Nonzero matrix that is 2 by 2

What you speak of is called a nilpotent matrix.

The matrices have the property : Ak=0\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:

\(\displaystyle Let ~A=[0100]A=\left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\\end{array} } \right]\)

or

\(\displaystyle Let ~A=[0010]A=\left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\\end{array} } \right]\)

are two solutions.

Obviously the 1 can be any constant you want.
 
Hello, msjoharia!

Here are a few more . . .

Find a nonzero 2 x 2 matrix A such that AA has all zero entries.
Click to expand...

\(\displaystyle \text{Let }\,A \:=\:\begin{pmatrix}a^&b\\c&d\end{pmatrix}\;\text{such that }\,A\!\cdot\! A \:=\:0\)

We have:  (abcd)(abcd) = (0000)\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}


Hence:  {a2+bc = 0[1]ab+bd = 0[2]ac+cd = 0[3]bc+d2 = 0[4]}\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}


From [2]:  b(a+d) = 0\displaystyle \text{From [2]: }\:b(a+d) \:=\:0
From [3]:  c(a+d) = 0\displaystyle \text{From [3]: }\:c(a+d) \:=\:0

If b=0 or c=0, then all the elements are zero.\displaystyle \text{If }b = 0\text{ or }c = 0\text{, then all the elements are zero.}


So we have:  a+d = 0⇒d = −a\displaystyle \text{So we have: }\:a+d \:=\:0 \quad\Rightarrow\quad d \:=\:-a

Then [4] becomes:  bc+a2 = 0, identital to [1].\displaystyle \text{Then [4] becomes: }\:bc+a^2 \:=\:0\text{, identital to [1].}
. . and we have:  bc = −a2\displaystyle \text{and we have: }\:bc \:=\:-a^2

For simplicity, let b=±a  and  c=∓a\displaystyle \text{For simplicity, let }b = \pm a\,\text{ and }\,c = \mp a


Therefore:  A  =  (a±a∓a−a)  =  a(1±1∓1−1)   for a≠0\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 bc=−a2\displaystyle bc = -a^2, we can have: .{b=±a2c=∓1}\displaystyle \begin{Bmatrix}b &=& \pm a^2 \\ c &=& \mp1\end{Bmatrix}

. . Then:  A  =  (a±a2∓1−a)\displaystyle \text{Then: }\:A \;=\;\begin{pmatrix}a & \pm a^2 \\ \mp1 & -a \end{pmatrix}



If a is composite:  a = p ⁣⋅ ⁣q,  there are more variations.\displaystyle \text{If }a\text{ is composite: }\,a \,=\,p\!\cdot\!q,\,\text{ there are more variations.}

. . For example:  A  =  (pq±p2q∓q−pq)  =  q(p±p2∓1−p)\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}
 
Would a matrix [ 5 -25 ] [ 5 5]
[ 2 -10 ][ 1 1]


work?


When I mulitplied I got zero as the result
 
Last edited:
Hello, msjoharia!


Would a matrix:  [5-2555] [5511]  work?\displaystyle \text{Would a matrix: }\,\begin{bmatrix}5&\text{-}25 \\ 5&5\end{bmatrix}\,\begin{bmatrix}5&5\\1&1\end{bmatrix}\:\text{ work?}

When I multiplied, I got zero as the result.\displaystyle \text{When I multiplied, I got zero as the result.}
Click to expand...

Yes . . .

. . if the question was: "Find two different matrices A\displaystyle A and B\displaystyle B whose product is zero."

But that wasn't the question, was it?
 
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