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

[lfdahl]

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 $n \geq 2$.

 

[Opalg]

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]

 

[I like Serena]

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 $n \geq 2$.

Spoiler

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 $S$.

 

[lfdahl]

[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)

 

[lfdahl]

Spoiler

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)