将军们，线性代数求救！！！

[verdelite]

Consider a 3X3 matrix A and a vector v in R^3 such that A^3v = 0 , but A^2v != 0 .

a. Show that the vectors {A^2v, Av, v} form a basis of R3.
Hint: It suffices to show linear independence.
Consider a relation 2
c1 A^2 v +c2 Av +c3 v = 0 and multiply by A^2 to show that c3 = 0 .

b. Find the matrix of the transformation T(x) = Ax with respect the basis {A^2v,Av, v}.

a好做，b怎么做？

b. 因为 {A^2v,Av, v}相互独立，所以它们组成个basis。用这个basis表述x，我们可以写x=c1 A^2 v +c2 Av +c3 v
那么Ax=A(c1 A^2 v +c2 Av +c3 v)=c2 A^2v +c3 Av；后面就好写了自己完成即可

[meiyoumajia]

我有20年没搞这种变换了。真是不行了。很快写出来了很简单的答案，但就是由于太简单而确定了几次才相信自己。

在以A^2v,Av, 和v为基表示矢量x的情况下转换矩阵B（不再是A）的形式是
0 0 0
0 0 1
0 1 0

是由于以它们为基时那个变换的结果A*矢量x就是
0
x3
x2
（x1，x2，和x3）是x的新表示

这样的v，A似乎很多。不知是否有什么类型的应用。

[YWY]

我有20年没搞这种变换了。真是不行了。很快写出来了很简单的答案，但就是由于太简单而确定了几次才相信自己。

在以A^2v,Av, 和v为基表示矢量x的情况下转换矩阵B（不再是A）的形式是
0 0 0
0 0 1
0 1 0

是由于以它们为基时那个变换的结果A*矢量x就是
0
x3
x2
（x1，x2，和x3）是x的新表示

这样的v，A似乎很多。不知是否有什么类型的应用。

T(x) = Ax定义一个线性变换，这个变换在{A^2v, Av, v}这组基之下的表示矩阵B（不再是A）等于
0 1 0
0 0 1
0 0 0

[meiyoumajia]

T(x) = Ax定义一个线性变换，这个变换在{A^2v, Av, v}这组基之下的表示矩阵B（不再是A）等于
0 1 0
0 0 1
0 0 0

对
转换后的矢量在新基下的表示是
x2
x3
0

[meiyoumajia]

这样的转换A对任何矢量最多用三次就得到0矢量。
这是一种什么类型的转换呢？还想象不出来。

对更高维，不一定总是有此类变换？

[YWY]

ChatGpt竟然作对了！

a. To show that the vectors {A^2v, Av, v} form a basis of R^3, we need to show that they are linearly independent and span R^3.
First, let's show linear independence. Suppose we have a relation:
c1A^2v + c2Av + c3v = 0
Multiplying both sides by A^2, we get:
c1A^4v + c2A^3v + c3A^2v = 0
Since A^3v = 0, this simplifies to:
c1A^2v = 0
Since A^2v is not zero (by assumption), we have c1 = 0.

Now, the relation reduces to:

c2Av + c3v = 0

Multiplying both sides by A, we get:

c2A^2v + c3Av = 0

Using the previous result (c1 = 0), we can simplify this to:

c3Av = 0

Since v is not zero (by assumption), we have c3 = 0.

Thus, we have shown that the vectors {A^2v, Av, v} are linearly independent.

Next, let's show that they span R^3. We know that A^2v is not zero (by assumption). Also, since A^3v = 0, we have:

A^3v = A^2(Av) = 0


Thus, Av is in the null space of A^2. Since A^2 is a 3x3 matrix, its null space has dimension at least 1 (by the rank-nullity theorem). Since A^2v is not zero, we have a basis for the null space of A^2 consisting of the vector A^2v and (possibly) additional vectors. Since Av is in the null space of A^2, it must be a linear combination of the basis vectors. But the only basis vector we have is A^2v, so Av must be a scalar multiple of A^2v.

Finally, we know that {A^2v, Av, v} has three vectors, which is the same as the dimension of R^3. Since they are linearly independent and span R^3, they form a basis of 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 images of the basis vectors under T and express them as linear combinations of the basis vectors.

We have:

T(A^2v) = A(A^2v) = A^3v = 0
T(Av) = A(Av) = A^2v
T(v) = Av

Expressing T(Av) and T(v) as linear combinations of {A^2v, Av, v}, we get:

T(Av) = A^2v = 1(A^2v) + 0(Av) + 0(v)
T(v) = Av = 0(A^2v) + 1(Av) + 0(v)

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

[T] = [0 1 0; 0 0 1; 0 0 0]

(Note that this is the matrix of a linear transformation, not a matrix multiplication.)

the first step should yield c_3 = 0.

[rgg]

我有20年没搞这种变换了。真是不行了。很快写出来了很简单的答案，但就是由于太简单而确定了几次才相信自己。

在以A^2v,Av, 和v为基表示矢量x的情况下转换矩阵B（不再是A）的形式是
0 0 0
0 0 1
0 1 0

是由于以它们为基时那个变换的结果A*矢量x就是
0
x3
x2
（x1，x2，和x3）是x的新表示

这样的v，A似乎很多。不知是否有什么类型的应用。

这是Nilpotent Operator. 在线性代数的jordan 标准型里， 有限维李代数表示大概都会碰到。

[wind]

这种作业题还是该自己做。。。

[TheMatrix]

这是Nilpotent Operator. 在线性代数的jordan 标准型里， 有限维李代数表示大概都会碰到。

嗯，我想起来了。以前学李代数确实一上来就有nilpotent element。

我以前对线性代数重视不够。现在看来，加上各种表示论，线性代数可以说涵盖了一切数学结构。

[Caravel]

嗯，我想起来了。以前学李代数确实一上来就有nilpotent element。

我以前对线性代数重视不够。现在看来，加上各种表示论，线性代数可以说涵盖了一切数学结构。

对，幂零矩阵，属于一类 exponential可以明确写出的矩阵。前段时间我还准备发个帖子，后来救忘记了

[meiyoumajia]

我不懂。这种“降维”变换很有用，可以解决一些多维空间特殊子空间问题或相应的其它数学性等效的问题？
对所有的n都有A存在。前面的老问题：但对有些n没有此类的实数矩阵吧？

[YWY]

我不懂。这种“降维”变换很有用，可以解决一些多维空间特殊子空间问题或相应的其它数学性等效的问题？
对所有的n都有A存在。前面的老问题：但对有些n没有此类的实数矩阵吧？

n = 4: for following matrix A satisfies A^4 = 0 but A^3 is not 0

0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0

You can do this for every n.

[meiyoumajia]

n = 4: for following matrix A satisfies A^4 = 0 but A^3 is not 0

0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0

You can do this for every n.

对。我真糊涂。

用原问题得到的结果，逆回去看。这样的A（也就是前面得出的那种B）一次次被用在具有下面性质的表示的任何矢量x
x1
x2
。。。
xn
（特别地取所有xi都不为0）
上，从“下往上”（从第n维直到第一维）每次降了一维

因此对这种A，满足A^k=0的最小k是n。A^k是这样：与对角线距离为k的线上元素都为1，而其它元素都为0。

[wind]

看看Jordan标准型的几何证明就明白怎么回事了。

[meiyoumajia]

我对Jordan标准型有印象。非数学专业的人应该都学过。

但我对零幂矩阵没有印象。也许学过，但可能对专业不重要而没有深究？它似乎对线性变换和其它问题有更深的理解特别重要。

感觉这个版上开始有了在学人员/年轻人或者父母。如果我们帮助做作业题可能就对班上的教师们冒犯了。：-）

[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。