将军们，线性代数求救！！！

[MaLaRabbit]

a. To show that {A^2v, Av, v} form a basis of R^3, we need to show that they are linearly independent and span R^3.

First, we will show that they are linearly independent. Suppose there exist constants c1, c2, c3 such that:

c1A^2v + c2Av + c3v = 0

Multiplying both sides by A^2, we get:

c1A^4v + c2A^3v + c3Av = 0

Since A^3v = 0, this simplifies to:

c1A^4v + c3Av = 0

Multiplying both sides by A, we get:

c1A^5v + c3A^2v = 0

Since A^2v is nonzero (given that A^2v ≠ 0), we can divide both sides by A^2v, giving:

c1A^3v + c3 = 0

But we know that A^3v = 0, so c3 = 0. Therefore, the original equation becomes:

c1A^2v + c2Av = 0

Since A^2v and Av are linearly independent (given that A^2v ≠ 0), we must have c1 = c2 = 0. Therefore, {A^2v, Av, v} are linearly independent.

To show that they span R^3, note that any vector x in R^3 can be written as:

x = aA^2v + bAv + cv

for some constants a, b, c. Therefore, {A^2v, Av, v} span R^3.

b. To find the matrix of the transformation T(x) = Ax with respect to the basis {A^2v, Av, v}, we need to find the coordinates of T(A^2v), T(Av), and T(v) with respect to this basis.

Since T(A^2v) = A(A^2v) = A^3v = 0, its coordinates with respect to the basis are (0, 0, 0).

Similarly, T(Av) = A(Av) = A^2v, which has coordinates (1, 0, 0) with respect to the basis.

Finally, T(v) = Av, which has coordinates (0, 1, 0) with respect to the basis.

Therefore, the matrix of T with respect to the basis {A^2v, Av, v} is:

[0 1 0]
[0 0 1]
[0 0 0]

[Pegasi]

这个是出过的题，ChatGPT背下来了

ChatGpt竟然作对了！

[YWY]

a. To show that {A^2v, Av, v} form a basis of R^3, we need to show that they are linearly independent and span R^3.
First, we will show that they are linearly independent. Suppose there exist constants c1, c2, c3 such that:
c1A^2v + c2Av + c3v = 0
Multiplying both sides by A^2, we get:
c1A^4v + c2A^3v + c3Av = 0

Since A^3v = 0, this simplifies to:

c1A^4v + c3Av = 0

Multiplying both sides by A, we get:

c1A^5v + c3A^2v = 0

Since A^2v is nonzero (given that A^2v ≠ 0), we can divide both sides by A^2v, giving:

c1A^3v + c3 = 0

But we know that A^3v = 0, so c3 = 0. Therefore, the original equation becomes:

c1A^2v + c2Av = 0

Since A^2v and Av are linearly independent (given that A^2v ≠ 0), we must have c1 = c2 = 0. Therefore, {A^2v, Av, v} are linearly independent.

To show that they span R^3, note that any vector x in R^3 can be written as:

x = aA^2v + bAv + cv

for some constants a, b, c. Therefore, {A^2v, Av, v} span R^3.

b. To find the matrix of the transformation T(x) = Ax with respect to the basis {A^2v, Av, v}, we need to find the coordinates of T(A^2v), T(Av), and T(v) with respect to this basis.

Since T(A^2v) = A(A^2v) = A^3v = 0, its coordinates with respect to the basis are (0, 0, 0).

Similarly, T(Av) = A(Av) = A^2v, which has coordinates (1, 0, 0) with respect to the basis.

Finally, T(v) = Av, which has coordinates (0, 1, 0) with respect to the basis.

Therefore, the matrix of T with respect to the basis {A^2v, Av, v} is:

[0 1 0]
[0 0 1]
[0 0 0]

在c1A^4v + c2A^3v + c3Av = 0这一步chatGPT操作有误，在之后的处理上，chatGPT也显得很文科（比如两边除以A^2v的操作）。。。

总体来说，现阶段的chatGPT，文科强于理科，讲段子能力高于数学证明的能力；这可能是因为讲段子的要求比较宽松，而理科证明需要更加严谨。