(\displaystyle f_3=\left[\begin{array}{c}-1\\1\\2\end{array}\right]\). We will now express \(e_2\) and \(e_3\) in terms of \(e_1\), \(f_2\), and \(f_3\). We will first find the scalars \(\alpha_1\), \(\alpha_2\), and \(\alpha_3\) such that \(e_2=\alpha_1 e_1+\alpha_2f_2+\alpha_3f_3\). The last equation is equivalent to the following system: \begin{eqnarray*} \begin{array}{ccccc} \alpha_1&&-\alpha_3&=&0\\ &\alpha_2&+\alpha_3&=&1\\ &\alpha_2&+2\alpha_3&=&0. \end{array} \end{eqnarray*} From the last two equations we get \(\alpha_3=-1\). Now the first equations implies \(\alpha_1=-1\), and the second that \(\alpha_2=2\). Thus \[L(e_2)=L(-e_1+2f_2-f_3)=-\left[\begin{array}{c}3\\2\end{array}\right]+2\left[\begin{array}{c}3\\3\end{array}\right]- \left[\begin{array}{c}1\\4\end{array}\right]=\left[\begin{array}{c}2\\0\end{array}\right].\]

It remains to find \(L(e_3)\). In order to do so we first need to find \(\beta_1\), \(\beta_2\), and \(\beta_3\) such that \(e_3=\beta_1e_1+\beta_2f_2+\beta_3f_3\). The last equation can be written as: \begin{eqnarray*} \begin{array}{ccccc} \beta_1&&-\beta_3&=&0\\ &\beta_2&+\beta_3&=&0\\ &\beta_2&+2\beta_3&=&1. \end{array} \end{eqnarray*} From the last two equations we get \(\beta_3=1\). The first equation implies that \(\beta_1=1\), and the second that \(\beta_2=-1\). Therefore \[L(e_3)=L(e_1-f_2+f_3)=\left[\begin{array}{c}3\\2\end{array}\right]-\left[\begin{array}{c}3\\3\end{array}\right]+ \left[\begin{array}{c}1\\4\end{array}\right]=\left[\begin{array}{c}1\\3\end{array}\right].\] Thus the matrix of the transformation \(L\) is \(\displaystyle \left[\begin{array}{ccc}3&2&1\\2&0&3\end{array}\right]\).

Assume that \(u,v\in\mathbb R^m\) and that \(\alpha\) and \(\beta\) are two real numbers. We need to prove that \(M(\alpha u+\beta v)=\alpha M(u)+\beta M(v)\).

Using the definition of \(M\) and the linearity of \(K\) and \(L\) we obtain: \[M(\alpha u+\beta v)=K(L(\alpha u+\beta v))=K(\alpha L(u)+\beta L(v))=\alpha K(L(u))+\beta K(L(v))=\alpha M(u)+\beta M(v).\] This completes the proof of the theorem.

Assume that \(u,v\in\mathbb R^m\) and that \(\gamma\) and \(\delta\) are two real numbers. We need to prove that \(M(\gamma u+\delta v)=\gamma M(u)+\delta M(v)\).

Using the definition of \(M\) and the linearity of \(K\) and \(L\) we obtain: \begin{eqnarray*}M(\gamma u+\delta v)&=&\alpha L(\gamma u+\delta v)+\beta K(\gamma u+\delta v)=\alpha \left(\gamma L(u)+\delta L(v)\right) + \beta \left(\gamma K(u)+\delta K(v)\right) \\&=&\gamma\left(\alpha L(u)+\beta K(u)\right)+\delta \left(\alpha L(v)+\beta K(v)\right)= \gamma M(u)+\delta M(v) .\end{eqnarray*} This completes the proof of the theorem.

Let \(e_1\), \(\dots\), \(e_m\) be the vectors of the standard basis of \(\mathbb R^m\). Then \(e_j\) is the column vector with \(m\) entries, such that the \(j\)-th entry is \(1\) and every other entry is \(0\). We have that \(\displaystyle Q(e_j)=\left[\begin{array}{c}q_{1j}\\ q_{2j}\\\vdots\\ q_{pj}\end{array}\right]\).

By the definition of \(Q\) we have \(Q(e_j)=K(L(e_j))\). We have \(\displaystyle L(e_j)=\left[\begin{array}{c}l_{1j}\\ l_{2j}\\\vdots\\ q_{nj}\end{array}\right]\). Let us denote by \(f_1\), \(\dots\), \(f_n\) the standard basis of \(\mathbb R^n\). Then we can write \(\displaystyle L(e_j)=\sum_{s=1}^n l_{sj}f_s\). Using the linearity of \(K\) we conclude \[Q(e_j)=K(L(e_j))=\sum_{s=1}^n l_{sj}K(f_s)=\sum_{s=1}^n l_{sj} \left[\begin{array}{c} k_{1s}\\k_{2s}\\\vdots\\ k_{ps} \end{array}\right]=\left[\begin{array}{c}\sum_{s=1}^n l_{sj}k_{1s}\\ \sum_{s=1}^n l_{sj}k_{2s}\\ \vdots\\ \sum_{s=1}^n l_{sj}k_{ps}\end{array}\right]. \] Thus the \(i\)-th entry of \(Q(e_j)\) is \(\displaystyle q_{ij}=\sum_{s=1}^n l_{sj}k_{is}=\sum_{s=1}^nk_{is}l_{sj}\).

Our goal is to find scalars \(\alpha\) and \(\beta\) such that \[\left[\begin{array}{c} 17\\ -24\end{array}\right]= \alpha\left[\begin{array}{c} 1\\ -1\end{array}\right]+\beta\left[\begin{array}{c}2\\ -3\end{array}\right].\] Solving the system of equations \begin{eqnarray*} 1\alpha+2\beta&=&17\\ -1\alpha+-3\beta&=&-24 \end{eqnarray*} yields \(\alpha=3\) and \(\beta=7\). Thus \[ L\left[\begin{array}{c} 17\\ -24\end{array}\right]=3L\left[\begin{array}{c} 1\\ -1\end{array}\right]+7L\left[\begin{array}{c} 2\\ -3\end{array}\right]=3\left[\begin{array}{c} 6\\ -1\end{array}\right]+7\left[\begin{array}{c} 7\\ 5\end{array}\right]=\left[\begin{array}{c} 67\\ 32\end{array}\right], \] hence \(x=67\) and \(y=32\). Thus \(x+y=67+32=99\).

According to the formula for matrix multiplication we have \[c_{23}=a_{21}b_{13}+a_{22}b_{23}+a_{23}b_{33},\] where \(a_{ij}\) is the entry of the matrix \(A\) at the position \((i,j)\), and \(b_{kl}\) is the entry of the matrix \(B\) at the position \((k,l)\). Since \(a_{21}=5\), \(b_{13}=6\), \(a_{22}=7\), \(b_{23}=9\), \(a_{23}=-11\), and \(b_{33}=-5\), we get: \(c_{23}=148\).

For each \(i\in\{1,2,3,4\}\) and each \(j\in\{1,2,3\}\) we have \(\displaystyle m_{ij}=\sum_{s=1}^2 k_{is}l_{sj}\). Therefore we have \(\displaystyle \left[\begin{array}{ccc} m_{11} &m_{12}&m_{13}\\ m_{21} &m_{22}&m_{23}\\ m_{31} &m_{32}&m_{33}\\ m_{41} &m_{42}&m_{43}\\ \end{array}\right] = \left[\begin{array}{ccc} 37 &-12&50\\ 27 &1&18\\ -30 &1&-24\\ -62 &-3&-40\\ \end{array}\right], \)

which implies that \(m_{11}+m_{22}+m_{33}-m_{41}-m_{42}-m_{43}=119\).

We need to find the scalars \(\alpha_1\), \(\alpha_2\), and \(\alpha_3\) such that \(\displaystyle \left[\begin{array}{c}1\\0\\0\end{array}\right]=\alpha_1\left[\begin{array}{c} -1\\ 2\\1\end{array}\right]+ \alpha_2\left[\begin{array}{c} 1\\ 0\\1\end{array}\right]+ \alpha_3\left[\begin{array}{c} 0\\ 1\\2\end{array}\right]\). The previous equation is equivalent to the system: \[\begin{array}{ccccc} -\alpha_1&+\alpha_2&&=&1\\ 2\alpha_1&&+\alpha_3&=&0\\ \alpha_1&+\alpha_2&+2\alpha_3&=&0.\end{array} \] We put the box around the variable \(\alpha_2\) in the first equation and obtain the system: \[\begin{array}{ccccc} -\alpha_1&+\begin{array}{|c|}\hline \alpha_2\\ \hline\end{array}&&=&1\\ 2\alpha_1&&+\alpha_3&=&0\\ \alpha_1&+\alpha_2&+2\alpha_3&=&0.\end{array} \] We now eliminate \(\alpha_2\) from the second and third equation and obtain the system: \[\begin{array}{cccc} 2\alpha_1&+\alpha_3&=&0\\ 2\alpha_1&+2\alpha_3&=&-1.\end{array} \]

We now put the box around the variable \(\alpha_1\) in the first equation of the last system: \[\begin{array}{cccc} \begin{array}{|c|}\hline 2\alpha_1\\ \hline\end{array}&+\alpha_3&=&0\\ 2\alpha_1&+2\alpha_3&=&-1.\end{array} \] Eliminating \(\alpha_1\) from the last equation gives us \(\alpha_3=-1\). Now the first equation yields \(\alpha_1=\frac12\), and the first equation with box implies \(\alpha_2=\frac32\). Thus \[L\left[\begin{array}{c}1\\0\\0\end{array}\right]=\frac12L \left[\begin{array}{c} -1\\ 2\\1\end{array}\right]+ \frac32L\left[\begin{array}{c} 1\\ 0\\1\end{array}\right]-L\left[\begin{array}{c} 0\\ 1\\2\end{array}\right]=\frac12 \left[\begin{array}{c}-11\\30\end{array}\right]+\frac32 \left[\begin{array}{c}-1\\-10\end{array}\right]- \left[\begin{array}{c}-14\\3\end{array}\right]= \left[\begin{array}{c}7\\-3\end{array}\right].\] In an analogous way we obtain \[L\left[\begin{array}{c}0\\1\\0\end{array}\right]=\left[\begin{array}{c}2\\17\end{array}\right] \quad\quad\quad\mbox{and} \quad\quad\quad L\left[\begin{array}{c}0\\0\\1\end{array}\right]=\left[\begin{array}{c}-8\\-7\end{array}\right]. \] Thus the matrix of the transformation is \(\displaystyle \left[\begin{array}{ccc} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\end{array}\right]= \left[\begin{array}{ccc} 7&2&-8\\ -3&17&-7\end{array}\right]\) and \(a_{11}+a_{22}+a_{13}=16\).