Null space of [66323633063]\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right]

The calculator will find the null space of the 22x33 matrix [66323633063]\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right], with steps shown.
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Find the null space of [66323633063]\left[\begin{array}{ccc}\frac{\sqrt{6}}{6} & \frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3}\end{array}\right].

Solution

The reduced row echelon form of the matrix is [102011]\left[\begin{array}{ccc}1 & 0 & - \sqrt{2}\\0 & 1 & 1\end{array}\right] (for steps, see rref calculator).

To find the null space, solve the matrix equation [102011][x1x2x3]=[00].\left[\begin{array}{ccc}1 & 0 & - \sqrt{2}\\0 & 1 & 1\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].

If we take x3=tx_{3} = t, then x1=2tx_{1} = \sqrt{2} t, x2=tx_{2} = - t.

Thus, x=[2ttt]=[211]t.\mathbf{\vec{x}} = \left[\begin{array}{c}\sqrt{2} t\\- t\\t\end{array}\right] = \left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right] t.

This is the null space.

The nullity of a matrix is the dimension of the basis for the null space.

Thus, the nullity of the matrix is 11.

Answer

The basis for the null space is {[211]}{[1.41421356237309511]}.\left\{\left[\begin{array}{c}\sqrt{2}\\-1\\1\end{array}\right]\right\}\approx \left\{\left[\begin{array}{c}1.414213562373095\\-1\\1\end{array}\right]\right\}.A

The nullity of the matrix is 11A.