2

I am trying to answer the end of the following problem:

enter image description here

Specifically: If $A^2=-I_{2n}$, then $A$ is similar to $\begin{bmatrix} 0& -I_n \\ I_n & 0 \end{bmatrix}$.

I tried the following: Applying $f$ in the given basis, we have:

$$\begin{array}{rcl} f(v_1) & = & 0 v_1 + \dots + 0 v_n + f(v_1)+ \dots + 0f(v_n) \\ & \vdots & \\ f(v_n) & = & 0 v_1 + \dots + 0 v_n+0f(v_1)+ \dots + f(v_n) \\ f(f(v_1)) & = & - v_1 + \dots + 0 v_n+0f(v_1)+ \dots + 0f(v_n)\\ & \vdots & \\ f(f(v_n)) & = & 0 v_1 + \dots + - v_n+0f(v_1)+ \dots + 0f(v_n)\\ \end{array}$$

Considering $A^t$ as the matrix of coefficients of the previous system we have that $A=\begin{bmatrix}0& -I_n \\ I_n & 0 \end{bmatrix}$. Applying $f$ again, we see that $A^2=-I_{2n}$. Here we can see that if we reorder the basis - say switching $v_i$ by $v_j$ - we produce a similarity between $A$ and the new matrix of coefficients $B$. Where $B$ is $A$ with rows $i,j$ exchanged and $i,j$ columns exchanged. This seems to suggest that whatever other matrices $B$ that obey $B^2=-I_{2n}$ can be obtained with $A$ by adequate elementary operations on rows and columns.

The trouble for me is the following: This construction where we obtained $A$ seems to rely on $f$ and the basis $\{v_1,\dots,v_n, f(v_1),\dots,f(v_n)\}$ how can we guarantee that there isn't a different linear transformation and a different basis where we obtain matrix $B$ such that $B^2=-I_{2n}$ but $B$ is not similar to $A$?

Ѕᴀᴀᴅ
  • 35,369
Red Banana
  • 24,885
  • 21
  • 101
  • 207
  • 1
    You have it backwards. Given $A$, you get a linear map $f_A : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$. What does your previous result tell you about the relationship between $A$ and the block matrix? – Michael Albanese Mar 20 '25 at 05:53
  • @MichaelAlbanese I guess I get it. We're need to use this theorem, right? – Red Banana Mar 20 '25 at 06:26
  • 1
    Actually, the converse of that theorem. – Michael Albanese Mar 20 '25 at 06:31
  • @MichaelAlbanese This one, right? But in this theorem, we have $f$ represented as $A$ relative to the ordered basis given in the problem. But couldn't there be an $A$ such that $A^2=-I_{2n}$ which is not represented by $f$ in any basis? – Red Banana Mar 20 '25 at 06:48
  • 1
    Yes, that theorem. As I said in my first comment, given $A$, you get a linear map $f_A : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ (namely $x \mapsto Ax$) which is represented by $A$ in the standard basis. – Michael Albanese Mar 20 '25 at 06:56
  • 1
    The big guns answer to your final question is: by the Rational Canonical Form Theorem, if a real $2n \times 2n$ matrix $C$ satisfies the equation $X^2+1$ then $C$ is similar to the direct sum of $n$ copies of $\begin{bmatrix} 0 & -1\1 &0\end{bmatrix}$. – ancient mathematician Mar 20 '25 at 08:01
  • @MichaelAlbanese I guess I get it, considering:

    $$M=\begin{bmatrix} 0&-I_n \ I_n& 0 \end{bmatrix}$$

    We have the map $f_M$ defined as $x\mapsto Mx$ which is represented as $M$ in the standard basis. Using theorem $7.5$, the matrix of this mapping relative to another basis will be $A=P^{-1}MP$ for some suitable $P$ and hence $A,M$ are similar. We know that $M^2=-I_{2n}$, we need to check if that is also the case for $A$, with a simple calculation, we see that:

    $$A^2=(P^{-1}MP)(P^{-1}MP) = P^{-1}M^2P=P^{-1}(-I_{2n})P=-I_{2n}$$

    Is this correct?

     – Red Banana Mar 20 '25 at 15:51
  • 1
    a different straightforward argument is: $A$ is similar to $U\in O_{2n}(\mathbb R)$ per https://math.stackexchange.com/questions/3792992/a-is-real-matrix-and-for-some-k-geq-2-ak-is-similar-to-an-orthogonal-matr/ [basically a finite group rep argument]. And $U^2=-I\implies U=-U^T$ and so $U$ is similar to $\begin{bmatrix}0& -I_n \ I_n & 0 \end{bmatrix}$ since skew orthogonal matrices are similar per https://math.stackexchange.com/questions/4556807/orthogonal-skew-symmetric-matrices-are-orthogonally-conjugate/ [or just run Real Schur (block) Triangularization on each] – user8675309 Mar 20 '25 at 16:36
  • 1
    @RedBanana: If you try to approach it that way, you would have to justify why there is a basis of $\mathbb{R}^{2n}$ in which the matrix of $f_M$ relative this basis is $A$, you haven't done that. Instead, you know by the exercise that there is a basis of $\mathbb{R}^{2n}$ in which the matrix of $f_A$ relative to this basis is $M$, so $M = P^{-1}AP$. – Michael Albanese Mar 20 '25 at 21:18
  • @MichaelAlbanese Sorry if I'm extremely confused, I'll try to formulate my doubt the best I can: $[1]$ We have a linear map $f$ such that $f^2=-id_v$, we also have the basis $B={v_1,\dots,v_n,f(v_1),\dots , f(v_n)}$ where $f$ represented in this basis is $M=\begin{bmatrix} 0&-I_n \ I_n& 0 \end{bmatrix}$. $[2]$ We also have an arbitrary matrix $A$ such that $A^2=-I_{2n}$. I understand the first part but I don't see how the second part enters into picture, from what you wrote it seems we can use theorem $7.5$ to obtain the relation $M=P^{-1}AP$. – Red Banana Mar 21 '25 at 23:22
  • @MichaelAlbanese In my mind, using theorem $7.5$ we get another matrix (say, H) representing $f$ in another basis and $M=P^{-1}HP$. I guess my doubt is: Given any $A$ such that $A^2=-I_{2n}$ why can we make a change of basis such that $H=A$? – Red Banana Mar 21 '25 at 23:22
  • 1
    Let me reformulate [1]. For any finite-dimensional real vector space $V$, and any choice of linear map $f : V \to V$ such that $f^2 = -\operatorname{id}_V$, there is a basis $B$ of $V$ in which $f$ is represented by $M$. In particular, for $V = \mathbb{R}^{2n}$ and $f = f_A$ where $f_A : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$, $x \mapsto Ax$, there is a basis $B$ of $\mathbb{R}^{2n}$ in which $f_A$ is represented by $M$. – Michael Albanese Mar 21 '25 at 23:32
  • @MichaelAlbanese I guess I understand it now: We have $f_A$ defined as $x\mapsto Ax$, when represented in the standard basis we get the matrix $A$. By the theorem $7.5$, we choose the basis $B={v_1,\dots,v_n,f(v_1)\dots,f(v_n)}$ and the matrix representing $f_A$ in this basis is $P^{-1}AP$. But $f$ is a mapping such that $f^2=-id$ and as we have seen, any map like this is represented in the basis $B$ by $M=\begin{bmatrix} 0&-I_n \ I_n& 0 \end{bmatrix}$ and then $M=P^{-1}AP$. Hence, $A$ must be similar to $M$. Is this correct? – Red Banana Mar 22 '25 at 05:36
  • 1
    Yes, that's it! – Michael Albanese Mar 22 '25 at 06:31

2 Answers2

2

Let $\{e_k\}_{k=1}^n$ be linearly independent elements and $f_k=Ae_k.$ As $A$ is invertible the elements $\{f_k\}_{k=1}^n$ are linearly independent. We may consider $A$ as acting on $\mathbb{C^{2n}}.$ We have $$A(e_k-if_k)=f_k+ie_k=i(e_k+if_k)\\ A(e_k+if_k)=f_k-ie_k=-i(e_k+if_k)$$ The collections $\{e_k-if_k\}_{k=1}^n$ and $\{e_k+if_k\}_{k=1}^n,$ correspond to different eigenvalues hence the set $\{e_k\pm if_k\}_{k=1}^n$ is linearly independent, hence form a basis in $\mathbb{C}^{2n}.$ Hence the set $$\{e_k\}_{k=1}^n\cup \{f_k\}_{k=1}^n\quad (*)$$ forms a basis in $\mathbb{C}^{2n}.$ As the vectors have real coefficients, the collection in $(*)$ forms a basis in $\mathbb{R}^{2n}.$

The formulas $Ae_k=f_k$ and $Af_k=-e_k$ yield that $A$ has the desired form with respect to the basis $\{f_k\}_{k=1}^n\cup\{e_k\}_{k=1}^n.$

Ryszard Szwarc
  • 44,675
1

This answer summarises my comments under the question. In particular, the method below establishes the similarity of $A$ and $\begin{bmatrix} 0 & -I_n\\ I_n & 0\end{bmatrix}$ using the previous parts of the problem.

You've already shown that for any finite-dimensional real vector space $V$, and any linear map $f : V \to V$ such that $f\circ f = -\operatorname{id}_V$, there is a basis of $V$ of the form $\mathcal{B} = \{v_1, \dots, v_n, f(v_1), \dots, f(v_n)\}$. The matrix of $f$ with respect to this basis is

$$M:= \begin{bmatrix} 0 & -I_n\\ I_n & 0\end{bmatrix}.$$

Now, given a matrix $2n\times 2n$ matrix $A$ such that $A^2 = -I_{2n}$, consider the linear map $f_A : \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ given by $f_A(v) = Av$. The matrix of $f_A$ with respect to the standard basis of $\mathbb{R}^{2n}$ is $A$.

Note that $f_A\circ f_A = -\operatorname{id}_{\mathbb{R}^{2n}}$ since $A^2 = -I_{2n}$, so by the above, there is a basis of $\mathbb{R}^{2n}$ of the form$\mathcal{B} = \{v_1, \dots, v_n, f_A(v_1), \dots, f_A(v_n)\} = \{v_1, \dots, v_n, Av_1, \dots, Av_n\}$. The matrix of $f_A$ with respect to the basis $\mathcal{B}$ is $M$.

The two matrix representations of $f_A$ are related by a change of basis matrix (see for example the theorem in this comment). Let $P$ be the unique matrix which satisfies $Pe_i = v_i$ for $i \in \{1, \dots, n\}$ and $Pe_i = Ae_{i-n}$ for $i \in \{n+1, \dots, 2n\}$, then $M = P^{-1}AP$ so $A$ is similar to $M$.

The converse statement (if $A$ and $M$ are similar, then $A^2 = -I_{2n}$) is immediate: $$A^2 = (PMP^{-1})^2 = PM^2P^{-1} = P(-I_{2n})P^{-1} = -PP^{-1} = -I_{2n}.$$

Michael Albanese
  • 104,029