Answer
See below
Work Step by Step
Given $A=\begin{bmatrix}
1 & 5\\
2 & 4\\
3 & 3\\
4 & 2\\
5 & 1
\end{bmatrix}$
Basic for rowspace$(A)$ is $\{(1,5),(2,4)\}$
According to Gram-Schmidt process, we have:
$v_1=x_1=(1,5)\\
v_2=x_2-\frac{}{||v_1||^2}v_1\\
=(2,4)-\frac{)}{||(1,5)||^2}(1,5)\\
=(2,4)-\frac{2.1+4.5}{1^2+5^2}(1,5)\\
=(2,4)-\frac{11}{13}(1,5)\\
=\frac{3}{13}(5,-1)$
Hence, an orthogonal basis for rowspace $(A)$ is:
$\{(1,5),\frac{3}{13}(5,-1)\}$
Determine basis for colspace $(A)$:
$\begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
5 & 4 & 1 & 2 & 1
\end{bmatrix}\approx \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
0 & -6& -12 & -18& -24
\end{bmatrix} \approx \begin{bmatrix}
1 & 2 & 3 & 4 & 5\\
0& 1 & 2 & 3 & 4
\end{bmatrix}$
Basis for colspace $(A):\{(1,2,3,4,5);(5,4,3,2,1)\}$
Assume $y_1=(1,2,3,4,5)\\
y_2=(5,4,3,2,1)$
Apply Gram- Schmidt
$v_1=y_1=(1,2,3,4,5)\\
v_2=y_2-\frac{}{||v_1||^2}v_1\\
=(5,4,3,2,1)-\frac{}{||(1,2,3,4,5)||^2}(1,2,3,4,5)\\
=(5,4,3,2,1)-\frac{5.1+4.2+3.3+2.4+5.1}{1^2+2^2+3^2+4^2+5^2}(1,2,3,4,5)\\
=(5,4,3,2,1)-\frac{5+8+9+8+5}{1+4+9+16+25}(1,2,3,4,5)\\
=(5,4,3,2,1)-\frac{7}{11}(5,4,3,2,1)\\
=\frac{6}{11}(8,5,2,-1,-4)$
Consequently, an orthogonal basis for colspace $(A)$ is
$\{(1,2,3,4,5),\frac{6}{11}(8,5,2,-1,-4)\}$
