Can anyone give me an example of a normal matrix that isn't unitary? Here, $A \in \mathbb{C^{n \times n}}$.
124161
I have to find out a normal transformation that is neither hermitian nor unitary.  gives me the answer. However, I would like to know how to find it out mathematically, not just guess and test. Can I go from the properties of eigenvalues of hermitian and unitary to get the answer? For example: eigenvalues of a hemitian must be real, then I choose (i,-i,0) as eigenvalues of the required matrix. Those eigenvalues satisfy the condition that the required matrix is not unitary whose eigenvalues are |1|. From those, I have characteristic equation has the form of x(x^2+1)= 0 which leads to the form of the required matrix has the trace =0 and det = 0.  Can I do it?
95653
In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?  I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated
99453
Let $A$ be a normal matrix and $A^{247} = A^{246}$. Then I want to prove:  $A^2 = A$. $A$ is Hermitian.  I'm not sure on how to prove this. If $A$ is normal, then $A$ is unitarily diagonalizable. Then $A^{247}=A^{246}$ implies $U^\ast A^{247}U = U^\ast  A^{246}U$. Then if $D = U^\ast AU$, can I say that $D^{247}=D^{246}$ ? I don't know how to continue or if my thinking was correct. Please help.
90417
Given a matrix $A \in R^{n \times n}$ which is normal ($AA^H=A^HA$ where $A^H$ is hermitian of $A$) and nilpotent ($A^k=0$ for some $k$). Now we need to show that $A=0$. (This is essentially exercise 5(b) in sec. 80 on p.162 of Paul R. Halmos' Finite-Dimensional Vector Spaces.) I tried to show in the following way, we know that, $AA^H=A^HA$ pre-multiply by $A^{k-1}\implies A^kA^H=A^{k-1}A^HA$ Now, we have $0 = A^{k-1}A^HA$, since $A$ nilpotent. I am not sure how to proceed from here to show $A=0$. Can someone help me in this problem?
100014
Is it true that   $ \left( \begin{array}{ccc} A &amp; B \\ C &amp; D \\\end{array} \right)$  is non-invertible?  Assume that the matrix is over a field.
63150
$A$ is defined as an $m\times m$ matrix which is not invertible. How can i show that there is an $m\times m$ matrix $B$ where $AB = 0$ but $B$ is not equal to $0$?   For the solution of this question I think giving an example is not enough because it is too easy to solve this by giving an example, so how can I show that $B$ is not the $0$ matrix?
7464
Given a set of vectors $v_1 \cdots v_n$, the $n\times n$ Gram matrix $G$ is defined as $G_{i,j}=v_i \cdot v_j$  Due to symmetry in the dot product, $G$ is Hermitian.  I'm trying to remember why $|G|=0$ iff the set of vectors are not linearly independent.
103587
Let $A\in\mathbb{M}_n(\mathbb{C})$ be a matrix of order $n\geq 2$. Let $p_A(x) = (x-\lambda_1)\cdots (x-\lambda_n)$ be the characteristic polynomial of $A$ such that all $\lambda_i$ are positive real numbers.  Suppose that $A$ is a normal matrix. Prove that one can write $A=G^4$ for some self-adjoint matrix $G$.  My thinking is that since $A$ is normal, by Principal Axis Theorem, we have $A=UDU^\ast$, where $U$ is a unitary matrix and $D$ a diagonal matrix. However, I fail to see how to proceed. Could anyone help me out?
48440
Suppose the matrices $A$ and $B$ commute. Do there exists sequences $A_n$ and $B_n$ of matrices such that   $A_n \rightarrow A$, $B_n \rightarrow B$.  Each $A_n$ is diagonalizable  and the same for each $B_n$.  For every $n$, $A_n$ commutes with $B_n$.    Moreover, it would be nice if the following property was additionally satisfied: if $A,B$ are real, then $A_n,B_n$ can be chosen to be real as well.
85108
Let $A$ is a normal matrix: $A^*\! A = A A^*\!\!$,$\,$ and $AB = BA$. Prove that $A^*\!B=BA^*\!\!$.  I can prove that if $\det A\ne 0$ by multiplication $AB=BA$ by $A^*$ left and right and using some manipulation. But I have no idea what to do if $\det A = 0$.
121588
I have matrices $A,B$ of dimension $n$ with real coefficients which satisfy the following: $A^2-B^2=c(AB-BA)$ where $c$ is a real number. If $c\neq0$ , prove that $(AB-BA)^n = 0$.   So far, I've been able to show that $AB-BA$ is singular. Can someone help?
28670
I have problems to show that if $A$ is a $2 \times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.
92364
I'm reading Sheldon Axler's "Linear Algebra Done Right", and I have a question about one of the examples he gives on page 130. Let $T$ be a linear operator on $F^2$ whose matrix (with respect to the standard basis) is $$\begin{bmatrix} 2 &amp; -3 \\ 3 &amp; 2\end{bmatrix}$$ I can see why this operator is not self-adjoint, but I can't see why it is normal. I have to find $T^*$ and show that $TT^* = T^*T$. How would you do this?
89772
Can someone give me an example of noncomplete inner product space $H$, its closed linear subspace of $H_0$ and element $x\in H$ such that there is no orthogonal projection of $x$ on $H_0$. In other words I need to construct a counterexample to theorem about orthogonal projection when inner product space is not complete.
57037
This problem seems to be easy yet I have no idea to deal with.  If $\mathbb{M} ^n$ is the set of all real square matrices of order n, identified here with $\mathbb{R} ^{n^2}$ equipped with its usual Euclidean norm which induce an usual topology on the space $\mathbb{M}^n$. Show that:   The set of all invertible real matrices of order n is open in $\mathbb{M} ^n$ with respect to the Euclidean norm topology; The set $\mathbb{S} ^n$ which consists of all real symmetric matrices of order n is closed in $\mathbb{M} ^n$ with respect to the Euclidean norm topology.   For question #1, by the hint which is given by @Brent Kerby, that $det:\mathbb{M}^n \rightarrow \mathbb{R}$ is continuous.
17031
Let $A$ be a matrix in the special orthogonal group, $A \in \mathrm{SO}_n$.  This means that $A$ is real, $n \times n$, $A^t A = I$ and $\det(A)=1$, that is, the column vectors of $A$ make a positively-oriented orthonormal basis for $\mathbb R^n$. Decompose $A$ as a block matrix $$A = \begin{pmatrix} B &amp; C \\ D &amp; E\end{pmatrix}$$ where $B$ is $k \times k$ and $E$ is $(n-k)\times (n-k)$. I'm looking for a basic linear-algebra argument that $\det(B) = \det(E)$, ideally something that could be presented in a 2nd or 3rd year undergraduate course.  So I do not want people to invoke anything like tensor products or differential forms.
43480
Consider a normal operator on a complex Hilbert space with its spectrum contained in the real line. Show that the operator is hermitian without using the spectral theorem.  What I have tried- Since the spectrum is inside the real line, the boundary of the spectrum as a subset of the complex numbers is the spectrum itself. Hence the spectrum is equal to the approximate point spectrum. I have been unable to proceed further.