# Rowspace and kernel

by UrbanXrisis
Tags: kernel, rowspace
 P: 1,200 $$A = \left(\begin{array}{cccc}-1 &6&5&9 \\ -1&0&1&3 \end{array}\right)$$ Find orthonormal bases of the kernel, row space. To find the bases, I did reduced the array to its RREF. $$A = \left(\begin{array}{cccc}1 & 0&-1&-3\\ 0&1&2/3&1 \end{array}\right)$$ Then the orthonormal bases would just be that divided by the length. $$||v_1||=\sqrt{1+1+3^2}=\sqrt{11}$$ $$||v_2||=\sqrt{1+(2/3)^2+1}=\sqrt{2.44444}$$ so that means, the orthonormal bases would be: $$A = \left(\begin{array}{cccc} \frac{1}{ \sqrt{11}} & 0&\frac{-1}{ \sqrt{11}}&\frac{-3}{ \sqrt{11}} \\0 & \frac{1}{ \sqrt{2.44444}} & \frac{.66666}{ \sqrt{2.44444}} &\frac{1}{ \sqrt{2.44444}}\end{array}\right)$$ what exactly is the orthonormal bases of the kernel? Also, isnt the row space the same as the vectors of the bases? I think I also did something wrong in my calculations
 P: 1,200 is what i did above the orthonormal row space? that is wrong as well, i dont know why.... however: using the Gram-Schmidt process, i still get an error: $$A = \left(\begin{array}{cccc}-1 &6&5&9 \\ -1&0&1&3 \end{array}\right)=\left(\begin{array}{cc}W_1 &W_2 \end{array}\right)$$ want to find an orthonormal basis $$R={U_1 ,U_2}$$ $$U_1=\frac{W}{||W_1||}$$ $$||W_1||=\sqrt{11}$$ $$U_1=\frac {\left(\begin{array}{cccc}-1 &6&5&9 \end{array}\right)}{\sqrt{11}}$$ $$U_2=\frac{W_2- U_1}{||W_2- U_1||}$$ where $$W_2=\left(\begin{array}{cccc} -1&0&1&3 \end{array}\right)$$ $$U_1=\frac {\left(\begin{array}{cccc}-1 &6&5&9 \end{array}\right)}{\sqrt{11}}$$ $$U_1=\left(\begin{array}{cccc}-1/\sqrt{11} &6\sqrt{11}&5\sqrt{11}&9\sqrt{11} \end{array}\right)$$ is the the correct set up to get an orthonormal basis? Also, to get an orthonormal row space, what would I have to do?