Nonzero matrix that is 2 by 2
- Thread starter msjoharia
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renegade05
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What you speak of is called a nilpotent matrix.
The matrices have the property : \(\displaystyle A^k = 0\) where A is a square matrix and k is a positive integer.
A matrix is nilpotent if it is a triangular matrix. Meaning all the entries below or above a diagonal of zeros is also zero.
So for your problem:
\(\displaystyle Let ~\[A=\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\\end{array} } \right]\]\)
or
\(\displaystyle Let ~\[A=\left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\\end{array} } \right]\]\)
are two solutions.
Obviously the 1 can be any constant you want.
The matrices have the property : \(\displaystyle A^k = 0\) where A is a square matrix and k is a positive integer.
A matrix is nilpotent if it is a triangular matrix. Meaning all the entries below or above a diagonal of zeros is also zero.
So for your problem:
\(\displaystyle Let ~\[A=\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\\end{array} } \right]\]\)
or
\(\displaystyle Let ~\[A=\left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\\end{array} } \right]\]\)
are two solutions.
Obviously the 1 can be any constant you want.
Hello, msjoharia!
Here are a few more . . .
\(\displaystyle \text{Let }\,A \:=\:\begin{pmatrix}a^&b\\c&d\end{pmatrix}\;\text{such that }\,A\!\cdot\! A \:=\:0\)
\(\displaystyle \text{We have: }\:\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix} \:=\:\begin{pmatrix}0&0\\0&0\end{pmatrix}\)
\(\displaystyle \text{Hence: }\:\begin{Bmatrix} a^2 + bc \:=\:0 & [1] && ab + bd \:=\: 0 & [2] \\ \\ ac+cd \:=\: 0 & [3] && bc+d^2 \:=\: 0 & [4] \end{Bmatrix}\)
\(\displaystyle \text{From [2]: }\:b(a+d) \:=\:0\)
\(\displaystyle \text{From [3]: }\:c(a+d) \:=\:0\)
\(\displaystyle \text{If }b = 0\text{ or }c = 0\text{, then all the elements are zero.}\)
\(\displaystyle \text{So we have: }\:a+d \:=\:0 \quad\Rightarrow\quad d \:=\:-a\)
\(\displaystyle \text{Then [4] becomes: }\:bc+a^2 \:=\:0\text{, identital to [1].}\)
. . \(\displaystyle \text{and we have: }\:bc \:=\:-a^2\)
\(\displaystyle \text{For simplicity, let }b = \pm a\,\text{ and }\,c = \mp a\)
\(\displaystyle \text{Therefore: }\:A \;=\;\begin{pmatrix}a & \pm a \\ \mp a & -a \end{pmatrix} \;=\;a\begin{pmatrix}1 & \pm 1 \\ \mp1 & -1 \end{pmatrix}\;\text{ for }a \ne 0\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Of course, there are many other variations.
From \(\displaystyle bc = -a^2\), we can have: .\(\displaystyle \begin{Bmatrix}b &=& \pm a^2 \\ c &=& \mp1\end{Bmatrix}\)
. . \(\displaystyle \text{Then: }\:A \;=\;\begin{pmatrix}a & \pm a^2 \\ \mp1 & -a \end{pmatrix}\)
\(\displaystyle \text{If }a\text{ is composite: }\,a \,=\,p\!\cdot\!q,\,\text{ there are more variations.}\)
. . \(\displaystyle \text{For example: }\:A \;=\;\begin{pmatrix}pq & \pm p^2q \\ \mp q & -pq\end{pmatrix} \;=\;q\begin{pmatrix}p & \pm p^2 \\ \mp1 & -p\end{pmatrix}\)
Here are a few more . . .
\(\displaystyle \text{Let }\,A \:=\:\begin{pmatrix}a^&b\\c&d\end{pmatrix}\;\text{such that }\,A\!\cdot\! A \:=\:0\)
\(\displaystyle \text{We have: }\:\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix} \:=\:\begin{pmatrix}0&0\\0&0\end{pmatrix}\)
\(\displaystyle \text{Hence: }\:\begin{Bmatrix} a^2 + bc \:=\:0 & [1] && ab + bd \:=\: 0 & [2] \\ \\ ac+cd \:=\: 0 & [3] && bc+d^2 \:=\: 0 & [4] \end{Bmatrix}\)
\(\displaystyle \text{From [2]: }\:b(a+d) \:=\:0\)
\(\displaystyle \text{From [3]: }\:c(a+d) \:=\:0\)
\(\displaystyle \text{If }b = 0\text{ or }c = 0\text{, then all the elements are zero.}\)
\(\displaystyle \text{So we have: }\:a+d \:=\:0 \quad\Rightarrow\quad d \:=\:-a\)
\(\displaystyle \text{Then [4] becomes: }\:bc+a^2 \:=\:0\text{, identital to [1].}\)
. . \(\displaystyle \text{and we have: }\:bc \:=\:-a^2\)
\(\displaystyle \text{For simplicity, let }b = \pm a\,\text{ and }\,c = \mp a\)
\(\displaystyle \text{Therefore: }\:A \;=\;\begin{pmatrix}a & \pm a \\ \mp a & -a \end{pmatrix} \;=\;a\begin{pmatrix}1 & \pm 1 \\ \mp1 & -1 \end{pmatrix}\;\text{ for }a \ne 0\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Of course, there are many other variations.
From \(\displaystyle bc = -a^2\), we can have: .\(\displaystyle \begin{Bmatrix}b &=& \pm a^2 \\ c &=& \mp1\end{Bmatrix}\)
. . \(\displaystyle \text{Then: }\:A \;=\;\begin{pmatrix}a & \pm a^2 \\ \mp1 & -a \end{pmatrix}\)
\(\displaystyle \text{If }a\text{ is composite: }\,a \,=\,p\!\cdot\!q,\,\text{ there are more variations.}\)
. . \(\displaystyle \text{For example: }\:A \;=\;\begin{pmatrix}pq & \pm p^2q \\ \mp q & -pq\end{pmatrix} \;=\;q\begin{pmatrix}p & \pm p^2 \\ \mp1 & -p\end{pmatrix}\)