91Ó°ÊÓ

The column space is spanned by the column vectors of the given matrix. So, our original basis vectors are $$\boldsymbol{u}_1=\begin{pmatrix}-1\\1\end{pmatrix} \text{ and } \boldsymbol{u}_2=\begin{pmatrix}3\\5\end{pmatrix}$$.

First, we normalize the first vector: $$\boldsymbol{v}_1 = \frac{\boldsymbol{u}_1}{\|\boldsymbol{u}_1\|} = \frac{1}{\sqrt{(-1)^2 + 1^2}}\begin{pmatrix}-1\\1\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}-1\\1\end{pmatrix}$$ For the second vector, we first find the orthogonal component to \(\boldsymbol{v}_1\): $$\boldsymbol{w}_2 = \boldsymbol{u}_2 - (\boldsymbol{u}_2 \cdot \boldsymbol{v}_1)\boldsymbol{v}_1 = \begin{pmatrix}3\\5\end{pmatrix} - \left(\begin{pmatrix}3\\5\end{pmatrix} \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}-1\\1\end{pmatrix}\right) \frac{1}{\sqrt{2}}\begin{pmatrix}-1\\1\end{pmatrix} = \begin{pmatrix}3\\5\end{pmatrix} - 4\frac{1}{\sqrt{2}}\begin{pmatrix}-1\\1\end{pmatrix}$$ Now we normalize the orthogonal component $$\boldsymbol{v}_2 = \frac{\boldsymbol{w}_2}{\|\boldsymbol{w}_2\|}= \frac{1}{\sqrt{3^2 + 1^2}}\begin{pmatrix}3\\1\end{pmatrix} = \frac{1}{\sqrt{10}}\begin{pmatrix}3\\1\end{pmatrix}$$ (b) For the given matrix \(A=\left(\begin{array}{rr}2 & 5 \\\ 1 & 10\end{array}\right)\):

The column space is spanned by the column vectors of the given matrix. So, our original basis vectors are $$\boldsymbol{u}_1=\begin{pmatrix}2\\1\end{pmatrix} \text{ and } \boldsymbol{u}_2=\begin{pmatrix}5\\10\end{pmatrix}$$.

First, we normalize the first vector: $$\boldsymbol{v}_1 = \frac{\boldsymbol{u}_1}{\|\boldsymbol{u}_1\|} = \frac{1}{\sqrt{2^2 + 1^2}}\begin{pmatrix}2\\1\end{pmatrix} = \frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix}$$ For the second vector, we first find the orthogonal component to \(\boldsymbol{v}_1\): $$\boldsymbol{w}_2 = \boldsymbol{u}_2 - (\boldsymbol{u}_2 \cdot \boldsymbol{v}_1)\boldsymbol{v}_1 = \begin{pmatrix}5\\10\end{pmatrix} - \left(\begin{pmatrix}5\\10\end{pmatrix} \cdot \frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix}\right) \frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}5\\10\end{pmatrix} - 15\frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix}$$ Now we normalize the orthogonal component $$\boldsymbol{v}_2 = \frac{\boldsymbol{w}_2}{\|\boldsymbol{w}_2\|}= \frac{1}{\sqrt{5^2 + 0^2}}\begin{pmatrix}5\\0\end{pmatrix} = \frac{1}{\sqrt{25}}\begin{pmatrix}5\\0\end{pmatrix} = \frac{1}{5}\begin{pmatrix}5\\0\end{pmatrix} = \begin{pmatrix}1\\0\end{pmatrix}$$ So, the orthonormal bases for the given matrices are: (a)$$R(A)={\left\{\frac{1}{\sqrt{2}}\begin{pmatrix}-1\\1\end{pmatrix},\frac{1}{\sqrt{10}}\begin{pmatrix}3\\1\end{pmatrix}\right\}}$$ (b)$$R(A)={\left\{\frac{1}{\sqrt{5}}\begin{pmatrix}2\\1\end{pmatrix},\begin{pmatrix}1\\0\end{pmatrix}\right\}}$$