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HW for Oct 15
A =
−Gauss-Jordan−−−−−−→ C =
RREF
C corresponds to the system x 1 = −x 3 − x 4 and x 2 = −x 3 + x 4.
AT^ −Gauss-Jordan−−−−−−→ D =
RREF
D corresponds to the system y 1 = (9/7)y 3 and y 2 = (1/7)y 3. m = # rows of A = 3. n = # columns of A = 4. r = rank(A) = # non-zero rows of C = # non-zero rows of D = 2. dim span(A) = dim rowspan(A) = r = 2. dim N (A) = n − r = 4 − 2 = 2. dim N (AT^ ) = m − r = 3 − 2 = 1. v 1 , v 2 , v 3 , v 4 ∈ R^3 are the columns of A and u 1 , u 2 , u 3 ∈ R^1 ×^4 are the rows of A. 2 = r < n = 4 =⇒ the columns of A are linearly dependent and the rows of A are non-spanning of R^1 ×^4. 2 = r < m = 3 =⇒ the rows of A are linearly dependent and the columns are non-spanning of R^3. Let
B 1 =
, B′
, and B 2 =
Then the columns of B 1 form a basis of span(A). Same for B 1 ′. The columns of B 2 form a basis for N (A). Also we have that
(−1)v 1 + (−1)v 2 + 1v 3 = ¯ 0 and [− 1 , − 1 , 1 , 0] 6 ∈ rowspan(A).
Let
B 3 =
[
]
, B 3 ′ =
[
]
, and B 4 =
Then the rows of B 3 form a basis of rowspan(A). Same for B 3 ′. The columns of B 4 form a basis for N (AT^ ). Also we have that
(9/7)u 1 + (1/7)u 2 + 1u 3 = ¯ 0 and [9/ 7 , 1 / 7 , 1]T^6 ∈ span(A).
Let T ∈ R^2 ×^2 be the transition matrix for the change of basis from [v 1 , v 2 ] (the columns of B 1 ) to [w 1 , w 2 ] (the columns of B 1 ′) of span(A). The first column of T is the solution to the system B′ 1 x = v 1 and the second to B 1 ′x = v 2. Hence
T =
[
]
A =
−Gauss-Jordan−−−−−−→ C =
RREF
C corresponds to the system x 1 = 0 and x 2 = 0.
AT^ −Gauss-Jordan−−−−−−→ D =
[
]
RREF
D corresponds to the system y 1 = (10/7)y 3 and y 2 = (9/7)y 3.
