MAT 240 ASSIGNMENT # 4

（1）设V = Sym2（F）（并假定char（F）6 = 2）。令β= {E11，E12 + E21，E22}并且
γ= {E11 + E12 + E21，E22 + E12 + E21，E11 + E22}。

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（a）求Q-坐标矩阵从β坐标到γ坐标的变化。
（b）找出Q-1
-将坐标矩阵从γ坐标更改为β坐标。
（c）找出[T]β和[T]γ。
（2）设A∈Mn×n（F）为一列c1，…的矩阵。 。 。 ，cn。
（a）证明A的列构成F的基础
ñ

（b）推论每个可逆矩阵Q都存在F的基数β，γ
ñ

（c）（T / F :)假设V是任何n维向量空间，T：V→V是同构。如果Q是可逆的，则存在V的基数β，γ，因此[T]
γ
β=Q。
（d）（T / F :)假设V是任何n维向量空间，而T：V→V是同构。如果Q是可逆的，则存在V的基数β，因此[T]β=Q。
（3）设A =

1 0 1 2
1 1 2 3
2 1 3 5

。求出R-A的RREF，并针对某些可逆矩阵Q表示R = QA。
（提示：跟踪行操作，并记住每个可逆矩阵都是一个乘积

（4）证明对于每个n×n矩阵A，存在一个可逆矩阵Q，以使QA为上三角。 （提示：对n和行运算/元素矩阵使用归纳法。）
(5) Define an equivalence relation on Mm×n(F) by A ∼ B if there exists invertible matrices
Q ∈ Mm×m(F), P ∈ Mn×n(F) so that A = QBP.
(a) Prove that this is an equivalence relation.
(b) Express the number of distinct equivalence classes in Mm×n(F) as a function of m and n.
(And, as usual, prove your claim.)
(6) (a) Let T : V → W, S : W → X be isomorphisms. Prove that S ◦ T is an isomorphism and
find a formula for (S ◦ T)
−1
.
(b) Prove that the product of two invertible matrices of the same size is invertible using (a).
(c) Give a second proof of the fact that the product of two invertible matrices of the same
size is invertible, using elementary matrices.
(d) Give a third proof of the fact that the product of two invertible matrices of the same size
is invertible, using rank.
(7) Find an isomorphism T : Sk3(R) → W, where W = {(x, y, z, w) ∈ R
4
| x + y + z + w = 0}.
(8) Let T : R
2 → R
2 be a linear transformation so that T(1, 1) = (2, 2) and T(1, 0) = (−1, 0). Let
γ = {(1, 1),(1, 0)} and β be the standard basis.
(a) Find [T]γ.
(b) Find Q – the change of coordinate matrix from γ to β coordinates.
(c) Find Q−1 using row operations on (Q|I2).
(d) Find [T]β and deduce an explicit formula for T(x, y).
(e) Which coordinates are best suited for studying this transformation – β or γ? Why?
(9) Determine if the statements below are true or false. If true, give a proof. If false, explain why,
and/or provide a counterexample.
(a) There exists a m × n matrix A so that the system Ax = b has no solutions for all b ∈ F
m.
(b) Suppose that A ∈ Mn×n(F). If Ax = b has solutions for all b ∈ F
n
, then A is invertible.
(c) If A ∈ Mm×n(F) has rank n − 1, then Ax = b has no solutions for some b ∈ F
m.