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Find the possible conditions and possible cases in the real numbers for which A2 = 0

syncstar  Aug 15, 2017
edited by syncstar  Aug 15, 2017
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 #1
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\(\begin{bmatrix} a && b \\ c && d \end{bmatrix}*\begin{bmatrix} a && b \\ c && d \end{bmatrix}=\begin{bmatrix} a^2+b*c &&a*b+b*d \\ c*a+d*c && d^2+c*b \end{bmatrix}= \begin{bmatrix} a^2+bc && b*(a+d) \\ c*(a+d) && d^2+bc \end{bmatrix}=\begin{bmatrix} 0 && 0 \\ 0 && 0 \end{bmatrix}\)

(The multiplication of the matrices is derived from the formula: \((A*B)i,j =\sum_{k=1}^{n}a(i,k)*b(k,j)\)

 

so we have a set of equations:

1. a2+bc=0

2. b*(a+d)=0

3. c*(a+d)=0

4. d2+bc=0

 

using the first and the fourth equations, we can find that :(a2+bc)-(d2+bc)=(0)-(0)=0=a2-d2=(a-d)*(a+d). therefore, we can divide our answer to two cases:

 

first case- a=d

that means b*(a+d)=0=b*(2a) and c*(a+d)=0=c*(2a). now we have a new set of equations:

 

1. a2+bc=0

2. b*(2a)=0

3. c*(2a)=0

 

we can make it simpler:

 

1. a2+bc=0

2. b*a=0

3. c*a=0

 

now we can divide that case to two different cases:

 

a=0:

now we have 1 equation:

 

1. a2+bc=0=02+bc=bc=0. that means b or c must be equal to 0 (that or is a mathematical or: it means if both of them are 0 it works too)

 

that means the matrices that maintain the equations are of the following form:

 

\(\begin{bmatrix} 0 && b \\ c && 0 \end{bmatrix}\)(where c*b=0)

 

a is not 0:

that means we can divide the second and the third equations by a and get the set of equations:
 

1. a2+bc=0

2. b=0

3. c=0

 

that means a2+bc=a2+0*0=a2. but that means a=0, and that doesnt make any sense because we know that a CANNOT be 0. So there is no solution

 

second case- d=-a

now we can derive a simpler set of equations:

 

1. a2+bc=0

2. b*(a-a)=0=b*0=0

3. c*(a-a)=0=c*0=0

 

now we have one equation- a2+bc=0:

 

a2+bc=0/ -a2

bc=-a2

 

that gives us the next set of matrices: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

 

(where n is a natural number, couldnt find the plus minus sign). but as you can see, every matrice from the first set of matrices is also a member of the second set of matrices, therefore we can simply say that every matrice that is a member of the second set is a solution (keep in mind that every matrice from that set is a solution)

 

but we dont want A=0 as a solution, so we simply have to add that b*c is not 0.

 

solution: \(\begin{bmatrix} (-1)^n\sqrt{-bc} && b \\ c && -(-1)^n\sqrt{-bc} \end{bmatrix} \)

(where b*c is not 0)

Ehrlich  Aug 17, 2017
 #2
avatar+27 
0

The whole matrix (A) cant be 0. But "a"  of course it can, I guess.

a, b, c, d cant be 0 at the same time . Because Thats the whole matrix. 

 

Is there something to fix? 

 

Where you say that couldnt find the minus plus sign. Where is it should placed? 

syncstar  Aug 17, 2017
 #3
avatar+18715 
0

Matrix Problem: A \(\mathbf{\in \mu}\) (2x2).

Find the possible conditions and possible cases in the real numbers for which A2 = 0

In linear algebra, a nilpotent matrix is a square matrix A such that

\({\displaystyle A^{k}=0\,}\)

for some positive integer k.

 

If A is a nilpotent matrix then det(A) = 0

\(\begin{array}{|rcll|} \hline det(A) &=& \begin{vmatrix} a & b \\ c&d \end{vmatrix} = ad-bc = 0 \\\\ \Rightarrow \qquad \mathbf{bc} & \mathbf{=} & \mathbf{ad} \\ \hline \end{array}\)

 

\(A^2 = 0 \\ \begin{array}{|rcll|} \hline A^2 &=& \begin{pmatrix} a & b \\ c&d \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c&d \end{pmatrix} \\\\ &=& \begin{pmatrix} a^2+bc & b\cdot (a+d) \\ c\cdot (a+d) & d^2+bc \end{pmatrix} \quad & | \quad \mathbf{bc=ad} \\\\ &=& \begin{pmatrix} a^2+ad & b\cdot (a+d) \\ c\cdot (a+d) & d^2+ad \end{pmatrix} \\\\ &=& \begin{pmatrix} a\cdot (a+d) & b\cdot (a+d) \\ c\cdot (a+d) & d\cdot(a+d) \end{pmatrix} \\\\ A^2 &=& \underbrace{(a+d)}_{=0}\cdot \underbrace{\begin{pmatrix} a & b \\ c & d \end{pmatrix}}_{A\ne 0} = 0\\\\ a+d &=& 0 \\ \mathbf{a} &\mathbf{=}& \mathbf{-d} \\ \hline \end{array}\)

 

if \(A\ne 0 \) and \(A^2 = 0\), then \( a = -d\) and \(det(A) = 0\).

 

\(\text{Example } 1:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 0 & 1 \\ 0&0 \end{vmatrix} = 0\cdot 0 - 0 \cdot 1 = 0 \\ a &=& -d \\ 0 &=& 0\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 0&0 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

 

\(\text{Example } 2:\\ \begin{array}{|rcll|} \hline A &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \\ det(A) &=& \begin{vmatrix} 4 & 8 \\ -2 & -4 \end{vmatrix} = 4\cdot(-4) - (-2) \cdot 8 = -16+16 = 0 \\ a &=& -d \\ 4 &=& -(-4) \\ 4 &=&4\ \checkmark \\\\ A^2 &=& \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} \cdot \begin{pmatrix} 4 & 8 \\ -2 & -4 \end{pmatrix} = 0\ \checkmark \\ \hline \end{array}\)

 

 

laugh

heureka  Aug 18, 2017
edited by heureka  Aug 18, 2017
edited by heureka  Aug 18, 2017

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