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

• MHB
• lfdahl
In summary, there is no matrix $S$ such that $S^n= \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$for any integer $n \geq 2$.
lfdahl
Gold Member
MHB
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 $n \geq 2$.

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)

## What is the "Prove 2x2 Matrix Puzzle"?

The "Prove 2x2 Matrix Puzzle" is a mathematical puzzle that involves proving that there does not exist a matrix S that satisfies the equation S^n = 0 for any positive integer n.

## Why is this puzzle important?

This puzzle is important because it is a fundamental concept in linear algebra and matrix theory. It also has many practical applications in fields such as engineering, physics, and computer science.

## What is the difficulty level of this puzzle?

The difficulty level of this puzzle can vary depending on one's mathematical background and problem-solving skills. For those with a strong foundation in linear algebra and matrix theory, it may be considered a relatively easy puzzle. However, for those with less experience in these areas, it may be more challenging.

## What are some strategies for solving this puzzle?

One strategy for solving this puzzle is to approach it from a proof by contradiction standpoint. Assume that there exists a matrix S that satisfies the equation S^n = 0 and then show that this leads to a contradiction. Another strategy is to use properties of matrices and their operations to manipulate the equation and arrive at a contradiction.

## Are there any real-world applications of this puzzle?

Yes, there are many real-world applications of this puzzle. For example, it can be used in coding theory for error correction, in physics for studying transformations and symmetries, and in data compression algorithms. It also has connections to other areas of mathematics, such as group theory and abstract algebra.

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