Gram-Schmidt Process
Power Series

Example 1.

 

Suppose we are given the map f :\mathbb{R}^{2} \rightarrow \mathbb{R}^{2} defined by f\left (x ,y\right ) =\left (x +y ,x -y\right ). We would like to check if f is linear. Take two vectors \left (x_{1} ,x_{2}\right ) and \left (y_{1} ,y_{2}\right ) and write

(1)   \begin{align*}f\left (x_{1} ,x_{2}\right ) =\left (x_{1} +x_{2} ,x_{1} -x_{2}\right ) \\ f\left (y_{1} ,y_{2}\right ) =\left (y_{1} +y_{2} ,y_{1} -y_{2}\right ) \\ f\left (x_{1} ,x_{2}\right ) +f\left (y_{1} ,y_{2}\right ) =\left (x_{1} +x_{2} ,x_{1} -x_{2}\right ) +\left (y_{1} +y_{2} ,y_{1} -y_{2}\right ) \\ =\left (x_{1} +x_{2} +y_{1} +y_{2} ,x_{1} -x_{2} +y_{1} -y_{2}\right )\end{align*}

Next, write

(2)   \begin{align*}f\left (\left (x_{1} ,x_{2}\right ) +\left (y_{1} ,y_{2}\right )\right ) =f\left (x_{1} +y_{1} ,x_{2} +y_{2}\right ) \\ =\left (x_{1} +y_{1} +x_{2} +y_{2} ,x_{1} +y_{1} -x_{2} -y_{2}\right )\end{align*}

Now observe that

(3)   \begin{align*}\left (x_{1} +x_{2} +y_{1} +y_{2} ,x_{1} -x_{2} +y_{1} -y_{2}\right ) =\left (x_{1} +y_{1} +x_{2} +y_{2} ,x_{1} +y_{1} -x_{2} -y_{2}\right ) \\ f\left (x_{1} ,x_{2}\right ) +f\left (y_{1} ,y_{2}\right ) =f\left (\left (x_{1} ,x_{2}\right ) +\left (y_{1} ,y_{2}\right )\right )\end{align*}

Next we check the second condition and take a real number c . Write

(4)   \begin{align*}cf\left (x_{1} ,x_{2}\right ) =c\left (x_{1} +x_{2} ,x_{1} -x_{2}\right ) \\ =\left (cx_{1} +cx_{2} ,cx_{1} -cx_{2}\right ) \\ =f\left (cx_{1} ,cx_{2}\right ) \\ =f\left (c\left (x_{1} ,x_{2}\right )\right )\end{align*}

 

Therefore, f is a linear transformation.

 

Example 2.

 

Suppose f is a linear transformation and we know

(5)   \begin{align*}f\left (1 ,0 ,0\right ) =\left (1 ,2 ,3\right ) \\ f\left (0 ,1 ,0\right ) =\left (5 ,0 ,2\right ) \\ f\left (0 ,0 ,1\right ) =\left ( -1 , -1 ,1\right )\end{align*}

We would like to find f\left (2 ,3 ,4\right ) using this information. Since f is a linear transformation, we can write

(6)   \begin{align*}f\left (2 ,3 ,4\right ) =f\left (\left (2 ,0 ,0\right ) +\left (0 ,3 ,0\right ) +\left (0 ,0 ,4\right )\right ) \\ =f\left (2\left (1 ,0 ,0\right ) +3\left (0 ,1 ,0\right ) +4\left (0 ,0 ,1\right )\right ) \\ =2f\left (1 ,0 ,0\right ) +3f\left (0 ,1 ,0\right ) +4f\left (0 ,0 ,1\right ) \\ =2\left (1 ,2 ,3\right ) +3\left (5 ,0 ,2\right ) +4\left ( -1 , -1 ,1\right ) \\ =\left (13 ,0 ,16\right )\end{align*}

This tells us that knowing where each elements of the basis is mapped by a linear transformation is enough to find the image of any other vector under f .

 

Example 3.

 

We are given the matrix

(7)   \begin{equation*}A =\left (\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right ) \end{equation*}

and would like to find the eigenvalues and associated eigenvectors of this matrix. We write

(8)   \begin{align*}\det \left (A -\lambda I\right ) =det\left (\begin{array}{cc}6 -\lambda & -3 \\ -2 & 1 -\lambda \end{array}\right ) \\ =\left (6 -\lambda \right )\left (1 -\lambda \right ) -6 \\ =\lambda \left (\lambda -7\right )\end{align*}

The eigenvalues of A are 0 and 7 and are obtained by

(9)   \begin{equation*}\lambda \left (\lambda -7\right ) =0 \end{equation*}

Next, use the first eigenvalue to write and then reduce

(10)   \begin{align*}\left (\begin{array}{cc}6 -0 & -3 \\ -2 & 1 -0\end{array}\right ) =\left (\begin{array}{cc}6 & -3 \\ -2 & 1\end{array}\right ) \\ \rightarrow \left (\begin{array}{cc}6 & -3 \\ 0 & 0\end{array}\right )\end{align*}

Next,

(11)   \begin{equation*}6x -3y =0 \end{equation*}

implies \left (\begin{array}{c}t \\ 2t\end{array}\right ) =t\left (\begin{array}{c}1 \\ 2\end{array}\right ). This is the eigenvector associated to the eigenvalue 0. Next

(12)   \begin{align*}\left (\begin{array}{cc}6 -7 & -3 \\ -2 & 1 -7\end{array}\right ) =\left (\begin{array}{cc} -1 & -3 \\ -2 & -6\end{array}\right ) \\ \rightarrow \left (\begin{array}{cc}1 & 3 \\ 0 & 0\end{array}\right )\end{align*}

We have

(13)   \begin{equation*}x +3y =0 \end{equation*}

that is \left (\begin{array}{c} -3t \\ t\end{array}\right ) =t\left (\begin{array}{c} -3 \\ 1\end{array}\right ) is the second eigenvector of A and it’s associated to the eigenvalue 7.

 

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