Joy of Mathematics

Zero matrix

August 18, 2019 / Leave a comment

Let $A\in \mathcal{M}_n\left ( \mathbb{C} \right )$ with $n\geq 2$ . If $\det \left ( A+X \right )=\det A+\det X$ for every matrix $X \in \mathcal{M}_n\left ( \mathbb{C} \right )$ then prove that $A=\mathbb{O}_{n}$ .

Solution

Suppose that $A \neq 0$ , say $A_{ij} \neq 0$ for some $i,j$ . Let $P$ be any permutation matrix with $P_{ij}=1$ and let $Q$ be the matrix obtained from $P$ by changing its $ij$ -entry to $0$ . Finally let $X = xQ$ where $x \in \mathbb{C}$ .

We have that $\det X = 0$ and that $\det X = \det(A+X) - \det A$ is a polynomial in $x$ . Furthermore, the coefficient of $x^{n-1}$ of this polynomial is $\pm A_{ij}$ depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.

Isomorphic groups

August 18, 2019 / Leave a comment

Let $n>2$ . Define the group

$\mathcal{Q}_{2^n} = \langle x, y \mid x^2=y^{2^{n-2}} , y^{2^{n-1}} = 1, x^{-1} yx =y^{-1} \rangle$

Prove that $\displaystyle \mathcal{Q}_{2^n} / \mathcal{Z} \left ( \mathcal{Q}_{2^n} \right ) \simeq \mathcal{D}_{2^{n-1}}$ where $\mathcal{D}$ is the dihedral group.

Solution

Using $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$ . Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$ . Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n-1}-1\}$ . It is easy to prove inductively that $y^tx = xy^{-t}$ .

Let $\mathcal{Z} = \mathcal{Z}(\mathcal{Q}_{2^n})$ . We prove that $\mathcal{Z}= \{1,y^{2^{n-2}}\}$ . Obviously $1 \in \mathcal{Z}$ . Furthermore, $y^{2^{n-2}} \in \mathcal{Z}$ since

$y^{2^{n-2}}\left(x^ry^s\right) = xy^{-2^{n-2}}x^{r-1}y^s = \cdots = x^ry^{2^{n-2}}y^s =\left(x^ry^s\right)y^{2^{n-2}}$

If $y^k \in \mathcal{Z}$ (such that $0 \leq k < 2^{n-1}$ ) then $xy^k = y^kx = xy^{-k}$ hence $y^{2k} = 1$ and therefore $k = 0$ or $k = 2^{n-2}$ . If $xy^k \in \mathcal{Z}$ then $xy^{k+1} = yxy^{k} = xy^{k-1}$ hence $y^2 = 1$ which is a contradiction since $n \geq 2$ .

Therefore,

$\mathcal{Q}_{2^n}/\mathcal{Z} = \langle x,y| x^2 = y^{2^{n-2}}=1, yx = xy^{-1} \rangle$

which is precisely the dihedral group with $2^{n-1}$ elements.

Invertible matrix

August 15, 2019 / Leave a comment

Consider the matrices $A \in \mathcal{M}_{m \times n}$ and $B \in \mathcal{M}_{n \times m}$ . If $AB +\mathbb{I}_m$ is invertible prove that $BA+\mathbb{I}_n$ is also invertible.

Solution

So we have to answer the question if $-1$ is a zero of the essentially same characteristic polynomials. $AB$ and $BA$ have quite similar characteristic polynomials. In fact if $p(x)$ denotes the polynomial of $AB$ , then the polynomial of $BA$ will be $q(x)= x^{n-m} p(x)$ . It is easy to see that $-1$ cannot be an eigenvalue of the $AB$ matrix, otherwise it wouldn’t be invertible. Now, let us assume that $BA$ is not invertible. Then it must have an eigenvalue of $-1$ and let $\mathbf{x}$ be the corresponding eigenvector. Hence:

$\left ( BA \right )\mathbf{x}= -\mathbf{x} \Rightarrow AB \left ( A \mathbf{x} \right )= -A\mathbf{x}$

meaning that $AB$ has an eigenvalue of $-1$ which is a contradiction. The result follows.

Inequality of norm of matrix

August 15, 2019 / Leave a comment

Let $\left \| \cdot \right \|_2$ denote $\displaystyle \left \| A \right \|_2 = \sqrt{\sum_{i=1}^{n}\sum_{j=1}^{n}\left | a_{ij} \right |^2}$ . Consider $A \in \mathcal{M}_{n \times n}(\mathbb{C})$ and let $\ell_1, \dots, \ell_n$ be its eigenvalues. Prove that: