Science:Math Exam Resources/Courses/MATH221/April 2013/Question 02 (b)

MATH221 April 2013

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[hide] Question 02 (b)
Let $A=\left[{\begin{array}{ccccc}1&2&1&3&2\\3&4&9&0&7\\2&3&5&1&8\\2&2&8&-3&5\end{array}}\right]B=\left[{\begin{array}{ccccc}1&2&1&3&2\\0&1&-3&5&-4\\0&0&0&1&-7\\0&0&0&0&0\end{array}}\right]$ The matrices A and B are related by $B=MA$ , where M is an invertible 4x4 matrix. b) Find the dimension and a basis of the nullspace of A.

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[show] Hint
Again, is the nullspace preserved under left multiplication by invertible matrices? Can B be furthered simplified?

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[show] Solution
By the Rank-Nullity theorem, the dimension of the nullspace of A is equal to the number of column minus the rank of A (which we found in part (a)), thus it is 5 - 3 = 2. The nullspace is unaffected by left multiplication by an invertible matrix, that is $Bv=MAv=0$ if and only if $Av=0$ . Thus, we can use the matrix B to compute the nullspace. To find a basis of the nullspace, we will further simplify B by L₁ → L₁-2L₂, followed by L₂ → L₂-5L₃ and L₁ → L₁+7L₃ to get the matrix $\left[{\begin{array}{ccccc}1&0&7&0&-39\\0&1&-3&0&31\\0&0&0&1&-7\\0&0&0&0&0\end{array}}\right]$ Again, further row operations won't change the nullspace. Applying this matrix to a vector $\left[{\begin{array}{ccccc}x_{1}&x_{2}&x_{3}&x_{4}&x_{5}\\\end{array}}\right]$ and solving for zero we get the relations ${\begin{aligned}&x_{1}+7x_{3}-39x_{5}&=0\\&x_{2}-3x_{3}+31x_{5}&=0\\&x_{4}-7x_{5}=0\end{aligned}}$ We know that the dimension of the nullspace is 2, thus there must be 2 free variables, namely $x_{5},x_{3}$ . So, we can read off the basis as $\left[{\begin{array}{ccccc}39&-31&0&7&1\\\end{array}}\right],\left[{\begin{array}{ccccc}-7&3&1&0&0\\\end{array}}\right]$ .