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Math Help - Matrix rank and nullity

  1. #1
    Senior Member I-Think's Avatar
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    Matrix rank and nullity

    Consider

    A=\[ \left( \begin{array}{ccc}<br />
2 & 8 & 1 \\<br />
-3 & -12 & 0 \\<br />
5 & 20 & -2 \end{array} \right)\]

    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?
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  2. #2
    A Plied Mathematician
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    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}=<br />
\begin{bmatrix}2&8&1\\-3&-2&0\\5&20&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=<br />
\begin{bmatrix}2x+8y+z\\-3x-2y\\5x+20y-2z\end{bmatrix}.<br />

    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?
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