5
$\begingroup$

Let $A$ be an $n \times n$ matrix with real entries such that $A^2 + I = 0$ then $n$ is even. And if $n = 2k$, then $A$ is similar over the field of real numbers to a matrix of the block form

$$\begin{bmatrix} 0 & -I \\ I & 0 \\ \end{bmatrix}$$ where $I$ is the $k \times k$ identity matrix.

I have done the first part. Since $\det(A^2) = \det (-I)$ we must have $\det (-I)$ non negative and hence $n$ must be even.

Need Help in the second part.

Thank You.

$\endgroup$
  • $\begingroup$ I think that $A^2 + I = 0$, eigenvalues of $A$ are $\pm i$. Since $A$ is an $n \times n$ real matrix, $n$ is even. $\endgroup$ – bing Nov 10 '16 at 8:07
6
$\begingroup$

You can solve both parts of the question at the same time as follows: Given $A$, define a map $\mathbb C\times\mathbb R^n\to\mathbb R^n$ by $(u+iv,x)\mapsto (u+iv)\cdot x:=ux+vAx$. Evidently, this is linear over $\mathbb R$ in both components and satisfies $(1,x)\mapsto x$. Using that $A^2=-I$, one easily verifies that it is multiplicative, i.e. $z\cdot (w\cdot x)=(zw)\cdot x$. Hence is satisfies all properties required from scalar multiplication, and hence makes $\mathbb R^n$ into a complex vector space, so $n=2m$ for some $m\in\mathbb N$. Moreover, choosing a complex basis $\{x_1,\dots,x_m\}$ for this vector space, $\{x_1,\dots,x_m,i\cdot x_1,\dots,i\cdot x_m\}$ is a real basis for $\mathbb R^n$. By definition the matrix representation of $A$ with respect to this real basis is the block matrix you have indicated.

$\endgroup$
0
$\begingroup$

This is Exercise 16 of Section 7.2, "Cyclic Decomposition and the Rational Form," in Linear Algebra, second edition, by Hoffman and Kunze, so let us try to answer the question using the cyclic decomposition theorem, Theorem 3 on page 233.

The minimal polynomial for $A$ is $x^2 + 1$ or one of its factors, but over real numbers, the only factors are $x^2 + 1$ and $1$, so the minimal polynomial for $A$ is indeed $x^2 + 1$. In Theorem 3 with $W_0 = \{0\}$, $x^2 + 1$ is also the first annihilator $p_1$, and the remaining annihilators $p_2, \dots, p_r$ divide $p_1 = x^2 + 1$, so they are each $x^2 + 1$. Now by Theorem 4 on page 237, the characteristic equation for $A$ is the product of the annihilators $p_1 p_2 \cdots p_r = (x^2 + 1)^r = x^{2r} + \cdots$. Thus $n$ is even because $n$ is equal to the degree of the characteristic equation.

$\endgroup$
-1
$\begingroup$

I solve it another method : $det(A^2)=det(-I)=(-1)^n ....(1)$.

Now $\lambda$ is an eigenvalue of $A$ then ${\lambda}^2+I^2=0\Rightarrow \lambda =i,-i.$ So $detA=i(-i)=1\Rightarrow det(A^2)=(detA)^2=1 ....(2)$

So from $(1)$ and $(2)$ we get $n=2m$ for dome $m\in \mathbb{N} $

$\endgroup$
  • $\begingroup$ plz specify reason for downvote..(for any improvement) $\endgroup$ – RAM_3R Apr 21 '18 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.