Null space of $\left[\begin{array}{cc}2 & -10\\1 & -5\end{array}\right]$

Find the null space of $\left[\begin{array}{cc}2 & -10\\1 & -5\end{array}\right]$.

The reduced row echelon form of the matrix is $\left[\begin{array}{cc}1 & -5\\0 & 0\end{array}\right]$ (for steps, see rref calculator).

To find the null space, solve the matrix equation $\left[\begin{array}{cc}1 & -5\\0 & 0\end{array}\right]\left[\begin{array}{c}x_{1}\\x_{2}\end{array}\right] = \left[\begin{array}{c}0\\0\end{array}\right].$

If we take $x_{2} = t$, then $x_{1} = 5 t$.

Thus, $\mathbf{\vec{x}} = \left[\begin{array}{c}5 t\\t\end{array}\right] = \left[\begin{array}{c}5\\1\end{array}\right] t.$

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 $1$.

The basis for the null space is $\left\{\left[\begin{array}{c}5\\1\end{array}\right]\right\}$A.

The nullity of the matrix is $1$A.