# Matrix rank and nullity

• Mar 15th 2011, 11:52 PM
I-Think
Matrix rank and nullity
Consider

$A=$\left( \begin{array}{ccc} 2 & 8 & 1 \\ -3 & -12 & 0 \\ 5 & 20 & -2 \end{array} \right)$$

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

What is the rank and nullity of L_A?

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

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

Rank
${(1,4,0),(0,0,6)}$ should serve as a basis, so $rank(L_A)=2$?
• Mar 16th 2011, 02:21 AM
Ackbeet
Hmm. Right after the word "Attempt", you have an equation which is incorrect. You're doing pointwise multiplication of the first column times the coordinate vector (almost always treated as column vectors), and then the second column times the coordinate vector, etc. But that's not matrix multiplication. It's rows in the matrix dotted into the coordinate vector. That is,

$L_{A}\begin{bmatrix}x\\y\\z\end{bmatrix}=
\begin{bmatrix}2&8&1\\-3&-2&0\\5&20&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=
\begin{bmatrix}2x+8y+z\\-3x-2y\\5x+20y-2z\end{bmatrix}.
$

However, this operation isn't actually necessary. Just do row reduction on $A$ to find both the rank and the nulllity. What do you get?