Let $T = \lambda z + \mu \bar z$ where $\lambda, \mu \in \mathbb{C}$. Show that $T$ is bijective exactly when $\lambda \bar\lambda \ne \mu \bar\mu$.
122625
We consider $T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\lambda z+\mu\overline{z}$, where $\lambda ,\mu\in\mathbb{C}$. I want to prove that if $T$ is injective, then $\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}$.  Clearly, if $\mu =0$, then $\lambda\neq 0$ because $T$ is injective.  Now, if $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ and $\mu\neq 0$, then I wanted to find a $z\in\mathbb{C}\setminus\{0\}$ such that $z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}$, because we would have that:  $T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0$  ('Cause this would prove that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ implies $T$ is not injective.)  So, the question is, can we always find such a $z\in\mathbb{C}\setminus\{0\}$?
65235
I am stuck on this problem:  Define $f: N\times N \rightarrow N$ by   $f(i,j)=\frac{(i+j-1)(i+j-2)}{2}+i$.   How do you prove that $f$ is a bijection thus $N\times N$ and $N$ are numerically equivalent?  Work: First I tried to prove that $f$ is injective by setting $f(i,j)=f(x,y)$. However, from here I am stuck and I cannot reduce it properly. My work really does not get anywhere significant and its really messy.
70562
Let $C_c(\mathbb{R})$ be the following:  $$C_c = \{ f \in C(\mathbb{R}) \mid \exists \text{ } T &gt; 0 \text{ s.t. } f(t) = 0  \text{ for } |t| \geq T\}$$  Let $T_n \in L(C_c(\mathbb{R}))$ be a linear operator such that:  $$T_n u = \delta_n *u, \forall u \in C_c(\mathbb{R}),$$  where  $$ \delta_n(t)= \begin{cases}        n^2(t+1/n) &amp; -1/n \leq t \leq 0 \\       -n^2(t-1/n) &amp; 0 &lt; t \leq 1/n \\       0 &amp; \text{elsewhere}     \end{cases} $$  I have to prove that with respect to the 2-norm, the operator $T_n$ has $\|T_n\| = 1$ and I have succeeded in proving that $\|T_nu\|_2 \leq \|u\|_2, \forall u \in C_c(\mathbb{R}).$  Now, I remained  to prove that $\exists \text{ } u \in C_c(\mathbb{R}) \text{ s.t. } \|T_nu\|_2 = \|u\|_2$, but I really don't get how $u$ should be shaped in order to satisfy it.
92713
Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots. EDIT: ( Over $\Bbb C $ ) though it is obvious i am working over $\Bbb C $ as one of my statements is not true over $ \Bbb R $   I would like a better proof of this result what i did is below, there is the same question on here somewhere but only has an answer to one direction im looking for both.  I proved the result by using a different equivalent statement to diagonalizable looking for any complete proof that is shorter.  We notice that it is equivalent to prove that V has a basis consisting of eigenvectors of T iff the minimal polynomial of T has no repeated roots by theorem.  $(\Rightarrow ) $ First suppose that there is a basis $\beta = (v_1,\cdots , v_n ) $ consisting of eigenvectors of T. let $\lambda_1 , \cdots , \lambda_m $ be distinct eigenvalues of T. Then for each $ v_i $ there exists a $\lambda_k $ with $(T- \lambda_k I) v_i =0 $ it then follows that $(T- \lambda_1 I) \cdots (T- \lambda_mI) v_i =0 $ for each i as we can commute the operators. Since an operator that sends each vector in a basis to the $0$ vector is the $0$ operator we have that $(T- \lambda_1 I) \cdots (T- \lambda_mI) =0 $ Thus the polynomial $(z-\lambda_1) \cdots (z-\lambda_m ) $ when applied to T gives 0. but by theorem we know that that the minimal polynomial of T is a divisor of $(z-\lambda_1) \cdots (z-\lambda_m ) $ which has no repeated roots so the minimal polynomial cannot possibly have repeated roots the result follows.  $(\Leftarrow ) $ Let us assume that the minimal polynomial has no repeated roots; if we let $ \lambda_1 \cdots \lambda_m $ denote the distinct eigenvalues of T, this means the minimal polynomial of T is $(z-\lambda_1) \cdots (z-\lambda_m ) $ It follows that $(T- \lambda_1 I) \cdots (T- \lambda_mI) =0 $ Let $U_m $ be the subspace of a generalized eigenvectors corresponding to the eigenvalue $\lambda_m $. Since $ U_m $ is invariant under T by theorem we consider $ v\in U_m $ let $u= (T- \lambda_m I) v $ it follows that $u\in U_m $  Hence $$ (T|_{U_m} - \lambda_1 I ) \cdots (T|_{U_m} - \lambda_{m-1}I) u =   (T- \lambda_1 I) \cdots (T- \lambda_mI) v =0 $$  by theorem we have that $( T- \lambda_m I )|_{U_m} $ is nilpotent by previous question we have that 0 is the only eigenvalue of $( T- \lambda_m I )|_{U_m} $. Thus $T|_{U_m} - \lambda_jI $ is an invertable operator on $U_m $ for $j= 1, \cdots , m-1 $ it then follows by $$ (T|_{U_m} - \lambda_1 I ) \cdots (T|_{U_m} - \lambda_{m-1}I) u =   (T- \lambda_1 I) \cdots (T- \lambda_mI) v =0 $$ that $u=0$ in other words, $v$ is an eigenvector of T!   We have shown that every generalized eigenvector of T corresponding to the eigenvalue $\lambda_m $ is an eigenvector of T. However we choose $ \lambda_m $ arbitrarily we could of just of easily relabeled the eigenvalues so that any of them was called $ \lambda_m $. Therefore we have that every generalized eigenvector of T is actually an eigenvector of T. By theorem we have that there is a basis for V consisting of generalized eigenvectors of T but by above we have that there is a basis of V consisting of eigenvectors of T the desired result.
9025
Let $L:\mathbb{R}^n \rightarrow \mathbb{R}^m$ be a linear transformation such that $m&lt;n$. What does this tell you about injectivity and surjectivity?  I can't really create a (counter) argument against any of these alternatives:   $L$ is injective $L$ is surjective $L$ is not injective $L$ is not surjective   My intuition says that one could find examples where $L$ is both injective and/or surjective, but not for all inputs outputs.
22060
This is Exercise EP $8$ from Fernandez and Bernardes's book Introdução às Funções de uma Variável Complexa (in Portuguese).     If $z,w\in \mathbb{C}^{\times}$, then $|z+w|=|z|+|w|$ iff $w=tz$ for   some $t\gt 0.$   I supposed that $|z+w|=|z|+|w|$ and by squaring it I got  $\operatorname{Re}(z\bar{w})=|z||w|$ and I am stuck.  I would appreciate a hint in this question.
120297
Suppose $T:V\to W$ and that $V$ is finite-dimensional.  I want to prove that $$\text{Im }T'=(\ker T)^0$$ where $T'$ is the dual/transpose map and $(\ker T)^0$ is the annihilator of the kernel.  I know that $\phi \in V'$ is an annihilator of $\ker T$ if and only if $$\phi(v)=0 \space \forall v\in \ker T$$ if and only if $$\phi(v)=0 \space \forall v \in V \text{ such that } T(v)=0$$  Now I also know that $\phi \in \text{Im }T'\subset V'$ if and only if $$\phi = f \circ T \text{ for some } f \in W'$$ I can see that if $v\in \ker T$, then this implies $\phi(v) = f(T(v)) = 0$, so $\phi \in (\ker T)^0.$  However, I'm having trouble showing the other way, that if $\phi \in (\ker T)^0$, then $\phi \in \text{Im }T'.$  Could anybody help?
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.
49080
I am looking for a injective cochain map $\psi: C^*\rightarrow D^*$ such that the map $\Psi^i$ from $C^i$ to $D^i$ is injective, but the map $\psi^*$ from  $H^i(C^*)$ to $H^i(D^*)$ is not injective for any $i\geq0$.
17202
How can we prove that any bijection on any set is  a composition of 2 involutions ?  Since involutions are bijections mapping elements of a set to elements of the same set, I find it weird that this applies to any bijection.  Thanks for your help !
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!
97283
Let $T:\mathbb{R}^k\to\mathbb{R}^m$ is additive, i.e. $T(X_1+X_2)=T(X_1)+T(X_2)$. $T$ is locally bounded. Prove that $T$ is a linear transformation.  I know there is a relation between boundedness, convexity and scalar multiplicity. But not sure where to start or how to tie things together.
92272
Let $T$ be an invertible linear operator on the finite dimensional $F$-vector space $V$.  Prove that if $\lambda$ is an eigenvalue of $T$, then $\lambda^{-1}$ is an eigenvalue of $T^{-1}$ and that the eigenspaces $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$.  I have shown the first part of the proof, but it remains to show that $E_{\lambda}(T)=E_{\lambda^{-1}}(T^{-1})$ is true.  I'm thinking that this needs to be shown by proving the dimension of each of the eigenspaces, or can it be done through just by showing that all elements of one eigenspace are in the other, and vice versa.
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 $p_i(t)=0$?
75204
Problem:  Let T be a nonempty set and A(T) the set of all permuations of T. Show that A(T) is a group under the operation of composition of functions.  Permutation of the set T is the bijective function $f: T \rightarrow T $  For A(T) to be a group it must be closed, associate have an identity for every element and an inverse.  Since f is  bijection the composition of two bijective functions is itself bijective so A(T) is closed. Every bijective function has an inverse, and composition of functions is associative. What about the identity? Are we to take $ A(T) = S_n $ here, which would be assuming that $T$={$1, ..., n$}? In which case $ I = \begin{bmatrix}1&amp;2&amp;...&amp;n\\1&amp;2&amp;...&amp;n\end{bmatrix} $  Not sure what I'm supposed to be showing since $T$ and A(T) are arbitrarly specified here.
9279
Let's suppose that $|z|&lt;1$ and $|w|=1$. Show that the modulus of $\displaystyle \frac{z-w}{1-\bar{z}w}$ is always $1$. Some hint.
83448
I'm taking a course in Real Analysis and have come across the following question which I initially thought looked quite simple:  Let $X$ be an inner product space over $\mathbb{K}$ ($\mathbb{K} = \mathbb{C}$ or $\mathbb{R}$) Show that x and y are orthogonal if and only if $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$ for all $\mu, \lambda \in \mathbb{K}$.  I have proven the statement in one direction (starting with orthogonality and ending up with $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$). However, I seem to be getting stuck trying to go in the other direction. The following is what I have so far:  $||\lambda x  + \mu y||^2 = ||\lambda x||^2 + 2\bar{\lambda}\mu Re&lt;x,y&gt; + ||\mu y||^2$  Then we must have 2$\bar{\lambda}\mu Re&lt;x,y&gt; = 0$ for the desired equality to hold, and so $Re&lt;x,y&gt; = 0$.   However, for orthogonality I want $&lt;x,y&gt; = 0$ and can't see how I can make this next step.  Any help would be greatly appreciated.