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MATH 221 Practice Quiz 3 Key
- Define F : R^2 → R^2 by,
F
([
v 1 v 2
])
[
|v 2 | v 1
]
Prove or disprove: F is a linear transformation.
Solution. This is not linear. Let w~ =
[
]
and let r = −1. Then, r ~w =
[
]
, and,
F (r ~w) =
[
]
[
]
but, −F ( w~) = −
[
]
[
]
Thus, since F (− w~) 6 = −F ( w~), F is not a linear transformation. ♥
- Let F : R^3 → R^3 be defined by F (~v) = π~v Calculate the standard matrix for F.
Solution. The standard basis in R^3 is,
~e 1 =
(^) , ~e 2 =
(^) , ~e 3 =
Then,
F (~e 1 ) = π~e 1 =
π 0 0
F (~e 2 ) = π~e 1 =
π 0
F (~e 3 ) = π~e 1 =
π
Therefore, the standard matrix for F is,
A = [ F (~e 1 ) F (~e 2 ) F (~e 3 ) ] =
π 0 0 0 π 0 0 0 π
- Consider the following three matrices,
A =
,^ B^ =
, C =
[
]
(a) Each of these matrices are the standard matrices of a linear transformation. Let FA denote the linear transformation with standard matrix A, and similarly for FB and FC. i) Write down the domains and codomains for each of FA, FB , and FC.
Solution. Domain of FA is R^4 , codomain is R^4. Domain of FB is R^4 , codomain is R^3. Domain of FC is R^5 , cododomain is R^2. ♥
ii) Which of these linear transformations could possibly be onto? Justify your answer.
Solution. All 3 of these linear transformations could be onto because the dimension of their domain is at least as big as the dimension of their codomain. Alternatively, the standard matrices for each transformation have at least as many columns as rows, so it is possible to have a pivot in every row in an echelon form of these matrices, which would mean the linear transformations would be onto by The Span Theorem. ♥ iii) Which of these linear transformations could possibly be one-to-one? Justify your answer.
Solution. FA is the only one that could possibly be one-to-one because the other two matrices have the dimension of the codomain strictly less than the dimension of the domain. Alterna- tively, because the number of columns is strictly larger than the number of rows for each of B and C, it is impossible for an echelon form of either of these matrices to have a pivot in every column. Therefore, the corresponding linear transformations can not be one-to-one by The Linear Independence Theorem. ♥
(b) Calculate the following expressions if they are defined. If they are not defined, give a reason why.
i) C^2
Solution. C is 5 × 2, so has more rows than columns. Therefore, C^2 = CC is not defined.
ii) BT^ B − A
Solution. B is 3 × 4 so BT^ is 4 × 3, which means BT^ B is 4 × 4. Since A is also 4 × 4, this expression exists and is equal to,
BT B − A =
iii) I 3 − C−^1
Solution. This isn’t defined because C is not square, hence is not invertible. ♥
- An n × n matrix M is called antisymmetric if M T^ = −M ; that is the transpose of M is equal to the negative of M. (a) Let A and B be antisymmetric matrices. Show that AT^ − (rA + B) is an antisymmetric matrix for any r ∈ R.
Solution. We must show that (AT^ − (rA + B))T^ = −(AT^ − (rA + B)). By properties of transpose, we have,
(AT^ − (rA + B))T^ = (AT^ )T^ − (rA + B)T = (−A)T^ − ((rA)T^ + BT^ ) because A is antisymmetric = −AT^ − (rAT^ − B) becase B is antisymmetric = −AT^ − (r(−A) − B) because A is antisymmetric = −AT^ + (rA + B) = −(AT^ − (rA + B)) ♥
(b) By writing A =
[
a b c d
]
, determine conditions on a, b, c, d that guarantee that A is both antisym- metric and invertible.
Solution. For A to be antisymmetric, we must have,
AT^ = −A =⇒
[
a c b d
]
[
−a −b −c −d
]
Equating the entries in the matrix yields,
a = −a =⇒ a = 0, b = −c, d = −d =⇒ d = 0.
Therefore, a condition that guarantees that A is antisymmetric is that a = d = 0, and b = −c; so a general 2 × 2 antisymmetric matrix looks like, [ 0 b −b 0
]
For this to be invertible, we need 0 + b^2 6 = 0. The only solution to this equation is b = 0; therefore as long as b 6 = 0, A is invertible. Hence, the conditions that guarantee that A is antisymmetric and invertible is if a = d = 0 and b = −c with b 6 = 0. ♥
