Suppose:    ${A_j} \in {\mathbb{C}^{n \times n}},0&lt;{w_j}\in \mathbb{R} (j = 0,1,2....m)$ ${\rm{P(}}\lambda {\rm{) = }}{{\rm{A}}_m}{\lambda ^m} + .....{A_1}\lambda  + {A_0}$ is a matrix polynomial,  and $\lambda $ is a complex variable. ${\rm{q(}}\lambda {\rm{) = }}{{\rm{w}}_m}{\lambda ^m} + .....{w_1}\lambda  + {w_0}$ ${\sigma _{\varepsilon ,w}}(P) = \left\{ {\lambda  \in \mathbb{C}:\left\| {P{{(\lambda )}^{ - 1}}} \right\| \ge {{(\varepsilon q(\left| \lambda  \right|))}^{ - 1}}} \right\}$, where $\left\| . \right\|$ is any subordinate matrix norm.   Why does boundary of ${\sigma _{\varepsilon ,w}}(P)$ can be written in the form $S = \left\{ {\lambda  \in \mathbb{C}:\left\| {P{{(\lambda )}^{ - 1}}} \right\| = {{(\varepsilon q(\left| \lambda  \right|))}^{ - 1}}} \right\}$.
122355
Suppose:    $B_i \in \mathbb{C}^{n \times n}$, $0&lt;w_i\in \mathbb{R}$ $(i = 0,1,2,\ldots,m)$ ${\rm P}(x) ={\rm{B}_m} x ^m + \cdots + B_1 x  + A_0$ is a matrix polynomial,  and $x $ is a complex variable. ${\rm q}(x) =\rm{w}_m x^m + \cdots +w_1 x + w_0$ $U= \left\{ {x  \in \mathbb{C}:\left\| P(x)^{-1} \right\| \ge (\varepsilon q(|x|))}^{-1} \right\}$, where $\left\| \cdot \right\|$ is any subordinate matrix norm.   Why can the boundary of $U$ be written in the form $S = \left\{ x  \in \mathbb{C}:\left\| P(x )^{-1} \right\| = (\varepsilon q(|x|))^{-1} \right\}$?
28416
Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$?     deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field.  In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the  $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies  $$M \text{adj}(M) = \det(M).$$  If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus:     If $M$ is a counterexample, then $\text{adj}(M) = 0$.   When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case.   Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as $$M^k q(M) = 0$$  where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term.  Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence     If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent.    This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that $$q(M) = M - \lambda$$  where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case.   Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so     We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$.
61235
The following comes from page 3 of the paper &quot;Matrix Concentration for Products&quot; found here . Given two independent, complex, square matrices Y and Z: $$ YZ=\mathbb{E}(Y)Z+(Y-\mathbb{E}(Y))Z.$$ I understand that $$ \mathbb{E}((Y-\mathbb{E}(Y))Z|Z)=0. $$ Now, this is called an orthogonality property (Why?) and apparently used to show: $$ \mathbb{E}(\left\| YZ\right\|_2^2) =\mathbb{E}(\left\|(\mathbb{E}(Y))Z\right\|_2^2) + \mathbb{E}(\left\| (Y-\mathbb{E}(Y))Z\right\|_2^2) $$ where $\left\| .\right\|_2$ refers to the Schatten 2-norm or Frobenius-norm. Could someone clarify how exactly this property is used here? I assume the tower property comes into play here but I'm not sure how exactly. Thank you.
83113
I'm facing a problem finding a Jordan basis for this ($4 \times 4$) matrix: $$\left(\begin{matrix}3&amp;-1&amp;1&amp;7\\9&amp;-3&amp;-7&amp;-1\\0&amp;0&amp;4&amp;-8\\0&amp;0&amp;2&amp;-4\end{matrix}\right)$$  I know that the characteristic polynomial is $\lambda^4=0$ and it's minimal polynomial is: $x^2$. This means that $(A-0I)^2=0$. Also the Jordan form for this matrix is $$\left(\begin{matrix}0&amp;1&amp;0&amp;0\\0&amp;0&amp;1&amp;0\\0&amp;0&amp;0&amp;1\\0&amp;0&amp;0&amp;0\end{matrix}\right)$$ But I'm facing problems in the general procedure to compute the basis for such a matrix?
26879
I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.  $$\phi:a+bi \rightarrow \begin{pmatrix} a &amp; -b \\ b &amp; a \end{pmatrix}$$  This is what I have so far:  Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.  $$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$  So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.  Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$.   Surjective:  I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.  I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated!
42341
Theorem 13.9 in these  is the polynomial spectral mapping theorem:     Theorem 13.9 For a polynomial $p$ we have $\sigma(p(T)) = p(\sigma(T))$.     My questions:   First direction: Why would $q(T) (p(T) - p(\lambda) I)$ be an inverse for $T - \lambda I$? Update: @RobertIsrael said that instead $(p(T) - p(\lambda) I)^{-1} q(T)$ should be $(p(T) - p(\lambda) I)^{-1}$ be? According to the lines in the proof above, we have $$ (p(T) - p(\lambda) I)^{-1} = ((T - \lambda I) q(T))^{-1} \overset{?!}{=} q(T)^{-1} (T - \lambda I)^{-1}, $$ but as $\lambda \in \sigma(T)$ by assumption, $(T - \lambda I)^{-1}$ doesn't exist, right? For the second direction: Why does $x_i \in \sigma(T)$ imply that $p(x_i) - \mu = 0$? I thought that $x_i \in \sigma(T)$ only implies that $T - x_i I$ is not boundedly invertible. Since $\mu \in \sigma(p(T))$ we have that $p(T) - \mu I = c \prod_{k = 1}^{n} (T - x_i I)$ is not boundedly invertible but how does that help?   In both statements there seem to be a notion of "a product of operators is not invertible iff all factors aren't invertible", but how can that be true? Consider i.e. shift operators on a sequence spaces such that $ST = I$, i.e. the left and right shift operators, then the above mentioned notion doesn't hold, right?
81852
This question is in relation to the space $C(I)$, $I = [a, b]$. Define $P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}$ (any or all $a_i$ could be zero); clearly $P_n \subset C(I)$. The norm I'm using is $\lVert f\rVert_I = \sup_I |f(x)|$. Is $P_n$ closed under $\lVert\cdot\rVert_I$?   I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject  $P_n$ to $\mathbb{R}^{n+1}$, using coefficients as coordinates, and prove that sequences of degree-$n$ polynomials converge to degree-$n$ polynomials, but I can't prove that the metric $\lVert\cdot\rVert_I$ is equivalent to the standard metric on $\mathbb{R}^{n+1}$. Next, intuitively given a function $f \in C(I), \notin P_n$ I should be able to find some $\epsilon &gt; 0$ such that there are no low-degree polynomials "nearby", then that function was not a limit point of $P_n$ so $P_n$ is closed. Again I have no idea how to prove this. What should I do?
23108
This an exercise out of Spivak's "Calculus on Manifolds".  Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.  Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by  $$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.  A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if $\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.  The exercise as stated:  Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.  Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.  I'm having problems with the $(\Rightarrow)$ direction.  My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ &amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ &amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ &amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\ &amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*}  Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.  Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.  Thanks for any help you can give.
20860
I was supposed to find all possible Jordan canonical forms of a $5\times 5$ complex matrix with minimal polynomial $(x-2)^2(x-1)$ on a qualifying exam last semester.  I took the polynomial to mean that there were at least two 2's and one 1 on the main diagonal, and that the largest Jordan block with eigenvalue 2 is $2\times 2$ while the largest Jordan block with eigenvalue 1 is $1\times 1$.  Did I miss any matrices or interrupt the minimal polynomial incorrectly?  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;2  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;2  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;2  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;1  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;1  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}  \begin{pmatrix}   2 &amp;0  &amp;0  &amp;0  &amp;0\\   0 &amp;2  &amp;0  &amp;0  &amp;0\\   0 &amp;0  &amp;1  &amp;0  &amp;0\\   0 &amp;0  &amp;0  &amp;1  &amp;0\\   0 &amp;0  &amp;0  &amp;0  &amp;1 \end{pmatrix}
63340
For a given $n \times n$-matrix $A$, and $J\subseteq\{1,...,n\}$ let us denote by $A[J]$ its principal minor formed by the columns and rows with indices from $J$.  If the characteristic polynomial of $A$ is $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, then why $$a_k=(-1)^{n-k}\sum_{|J|=n-k}A[J],$$ that is, why is each coefficient the sum of the appropriately sized principal minors of $A$?
87268
Our teacher didn't explain us how to find it so I've had to look up a bit by myself.  I have this matrix :  $$A = \begin{pmatrix} 9 &amp; 4 &amp; 5 \\ -4 &amp; 0 &amp; -3 \\ -6 &amp; -4 &amp; -2 \end{pmatrix}$$  I've found its characteristic polynomial $p_A(\lambda) = -(\lambda -3)(\lambda - 2)^2$.  I've found its eigenspaces : $E_2 = span(\begin{pmatrix}-2 &amp; 1 &amp; 2\end{pmatrix}^T)$, $E_3 = span(\begin{pmatrix}-3 &amp; 2 &amp; 2\end{pmatrix}^T)$.  And I've found the generalized eigenspace of $\lambda = 2$ :  $\ker((A-2I_3)^2) = span(\begin{pmatrix}0 &amp; 1 &amp; 0\end{pmatrix}^T, \begin{pmatrix}1 &amp; 0 &amp; -1\end{pmatrix}^T)$  But I don't know what to do now. I've tried to see what result I would get with $$M = \begin{pmatrix} -3 &amp; -2 &amp; 0 \\ 2 &amp; 1 &amp; 1 \\ 2 &amp; 2 &amp; 0 \end{pmatrix}.$$  But $$M^{-1}AM = \begin{pmatrix} 3 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; -2 \\ 0 &amp; 0 &amp; 2 \end{pmatrix}.$$  So there it is. Any help ?
35946
Suppose that $T\in \mathcal L (V)$. prove that T is normal iff there exists a polynomal $f(x) \in \Bbb C [x] $ s.t $T^* = f(T)$  $(\Leftarrow) $ notice that $T^* T= f(T) T = T f(T) = T T^* $ as T commutes with a polynomail in T.  $(\Rightarrow ) $ Let $ v_1 , \cdots , v_n $ be an orthonormal basis for T.  $$ p(x) = \sum_{i=0}^{k} a_i x^{i} $$ $$ p(T) = a_0 I + a_1 T + \cdots + a_k T{k} $$ $$p(T)v_j = a_0 v_j + a_1 \lambda_{i} v_j  + a_2 \lambda_{i}^2 v_j \cdots + a_k \lambda_{i}^{k} v_j $$ $$ = p(\lambda_i) v_j $$ $$ =\bar {\lambda_{i}} v_{j} $$ $$= T^{*} v_j  $$
22385
I think the answer is yes.   Sketch of the proof Consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. Let $\{e_\lambda:\lambda\in\Lambda\}\subset\mathbb{R}$ be its Hamel basis. Then $\{(e_{\lambda_1},e_{\lambda_2}):\lambda_1,\lambda_2\in\Lambda\}\subset\mathbb{R}^2$ is a Hamel basis of $\mathbb{R}^2$. Consider set theoretic bijection $i:\Lambda\to\Lambda\times\Lambda$ define map $\varphi:\mathbb{R}\to\mathbb{R}^2$ by its action on element of Hamel basis of $\mathbb{R}$ by equality $\varphi(e_\lambda)=(e_{i(\lambda)_1},e_{i(\lambda)_2})$. This is an isomorphism between $\mathbb{Q}$-vector spaces $\mathbb{R}$ and $\mathbb{R}^2$.  Question Could you tell me is this proof correct, and if it is not, where is the mistake?  EDIT: Thanks to GAJO we found that this proof is wrong. The correct one basis is  $$ \{(e_\lambda,0):\lambda\in \Lambda\}\cup\{(0,e_\lambda):\lambda\in\Lambda\} $$ So we had to look at bijections of the form $i:\Lambda\to\Lambda\coprod\Lambda$ and the desired linear operator is defined by equalities $\varphi(e_\lambda)=(e_{i(\lambda)},0)$ if $i(\lambda)$ lies in the first copy $\Lambda$ and $\varphi(e_\lambda)=(0,e_{i(\lambda)})$ otherwise.
108649
Let $A$ and $B$ be $m \times n$ and $n \times m$ real matrices.  I proved that if $\lambda$ is a nonzero eigenvalue of the $m \times m$ matrix $AB$ then it is also an eigenvalue of the $n \times n$ matrix $BA$. But I am having trouble finding an example showing that this need not be true if $\lambda=0$  Thank you in advance
59989
Let     $$ \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} $$   be the integral closure of $ \mathbb{Z} $ in $ \mathbb{C} $. Prove that $ \mathbb{Z}^{'}_{\mathbb{C}} $ is not finitely generated as a $ \mathbb{Z} $-module.  My attempt: Suppose $ \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} $ for some $ \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} $. I know that given $ \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} $ with minimal polynomials of degrees $ m $ and $ n $ respectively, then we can construct a monic polynomial with integral coefficients $ P(z)=\prod(z-(\alpha_{i}+\beta_{j})) $ where $ \alpha_{i},\beta_{j} $ are the conjugates of $\alpha $ , $\beta $ and therefore the minimal polynomial of $ \alpha + \beta $ has degree $ \leq mn $.   I was thinking that then the degrees of the minimal polynomials of the elements of $ \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} $ could be somehow bounded by a certain constant, whereas $ \mathbb{Z}^{'}_{\mathbb{C}}$ contains elements whose degrees of the minimal polynomials are arbitrarily large, take $ 2^{\frac{1}{n}} $ for example,$ n\geq 2$. I think this works when we are asked to prove that the set of all algebraic numbers $ \overline{\mathbb{Q}} $ is not a finitely dimensional $\mathbb{Q}$-vector space, but I don't know how to proceed in my case.  I would really appreciate any ideas, hints or solutions. Thank you very much!
627
If $B\succeq A$, show that $\lambda_n(B)\ge\lambda_n(A)$, where $\lambda_n$ is the $n$th largest eigenvalue.  It is Theorem 6 in  but the proof is only given as "by characterization." Do they mean the min/max characterization of eigenvalues, $$\lambda_k=\min\max\frac{x^TAx}{x^Tx}$$ and if so, how is that applied?  Thanks.
124299
Let $\lambda_i \in \mathbb C$  $(i=1,2,\cdots,n)$ be $n$ different complex numbers and $p_i \in \mathbb C[t]$ $(i=1,2,\cdots,n)$ polynomials. If we have the relation $${e^{{\lambda _1}t}}{p_1}(t) + {e^{{\lambda _2}t}}{p_2}(t) +  \cdots  + {e^{{\lambda _n}t}}{p_n}(t) = 0$$ for all $t \geq 0$, can we conclude that all $p_i(t)=0$?