MHB Prove 2x2 Matrix Puzzle: No $S$ Exists for $S^n$

  • Thread starter Thread starter lfdahl
  • Start date Start date
  • Tags Tags
    Matrix Puzzle
AI Thread Summary
There is no 2x2 matrix \( S \) such that \( S^n = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \) for any integer \( n \geq 2 \). The matrix \( T \) has rank 1, and its image and null space are both 1-dimensional, leading to \( T^2 = 0 \). If \( T = S^n \), then \( S \) must also have rank 1, as it cannot be rank 2 or 0. This implies that the image and null space of \( S \) are equal, resulting in \( S^2 = 0 \), which contradicts \( T \neq 0 \). Therefore, \( T \) has no \( n \)th roots for \( n \geq 2 \.
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Prove, that there is no $2 \times 2$ matrix, $S$, such that

\[S^n= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\]

for any integer $n \geq 2$.
 
Mathematics news on Phys.org
lfdahl said:
Prove, that there is no $2 \times 2$ matrix, $S$, such that

\[S^n= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\]

for any integer $n \geq 2$.
[sp]Let $T = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$. Then $T$ has rank $1$. Considering $T$ as a linear operator on a $2$-dimensional space, its image and its null space are both equal to the $1$-dimensional subspace spanned by the first basis vector $\begin{bmatrix} 1\\ 0 \end{bmatrix}$ (and hence $T^2 = 0$, as you can easily check by squaring the matrix).

Suppose that $T = S^n$. Then $S$ cannot have rank $2$, because then it would be surjective and so every power of $S$ would also be surjective. Also, $S$ cannot have rank $0$, because then it would represent the zero operator and so $T$ would be $0$. So $S$ must have rank $1$.

If the vector $x$ is in the null space of $S$ then $Tx = S^nx = S^{n-1}Sx = 0$. So the null space of $S$ is contained in the null space of $T$. But since those spaces are both $1$-dimensional, they must be equal.

For any vector $y$, $Ty = S(S^{n-1}y)$. So the image of $S$ contains the image of $T$. But since those spaces are both $1$-dimensional, they must be equal.

Therefore the image and null space of $S$ are equal. So for any vector $y$, $S^2y = S(Sy) = 0$ and thus $S^2$ is the zero matrix. But if $S^2 = 0$ then $S^n = 0$ for all $n\geqslant 2$, contradicting the fact that $T \ne0$.

Thus $T$ has no $n$th roots, for $n\geqslant 2$.[/sp]
 
lfdahl said:
Prove, that there is no $2 \times 2$ matrix, $S$, such that

\[S^n= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\]

for any integer $n \geq 2$.

Suppose $S$ exists and has eigenvalues $\lambda_1$ and $\lambda_2$ (which could conceivably be complex).
Then $S^n$ has eigenvalues $\lambda_1^n$ and $\lambda_2^n$.
The given matrix is in Jordan Normal Form, showing that $\lambda_1^n=\lambda_2^n=0$, and therefore $\lambda_1=\lambda_2=0$.
So $S$ is either similar to the 0 matrix, or to the given nilpotent matrix.

If $S$ is similar to 0, then $S^n=0$, which is a contradiction.
And if $S$ is similar to the given nilpotent matrix, then $S^n=0$, which is also a contradiction.
Therefore there is no such $S$.
 
Opalg said:
[sp]Let $T = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$. Then $T$ has rank $1$. Considering $T$ as a linear operator on a $2$-dimensional space, its image and its null space are both equal to the $1$-dimensional subspace spanned by the first basis vector $\begin{bmatrix} 1\\ 0 \end{bmatrix}$ (and hence $T^2 = 0$, as you can easily check by squaring the matrix).

Suppose that $T = S^n$. Then $S$ cannot have rank $2$, because then it would be surjective and so every power of $S$ would also be surjective. Also, $S$ cannot have rank $0$, because then it would represent the zero operator and so $T$ would be $0$. So $S$ must have rank $1$.

If the vector $x$ is in the null space of $S$ then $Tx = S^nx = S^{n-1}Sx = 0$. So the null space of $S$ is contained in the null space of $T$. But since those spaces are both $1$-dimensional, they must be equal.

For any vector $y$, $Ty = S(S^{n-1}y)$. So the image of $S$ contains the image of $T$. But since those spaces are both $1$-dimensional, they must be equal.

Therefore the image and null space of $S$ are equal. So for any vector $y$, $S^2y = S(Sy) = 0$ and thus $S^2$ is the zero matrix. But if $S^2 = 0$ then $S^n = 0$ for all $n\geqslant 2$, contradicting the fact that $T \ne0$.

Thus $T$ has no $n$th roots, for $n\geqslant 2$.[/sp]

Hi, Opalg! - another excellent contribution from you! Thankyou for sharing your expertice in this challenge/puzzle forum!(Handshake)
 
I like Serena said:
Suppose $S$ exists and has eigenvalues $\lambda_1$ and $\lambda_2$ (which could conceivably be complex).
Then $S^n$ has eigenvalues $\lambda_1^n$ and $\lambda_2^n$.
The given matrix is in Jordan Normal Form, showing that $\lambda_1^n=\lambda_2^n=0$, and therefore $\lambda_1=\lambda_2=0$.
So $S$ is either similar to the 0 matrix, or to the given nilpotent matrix.

If $S$ is similar to 0, then $S^n=0$, which is a contradiction.
And if $S$ is similar to the given nilpotent matrix, then $S^n=0$, which is also a contradiction.
Therefore there is no such $S$.

Thanks a lot, I like Serena. You are fast on the trigger! - yet still hitting "bulls eye" with this very fine solution of yours. Thankyou for your participation!(Cool)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...

Similar threads

Replies
4
Views
1K
Replies
8
Views
3K
Replies
2
Views
2K
Replies
5
Views
2K
Replies
10
Views
3K
Back
Top