Consider

$\displaystyle A=\[ \left( \begin{array}{ccc}

2 & 8 & 1 \\

-3 & -12 & 0 \\

5 & 20 & -2 \end{array} \right)\] $

Consider the linear operator $\displaystyle L_A$ on $\displaystyle R^3$ which is left multiplication by A

What is the rank and nullity of L_A?

Attempt

$\displaystyle L_A (x,y,z)=$ $\displaystyle \[ \left( \begin{array}{ccc} 2x-3y+5x & 4(2x-3y+5z) & x-2z \end{array} \right)\] $

Nullity

So if $\displaystyle x=2z,y=3z, L_A=0$ so the null space of $\displaystyle L_A$ consists of vectors of the form $\displaystyle (2z,3z,z)$ so its basis should be $\displaystyle (2,3,1)$ and thus $\displaystyle nullity(L_A)=1$?

Rank

$\displaystyle {(1,4,0),(0,0,6)}$ should serve as a basis, so $\displaystyle rank(L_A)=2$?