- #1
UrbanXrisis
- 1,196
- 1
[tex]A = \left(\begin{array}{cccc}-1 &6&5&9 \\ -1&0&1&3 \end{array}\right)[/tex]
Find orthonormal bases of the kernel, row space.
To find the bases, I did reduced the array to its RREF.
[tex]A = \left(\begin{array}{cccc}1 & 0&-1&-3\\ 0&1&2/3&1 \end{array}\right)[/tex]
Then the orthonormal bases would just be that divided by the length.
[tex]||v_1||=\sqrt{1+1+3^2}=\sqrt{11}[/tex]
[tex]||v_2||=\sqrt{1+(2/3)^2+1}=\sqrt{2.44444}[/tex]
so that means, the orthonormal bases would be:
[tex]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)[/tex]
what exactly is the orthonormal bases of the kernel?
Also, isn't the row space the same as the vectors of the bases?
I think I also did something wrong in my calculations
Find orthonormal bases of the kernel, row space.
To find the bases, I did reduced the array to its RREF.
[tex]A = \left(\begin{array}{cccc}1 & 0&-1&-3\\ 0&1&2/3&1 \end{array}\right)[/tex]
Then the orthonormal bases would just be that divided by the length.
[tex]||v_1||=\sqrt{1+1+3^2}=\sqrt{11}[/tex]
[tex]||v_2||=\sqrt{1+(2/3)^2+1}=\sqrt{2.44444}[/tex]
so that means, the orthonormal bases would be:
[tex]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)[/tex]
what exactly is the orthonormal bases of the kernel?
Also, isn't the row space the same as the vectors of the bases?
I think I also did something wrong in my calculations
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