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Zero matrix

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.

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Isomorphic groups

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.

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Invertible matrix

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.

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Inequality of norm of matrix

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:

    \[\left \| A \right \|_2^2 \geq \sum_{i=1}^{n} \left| \ell_i \right|^2\]

Solution

Coming soon!

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Exponentiation of a matrix is normal

Give an example of a real matrix A that is not normal but e^A is. Exponentiation is to be taken with respect to the usual definition.

Solution

Coming soon!

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