## On idempotent matrices A, B such that A – B is invertible

Posted: May 27, 2019 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Recall that a matrix $A \in M_n(\mathbb{C})$ is said to be idempotent if $A^2=A.$

Problem. Let $A,B \in M_n(\mathbb{C})$ be two idempotent matrices such that $A-B$ is invertible and let $\alpha, \beta \in \mathbb{C}.$ Let $I \in M_n(\mathbb{C})$ be the identity matrix. Show that

i) if $\alpha \notin \{0,-1\},$ then $I+\alpha AB$ is not necessarily invertible

ii) if $\alpha \in \{0,-1\},$ then $I+\alpha AB$ is invertible

iii) $A+B-AB$ is invertible

iv) if $\alpha \beta \ne 0,$ then $\alpha A+\beta B$ is invertible.

Solution. i) Choose

$\displaystyle A=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \ \ \ \ \ B=\begin{pmatrix} -\frac{1}{\alpha} & -\frac{1}{\alpha} \\ 1+\frac{1}{\alpha} & 1+\frac{1}{\alpha}\end{pmatrix}.$

See that $A^2=A, \ B^2=B$ and $A-B$ is invertible but $\alpha AB+I$ is not invertible.

ii) It’s clear for $\alpha =0.$ For $\alpha =-1,$ suppose that $ABv=v$ for some $v \in \mathbb{C}^n.$ We need to show that $v=0.$ Well, we have $ABv=A^2Bv=Av$ and thus $Bv-v \in \ker A.$ But since $B^2=B,$ we also have $Bv -v \in \ker B$ and hence

$Bv -v \in \ker A \cap \ker B \subseteq \ker(A-B)=(0),$

because $A-B$ is invertible. So $Bv=v$ and therefore $Av=ABv=v.$ So $v \in \ker(A-B) =(0).$

iii) Let $A'=I-A, \ B'=I-B.$ See that $A',B'$ are idempotents and so, since $A'-B'=-(A-B)$ is invertible, $I-A'B'$ is invertible, by ii). The result now follows because

$I-A'B'=I-(I-A)(I-B)=A+B-AB.$

iv) Let $\displaystyle \gamma:=-\frac{\beta}{\alpha} \ne 0$ and suppose that $Av=\gamma Bv$ for some $v \in \mathbb{C}^n.$ We are done if we show that $v=0.$ Well, we have $BAv=\gamma Bv=Av=A^2v$ and thus $(A-B)Av=0$ implying that $Av=0$ because $A-B$ is invertible. Hence $\gamma Bv=Av=0,$ which gives $Bv=0$ because $\gamma \ne 0.$ Thus $(A-B)v=0$ and therefore $v=0. \ \Box$

## A finite p-subgroup of GL(n,k)

Posted: February 27, 2019 in Elementary Algebra; Problems & Solutions, Groups and Fields, Linear Algebra
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Problem. Let $k$ be a field, $n$ a positive integer. Let $G$ be a finite subgroup of $\text{GL}(n,k)$ such that $|G|>1$ and suppose also that every $g \in G$ is upper triangular and all the diagonal entries of $g$ are $1.$
Show that $\text{char}(k) > 0$ and $|G|$ is a power of $\text{char}(k).$

Solution. First, let’s put an order on the set

$S:=\{(i,j): \ \ 1 \le i < j \le n\}.$

We write $(i,j) < (i',j')$ if $i < i'$ or $i=i', \ j < j'.$ Now let $g=[g_{ij}]$ be any non-identity element of $G.$ Let $(r,s)$ be the smallest element of $S$ such that $g_{rs} \ne 0.$ See that, for any integer $m,$ the $(r,s)$-entry of $g^m$ is $mg_{rs}$ and the $(i,j)$-entries of $g^m,$ where $(i,j) \in S, \ (i,j) < (r,s),$ are all zero. But since $G$ is finite, there exists an integer $m > 1$ such that $g^m$ is the identity matrix and so we must have $mg_{rs}=0.$ Thus $m1_k=0$ (because $g_{rs} \ne 0$) and hence $p:=\text{char}(k) > 0.$ So the $(i,j)$-entries of $g^p,$ where $(i,j) \in S$ and $(i,j) \le (r,s),$ are all zero.
Now if $g^p$ is not the identity matrix, we can replace $g$ with $g^p$ and repeat our argument to find an element $(u,v) \in S, \ (u,v) > (r,s),$ such that all $(i,j)$-entries of $g^{p^2},$ where $(i,j) \in S, \ (i,j) \le (u,v),$ are zero. Then, again, if $g^{p^2}$ is not the identity matrix, we repeat the argument for $g^{p^2},$ etc. Since $g$ has only finitely many entries, there exists some positive integer $\ell$ such that all the $(i,j)$-entries of $g^{p^{\ell}},$ where $(i,j) \in S,$ are zero. That means $g^{p^{\ell}}$ is the identity matrix and hence $|g|$ is a power of $p.$ So we have shown that the order of every non-identity element of $G$ is a power of $p.$ Thus $|G|$ is a power of $p. \ \Box$

## An application of linear algebra in Calculus

Posted: September 16, 2018 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Definition. For an integer $n \ge 1,$ the $n\times n$ Hilbert matrix is defined by $H_n=[a_{ij}],$ where

$\displaystyle a_{ij}=\frac{1}{i+j-1}, \ \ 1 \le i,j \le n.$

It is known that $H_n$ is invertible and if $H_n^{-1}=[b_{ij}],$ then $\displaystyle \sum_{i,j}b_{ij}=n^2.$ We are going to use these two properties of Hilbert matrices to solve the following calculus problem.

Problem. Let $n \ge 1$ be an integer and let $f : [0,1] \longrightarrow \mathbb{R}$ be a continuous function. Suppose that $\displaystyle \int_0^1 x^kf(x) \ dx = 1$ for all  $0 \le k \le n-1.$ Show that $\displaystyle \int_0^1 (f(x))^2 dx \ge n^2.$

Solution. Since $H_n,$ the $n\times n$ Hilbert matrix, is invertible, there exist real numbers $p_0, p_1, \cdots , p_{n-1}$ such that

$\displaystyle \sum_{i=1}^n\frac{p_{i-1}}{i+j-1}=1, \ \ \ 1 \le j \le n.$

So the polynomial $\displaystyle p(x)=\sum_{k=0}^{n-1}p_kx^k$ satisfies the conditions

$\displaystyle \int_0^1x^k p(x) \ dx =1, \ \ \ 0 \le k \le n-1.$

Clearly $\displaystyle \sum_{k=0}^{n-1}p_k$ is the sum of all the entries of $H_n^{-1}$ and so $\displaystyle \sum_{k=0}^{n-1}p_k=n^2.$ Now let $f$ be a real-valued continuous function on $[0,1]$ such that

$\displaystyle \int_0^1x^kf(x) \ dx = 1, \ \ \ 0 \le k \le n-1.$

Let $p(x)$ be the above polynomial.Then since

$\displaystyle (f(x))^2-2f(x)p(x)+(p(x))^2 =(f(x)-p(x))^2 \ge 0,$

integrating gives

\displaystyle \begin{aligned} \int_0^1 (f(x))^2dx \ge 2\int_0^1f(x)p(x) \ dx -\int_0^1(p(x))^2dx=2\sum_{k=0}^{n-1}p_k \int_0^1 x^kf(x) \ dx - \\ \sum_{k=0}^{n-1}p_k\int_0^1x^kp(x) \ dx = 2\sum_{k=0}^{n-1}p_k-\sum_{k=0}^{n-1}p_k=\sum_{k=0}^{n-1}p_k =n^2. \ \Box \end{aligned}

## A property of 3×3 orthogonal matrices with determinant -1

Posted: May 22, 2018 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Problem. Let $A \in M_3(\mathbb{R})$ be orthogonal and suppose that $\det(A)=-1.$ Find $\det(A-I).$

Solution. Since $A$ is orthogonal, its eigenvalues have absolute value $1$ and it can be be diagonalized. Let $D$ be a diagonal matrix such that $PDP^{-1}=A$ for some invertible matrix $P \in M_3(\mathbb{C}).$ Then

$\det(D)=\det(A)=-1, \ \ \det(D-I)=\det(A-I).$

We claim that the eigenvalues of $A$ are $\{-1,e^{i\theta},e^{-i\theta}\}$ for some $\theta.$ Well, the characteristic polynomial of $A$ has degree three and so it has either three real roots or only one real root. Also, the complex conjugate of a root of a polynomial with real coefficients is also a root. So, since $\det(A)=-1,$ the eigenvalues of $A$ are either all $-1,$ which is the case $\theta=\pi,$ or two of them are $1$ and one is $-1,$ which is the case $\theta = 0,$ or one is $-1$ and the other two are in the form $\{e^{i\theta}, e^{-i\theta}\}$ for some $\theta.$ So

\displaystyle \begin{aligned} \det(D-I)=\det(A-I)=(-2)(e^{i\theta}-1)(e^{-i\theta}-1)=-4(1-\cos \theta). \ \Box \end{aligned}

Note that given $\theta,$ the matrix

$\displaystyle A=\begin{pmatrix} -1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{pmatrix}$

is orthogonal, $\det(A)=-1$ and $\det(A-I)=-4(1-\cos \theta). \ \Box$

## Limit of a sequence of 2 by 2 matrices

Posted: January 29, 2018 in Elementary Algebra; Problems & Solutions, Linear Algebra
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The following problem is from the American Mathematical Monthly. The problem only asks the reader to calculate $\displaystyle \lim_{n\to\infty} A_n^n,$ it doesn’t give the answer; I added the answer myself.

Problem (Furdui, Romania). Let $a,b,c,d$ be real numbers with $bc > 0.$ For an integer $n \ge 1,$ let

$A_n:=\begin{bmatrix} \cos \left(\frac{a}{n}\right) & \sin \left(\frac{b}{n}\right) \\ \\ \sin \left(\frac{c}{n}\right) & \cos \left(\frac{d}{n}\right) \end{bmatrix}.$

Let $\text{sgn(x)}$ be the sign function. Show that

$\displaystyle \lim_{n\to\infty} A_n^n = \begin{bmatrix} \cosh(\sqrt{bc}) & \text{sgn}(b)\sqrt{\frac{b}{c}} \sinh(\sqrt{bc}) \\ \\ \text{sgn}(b)\sqrt{\frac{c}{b}} \sinh(\sqrt{bc}) & \cosh(\sqrt{bc}) \end{bmatrix}.$

Solution. The characteristic polynomial of $A_n$ is

$\displaystyle p_n(x)=x^2-\left(\cos \left(\frac{a}{n}\right)+\cos\left(\frac{d}{n}\right) \right)x+\cos\left(\frac{a}{n}\right)\cos\left(\frac{d}{n}\right)-\sin \left(\frac{b}{n} \right)\sin \left(\frac{c}{n} \right).$

And roots of $p_n(x)$ are

$\displaystyle r_n=\frac{\cos \left(\frac{a}{n}\right)+\cos \left(\frac{d}{n}\right) + \sqrt{\Delta_n}}{2}, \ \ s_n=\frac{\cos \left(\frac{a}{n}\right) + \cos \left(\frac{d}{n}\right) - \sqrt{\Delta_n}}{2},$

where

$\displaystyle \Delta_n:=\left(\cos \left(\frac{a}{n}\right)-\cos \left(\frac{d}{n}\right)\right)^2+4 \sin \left(\frac{b}{n} \right) \sin \left(\frac{c}{n}\right).$

If $n$ is sufficiently large, which is the case we are interested in, then, since $b,c$ are either both positive or both negative (because $bc > 0$), $\displaystyle \sin \left(\frac{b}{n} \right) \sin \left(\frac{c}{n}\right) > 0$ and so $\Delta_n > 0.$ So, in this case, $r_n,s_n$ are distinct real numbers and hence $A_n$ is diagonalizable in $\mathbb{R}.$
Now $v \in \mathbb{R}^2$ is an eigenvector corresponding to $r_n$ if and only if $(A_n-r_nI)v=0$ if and only if $\displaystyle v \in \mathbb{R} \begin{bmatrix} 1 \\ \frac{r_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} \end{bmatrix}.$ Similarly, $v \in \mathbb{R}^2$ is an eigenvector corresponding to $s_n$ if and only if $(A_n-s_nI)v=0$ if and only if $\displaystyle v \in \mathbb{R} \begin{bmatrix} 1 \\ \frac{s_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} \end{bmatrix}.$ So if

$\displaystyle P:=\begin{bmatrix} 1 & 1 \\ \frac{r_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} & \frac{s_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} \end{bmatrix}, \ D:=\begin{bmatrix} r_n & 0 \\ 0 & s_n \end{bmatrix},$

then $A_n=PDP^{-1}$ and hence

$\displaystyle A_n^n=PD^nP^{-1}=\frac{\sin \left(\frac{b}{n}\right)}{s_n-r_n}\begin{bmatrix} 1 & 1 \\ \frac{r_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} & \frac{s_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} \end{bmatrix} \begin{bmatrix}r_n^n & 0 \\ 0 & s_n^n \end{bmatrix}\begin{bmatrix} \frac{s_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)} & -1 \\ \frac{\cos \left(\frac{a}{n}\right)-r_n}{\sin \left(\frac{b}{n}\right)} & 1 \end{bmatrix}.$

The rest of the solution is just Calculus and if you have trouble finding limits, see this post in my Calculus blog for details. We have

$\displaystyle \lim_{n\to\infty} \frac{r_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)}=-\lim_{n\to\infty} \frac{s_n-\cos \left(\frac{a}{n}\right)}{\sin \left(\frac{b}{n}\right)}=\text{sgn}(b)\sqrt{\frac{c}{b}}, \\ \lim_{n\to\infty} \frac{\sin \left(\frac{b}{n}\right)}{r_n-s_n}=\frac{\text{sgn}(b)}{2}\sqrt{\frac{b}{c}}, \lim_{n\to\infty}r_n^n=e^{\sqrt{bc}}, \ \ \lim_{n\to\infty}s_n^n =e^{-\sqrt{bc}}. \ \Box$

## Rank of Hermitian matrices

Posted: October 2, 2012 in Elementary Algebra; Problems & Solutions, Linear Algebra
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For $a \in \mathbb{C}$ let $\overline{a}$ denote the complex conjugate of $a.$ Recall that a matrix $[a_{ij}] \in M_n(\mathbb{C})$ is called Hermitian if $a_{ij}=\overline{a_{ji}},$ for all $1 \leq i,j \leq n.$ It is known that if $A$ is Hermitian, then $A$ is diagonalizable and every eigenvalue of $A$ is a real number. In this post, we give a lower bound for the rank of a Hermitian matrix. To find the lower bound, we first need an easy inequality.

Problem 1. Prove that if $a_1, \ldots , a_m \in \mathbb{R},$ then $(a_1 + \ldots + a_m)^2 \leq m(a_1^2 + \ldots + a_m^2).$

Solution.  We have $a^2+b^2 \geq 2ab$ for all $a,b \in \mathbb{R}$ and so

$(m-1)\sum_{i=1}^m a_i^2=\sum_{1 \leq i < j \leq m}(a_i^2+a_j^2) \geq \sum_{1 \leq i < j \leq m}2a_ia_j.$

Adding the term $\sum_{i=1}^m a_i^2$ to both sides of the above inequality will finish the job. $\Box$

Problem 2. Prove that if $0 \neq A \in M_n(\mathbb{C})$ is Hermitian, then ${\rm{rank}}(A) \geq ({\rm{tr}}(A))^2/{\rm{tr}}(A^2).$

Solution. Let $\lambda_1, \ldots , \lambda_m$ be the nonzero eigenvalues of $A.$ Since $A$ is diagonalizable, we have ${\rm{rank}}(A)=m.$ We also have ${\rm{tr}}(A)=\lambda_1 + \ldots + \lambda_m$ and ${\rm{tr}}(A^2)=\lambda_1^2 + \ldots + \lambda_m^2.$ Thus, by Problem 1,

$({\rm{tr}}(A))^2 \leq {\rm{rank}}(A) {\rm{tr}}(A^2)$

and the result follows. $\Box$

## A limit involving unitary matrices

Posted: April 17, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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Throughout $n \geq 1$ is an ineteger, $V = \mathbb{C}^n,$ and $A$ is an $n \times n$ matrix with entries in $\mathbb{C}.$

Definition 1. The complex conjugate of $A=[a_{ij}]$ is the matrix $\overline{A}=[\overline{a_{ij}}],$ where $\overline{a_{ij}}$ is the complex conjugate of $a_{ij}.$ Let $A^*$ be the transpose of $\overline{A}.$

Remark 1. Recall that the standard inner product over the $\mathbb{C}$-vector space $V = \mathbb{C}^n$ is defined by $\langle u,v \rangle = \sum_{i=1}^n u_i \overline{v_i},$ for all vectors $u,v \in V,$ where $u_i,v_i$ are the $i$-th entry of $u,v$ respectively.

Remark 2. A direct and easy computation shows that $\langle Au , v \rangle = \langle u , A^*v \rangle,$ for all $u,v \in V.$ Since obviously $(A^*)^*=A,$ we also have $\langle A^*u, v \rangle = \langle u, Av \rangle,$ for all $u, v \in V.$

Lemma. Let $T$ and $T^*$ be the linear transformations corresponding  to $A$ and $A^*$ respectively, i.e. $T(v) = Av$ and $T^*(v)=A^*v,$ for all $v \in V = \mathbb{C}^n.$ Then $\text{Im}(T) \oplus \ker(T^*)=V.$

Proof. We first remind the reader a couple of things from elementary linear algebra: if $W$ is a subspace of $V,$ then $W^{\perp} = \{v \in V : \ \langle v, w \rangle = 0, \ \forall w \in W \},$ where the inner product is the one defined in Remark 1. Then $W^{\perp}$ is a subspace of $V$ and $W \oplus W^{\perp} = V.$ So, to prove the lemma, we let $W = \text{Im}(T)$ and we will show that $W^{\perp} = \ker(T^*).$ This is easy to do: $v \in W^{\perp}$ if and only if $\langle v,w \rangle = 0,$ for all $w \in W.$ So, since $W=\{ T(u): \ u \in V \},$ we have $v \in W^{\perp}$ if and only if $\langle v, T(u) \rangle = 0,$ for all $u \in V.$ But, by Remark 2, we have $\langle v, T(u) \rangle = \langle T^*(v), u \rangle.$ So $v \in W^{\perp}$ if and only if $\langle T^*(v), u \rangle = 0,$ for all $u \in V.$ Thus $v \in W^{\perp}$ if and only if $T^*(v) = 0,$ i.e. $v \in \ker(T^*). \ \Box$

Definition 2. If $A$ is invertible and $A^{-1}=A^*,$ then $A$ is called unitary.

Remark 3. If $A$ is unitary, then $||Au|| =||u||,$ for all $u \in V = \mathbb{C}^n.$ The reason is that, by Remark 2 and the above definition, we have $||Au||^2=\langle Au, Au \rangle = \langle u, A^*Au \rangle = \langle u,u \rangle = ||u||^2.$ Using induction, we get $||A^iu||=||u||,$ for all integers $i$ and all $u \in V.$

Problem. Suppose that $A$ is unitary and $v \in V = \mathbb{C}^n.$ Evaluate $L=\displaystyle \lim_{m \to \infty} \frac{1}{m} \sum_{i=1}^m A^iv.$

Solution. Let $T$ and $T^*$ be the linear transformations corresponding to $A$ and $A^*$ respectively and let $I$ be the identity matrix. Apply the above lemma to the matrix $I - A$ to get $\text{Im}(\text{id}-T) \oplus \ker(\text{id}-T^*)=V,$ where $\text{id}$ is the identity transformation. So there exist (unique) $v_1 \in \text{Im}(\text{id}-T)$ and $v_2 \in \ker(\text{id}-T^*)$ such that $v=v_1+v_2.$  Let

$\displaystyle L_j = \lim_{m\to\infty} \frac{1}{m} \sum_{i=1}^m A^i v_j, \ j=1,2.$

Then $L=L_1+L_2.$ Let’s find $L_1$ first. Since $v_1 \in \text{Im}(\text{id}-T),$ we have $v_1 = u - T(u)=u-Au,$ for some $u \in V.$ Thus $\sum_{i=1}^m A^iv_1 = Au-A^{m+1}u$ and so, by Remark 3,

$|| \sum_{i=1}^m A^i v_1|| = ||Au - A^{m+1}u|| \leq ||Au|| + ||A^{m+1}u||=2||u||.$

Hence

$\displaystyle || \frac{1}{m} \sum_{i=1}^m A^iv_1 || \leq \frac{2||u||}{m}$

and therefore $L_1=0.$ We are now going to find $L_2.$ Since $v_2 \in \ker(\text{id}-T^*),$ we have $T^*(v_2)=v_2.$ Thus $A^{-1}v_2=A^*v_2=v_2$ and so $Av_2=v_2.$ Hence $A^iv_2=v_2$ for all integers $i$ and so $\displaystyle \frac{1}{m} \sum_{i=1}^m A^iv_2 = v_2.$ Thus $L_2=v_2$ and hence $L=L_1+L_2=v_2. \ \Box$