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(a) Let $A$ be an $n × n$ real matrix such that $(A + I)^4 = 0$ where $I$ denotes the identity matrix. Show that $A$ is non-singular. (b) Give an example of a nonzero $2×2$ real matrix $A$ such that $x′Ax = 0$ for all real vectors $x$.

(a) Note that $(A + I)^4 = 0=>A^4+4A^3+6A^2+4A+I=0$ Then I am stuck.

(b)$\begin{pmatrix} 0 & 1\\-1 & 0 \end{pmatrix}$ is an example. I get the matrix from intuitive idea. Is there another way to get such matrix.

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What is $x'$ for a vector $x$ ? –  Belgi Aug 5 '12 at 4:10
    
Here $x=\begin{pmatrix} {x_1} \\ {x_2} \end{pmatrix}$ and $x'=\begin{pmatrix} {x_1} & {x_2} \end{pmatrix}$ –  Argha Aug 5 '12 at 4:19
    
I never saw this notation before...I saw $x^t,x^T,x^{tr}$ –  Belgi Aug 5 '12 at 4:26
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3 Answers

up vote 2 down vote accepted

a) Notice that $p(A) = 0$ where $p(x) = \left(x+1\right)^4$. It follows that the minimal polynomial $\mu$ divides $p$. Since $\lambda$ is an eigenvalue if and only if $(x-\lambda)\mid \mu$ it follows that $-1$ is the only eigenvalue. More specifically, $0$ is not an eigenvalue and hence $A$ is invertible.

b) If we multiply the equation out we obtain $$x^\mathrm{T}Ax = 0$$ $$\begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} a & b\\c & d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix} = 0$$ $$ax^2 + dy^2 + (b+c)xy = 0$$ taking $x, y= 0$ gives $d, a = 0$ respectively. It is easily seen then that $$a = d = b+c = 0$$ is necessary and sufficient, i.e. all such examples are simply scalar multiples of the matrix you gave.

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You answer is very good for me.Thank you –  Argha Aug 5 '12 at 4:26
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For (a): If you factor $A(A^3+4A^2+6A+4I)=-I,$ and so $$\det(AB)=\det(A)\det(B)=\det(-I)=-1,$$ where $B=A^3+4A^2+6A+4I,$ hence we must have $\det(A)\neq 0.$

For (b): This is almost all the possibilities! Suppose $A$ is given by \begin{pmatrix} a & b\\c & d \end{pmatrix} Evaluating $v^tAv$ we get a quadratic form: $ax^2+(b+c)xy +by^2=0.$ In particular we can choose $v^t=(x,y)=(1,0)$, which shows $a=0.$ Similarly choosing $v^t=(0,1)$ shows $d=0,$ and choosing $v^t=(1,1)$ shows $b=-c.$ Thus $A$ must be \begin{pmatrix} 0 & b\\-b & 0 \end{pmatrix} for some scalar $b.$

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(a) If $A$ is singular there is a non-zero vector $v\in\mathbb{R}^n$ such that $Av=0$. Therefore, using the assumption $(A+I)^4=0$ we get $0=(A+I)^4v=A^4v+4A^3v+6A^2v+4Av+v=v$ which is a contradiction.

(b) Note that for all $x$, $x^\mathrm{T}Ax=x^\mathrm{T}A^\mathrm{T}x$ which implies $x^\mathrm{T}Ax=\frac{1}{2}x^\mathrm{T}(A^\mathrm{T}+A)x$. Now since $A^\mathrm{T}+A$ is a symmetric matrix, it is diagonalizable. Therefore, having $x^\mathrm{T}Ax=0$ for all $x$, is equivalent to all of the eigenvalues of $A^\mathrm{T}+A$ being zero, i.e., $A^\mathrm{T}+A=0$.

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+1 Nice concise and elementary answer that picks up where OP left it. –  Marc van Leeuwen Aug 5 '12 at 13:25
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