+0

+1
76
3
+27

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
Sort:

#1
+310
0

$$\begin{bmatrix} a && b \\ c && d \end{bmatrix}*\begin{bmatrix} a && b \\ c && d \end{bmatrix}=\begin{bmatrix} a^2+b*c &&a*b+b*d \\ c*a+d*c && d^2+c*b \end{bmatrix}= \begin{bmatrix} a^2+bc && b*(a+d) \\ c*(a+d) && d^2+bc \end{bmatrix}=\begin{bmatrix} 0 && 0 \\ 0 && 0 \end{bmatrix}$$

(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
+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
+18564
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 $$det(A) = 0$$.

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

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

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

19 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details