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Problems #1,11 on pg.236 are graded.
#1, pg.236: Assume that the matrix A is row equivalent to the matrix B. Determine the rank of A and the dimension of NulA without calculations. Find the bases for ColA, RowA and NulA:
A=[1−49−7−12−415−6107],
B=[10−150−25−60000].
Solution: By observation we know that rank(A)=2 because there are two nonzero rows (and hence a basis of Row A containing two elements --- this can also be seen by noting that there are only two pivot columns). The rank-nullity theorem now shows that
dimNulA=4−rankA=4−2=2.
Also by observation we can see that a basis for the row space of A is
Brow={[10−15]T,[0−25−6]T}
and a basis for the column space
Bcol={[1−15],[−42−6]}.
To find a basis for the nullspace, we will first compute the reduced echelon form of A (start computing this from B!) and we see
A∼[10−1501−5230000].
To find a basis for the nullspace we need to solve the equation A→x=→0 and our row reduction shows that solving this equation yields the augmented matrix
[10−15001−523000000].
From this we write the associated system of equations
{x1−x3+5x4=0x2−52x3+3x4=00=0
and so we see that all solutions →x of A→x=→0 can be expressed as
→x=[x1x2x3x4]=[x3−5x452x3−3x4x3x4]=x3[15210]+x4[−5−301].
Therefore we see that a basis for the nullspace of A is given by
BNul={[15210],[−5−301]}.
#11,pg.237: If the nullspace of an 8×5 matrix A is 3-dimensional, what is the dimension of the row space of A?
Solution: The rank-nullity theorem tells us that
rankA=5−dimNulA=5−3=2.
The other part of the rank-nullity theorem tells us that rankA=dimRowA and so we see
dimRowA=2.
