Let $T:V\rightarrow W$ a linear transformation, $i_w:W\rightarrow W^{**}$ and $i_v:V\rightarrow V^{**}$ canonical morphism of biduality.  Prove $i_w\circ T=T^{**}\circ i_v$  I' m very very confused with this exercise.  Suppose i need to prove this:  $i_w\circ T\subset T^{**}\circ i_v$ and $T^{**}\circ i_v\subset i_w\circ T $ but  i don't have idea of how to prove this. Can somone give me a hint?
96461
$\newcommand{\Hom}{\operatorname{Hom}}$Let $U,V$ be vector spaces. Denote the dual space of $U$ with $U^\intercal$ and the dual mapping of a linear map $\Phi$ be $\Phi^\intercal$. Define the double dual embedding on $U$ to be $\iota_U:U \to U^{\intercal\intercal}$. ($u \in U, f \in U^\intercal$) $$ \iota_U(u)(f) = f(u) $$     If $\Phi \in \Hom(V,U)$, then $\Phi^{\intercal\intercal} \circ \iota_V = \iota_U \circ \Phi:V \to U^{\intercal\intercal}$   Proof: Let $v \in V, f \in U^\intercal$. Then $$ \iota_U \circ \Phi(v)(f) = f(\Phi(v)) $$ but how do I show that $$ \Phi^{\intercal\intercal} \circ \iota_V(v)(f) = f(\Phi(v)) $$ ?
15674
Let $V$ be a real n dimensional vector space &amp; let $T:V→V$ be a linear transformation satisfying $T^2(v)=-v$ for all $v\in V$. Then how can we show that $n$ is even?  I am completely stuck on it. Can anybody help me please?
120708
Let $S: V \to V$ be a linear transformation. Show that $V \supseteq \operatorname{Im}(S) \supseteq \operatorname{Im}(S^2) \supseteq \cdots$ and $\{0\} \subseteq \ker(S) \subseteq \ker(S^2) \subseteq \cdots$. If $a_k = \operatorname{rank}(S^k)$, deduce that $a_k \geq a_{k+1}$ and that $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$.  I can show everything except for $a_k - a_{k+1} \geq a_{k+1} - a_{k+2}$, which eludes me. I attempted using the rank-nullity theorem, but no luck so far. Any ideas/hints?
116546
Studying from Roman's Advanced Linear Algebra, I want to prove that $$U^* \otimes V^* \cong (U \otimes V)^* \cong \hom(U,V;\mathbb{F})$$ The author proves that $U^* \otimes V^* \cong (U \otimes V)^*$ by showing that there is an unique linear transformation   $$\theta:U^* \otimes V^* \to (U \otimes V)^*$$  defined by $\theta(f \otimes g)=f \odot g$ where $(f \odot g)(u \otimes v)=f(u)g(v)$  I want to prove that $U^* \otimes V^* \cong \hom(U,V;\mathbb{F})$. I tried to use the universal property by fixing $f \in U^*, g\in V^*$ and defining $S: U^*\times V^* \to \hom(U,V;\mathbb{F})$ as $S(f,g)=F_{f,g}$, where $F_{f,g}(u,v)=f(u)g(v)$, but I got stuck.
15939
Let $(V,||\cdot||_V),(W,||\cdot||_W)$ be normed spaces over $\mathbb{C}$.  Let $T:V\rightarrow W$ be a linear transformation.  Assume $T$ is continuous.  Then, how do i prove that $\exists c&gt;0$ such that $\forall v\in V, ||Tv||_W&lt;c||v||_V$?
98145
It is given that $T$ is a linear transformation from $\mathbb R^n$ to $\mathbb R^n$ such that $T^2 = k T $  for some $k\in \mathbb R$.   Then, one or more of the options are true    $\|T(x)\| = |k| \|x\|$ for all $x\in \mathbb R^n$.  If $\|T(x)\| = \|x\|$ , for some non- zero vector $x\in \mathbb R^n$ then $ k =\pm 1 $.  $\|T(x)\|  &gt;  \|x\|$ for some non- zero vector $x\in \mathbb R^n$, then T is singular.  $ T = k I $ where I is an identity transformation.    Please suggest how to proceed.
93768
I am trying to proof the theorem that states that:  "A map to a smaller dimensional space is not injective"  So I first suppose that V and W are finite-dimensional vector spaces such that dimV > dimW and I let T be the linear transformation from V to W.  Given the above, you have that:  dim(null(T)) = dim(V) - dim(range(T)) >= dim(V) - dim(W) > 0  So I understand the first and second equalities. However, I don't understand why dim(V) - dim(W) is strictly greater than 0.
63551
Let $V$ be a finite-dimensional inner product space over $\mathbb{C}$ and $T: V \to V$ a linear transformation.   Show that if every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal.   We need to show that $\forall v \in V,\ TT^*v=T^*Tv$.  I started by picking $v$ to be an eigenvector of $T$, hence $Tv=\lambda_1 v$, and $T^*v=\lambda_2 v$. Therefore: $$TT^*v=T(\lambda_2 v)=\lambda_2 Tv=\lambda_1 \lambda_2 v=...=T^*Tv$$  But I'm stuck when $v$ is not an eigenvector.  (Also, this question is taken from a chapter about Jordan forms, but I can't see how it relates)
33147
I'm sure I've done this before in abstract algebra. Regardless it's escaped me now.   I have proved that for $T:U\rightarrow V$, with $dim(U)=m$ and $dim(V)=n$ that $rank(T)\le m$ which is obvious, but I have proved it none the less.  I want to.... this is where I get stuck. I'm not quite sure how to say it.  Suppose we have a $T$ for example that takes $U\subset\mathbb{R}^3\rightarrow V\subset\mathbb{R}^3$, we could still have a $T$ that maps a plane to something (a plane, line or point) in this case rather than $rank(T)\le 3$ I can state $rank(T)\le 2$.  (Correct my notation here, I don't like writing subset, I mean number of dimensions of the space!)  To show this I can say $F:P\subset\mathbb{R}^2\rightarrow U\subset\mathbb{R}^3$.  Then let $G:P\rightarrow V$ and that $G=TF$, now $rank(G)\le dim(P) = 2$  Now, if $U,P,Q,R$ are spaces of dimension $\le 2$ and P,Q,R are subspaces of U  $A:U\rightarrow P$  $B=A:P\rightarrow Q$  $C=A:Q\rightarrow R$  We can now say $A^2=BA$ then $A^3=CBA$  We know that for $A$ rank(A)$\le 2$, this thus provides an upper-bound for rank($A^3$)  I want to show now that if I keep applying a linear transformation that the rank is a monotonically decreasing ($\le$) sequence. I am unsure on a proof that the rank of a transformation cannot be greater than the domain's dimensions. Although this is trivial.  (NOTE:  I have proven that if a set of n vectors, R, span a vector space V, and you take a set W of m linearly independent vectors in V that m$\le$n)  If rank(A)=2 then rank($A^2$)=2 and so rank($A^3$)=2  if rank(A)=1 then rank($A^2$)=0 and rank($A^3$)=0  if rank(A)=0 then rank($A^2$)=0 and rank($A^3$)=0  But again, I can't prove this, or at least I am unsure of how to write it.  something to clarify:  For a map $T:U\rightarrow V$ how do we distinguish (using notation) whether or not we actually use all dim(U)?  For example we might have a T that maps a plane in R^3 to a line in R^3, this can be expressed as a composition of maps, one that takes a 2 dimensional vector space (coordinates of the plane in 3 space) to a 3 dimensional point in T's domain, which T then maps to a line.  We have rank - the dimensions of the image - to make this distinction on the map's target (the rank of T is 1 if it maps something to a line, but the dimensions of the target is 3)  How do we write this? What do I say to describe this?   How do I distinguish between the number of vectors in a basis and the number of dimensions the space that basis is in has?  Sorry for the length of this! I tried to talk though my problem in the hope of seeing it myself, no such luck, thanks.
58821
Let $T_1$ and $T_2$ be two Linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^n$.  Let $\{x_1,x_2,\cdots,x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1(x_i)\neq 0$ for every $1\leq i\leq n$ and that $x_i\perp Ker (T_2)$ for every $1\leq i\leq n$.   Which of the following is true?   $T_1$ is invertible $T_2$ is invertible Both $T_1$ and $T_2$ are invertible Neither $T_1$ nor $T_2$ is invertible.   As $T_1(x_i)\neq 0$ for each $1\leq i\leq n$ we do not have $T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 $ unless each $a_i=0$ i.e.,$a_1x_1+a_2x_2+\dots+a_nx_n=0$ i.e., $T_1$ is one one thus invertible.  I am not sure if $T_2$ is invertible or not.  we have $x_i\perp Ker (T_2)$ for all $x_i$. Would that be a good idea say something like  $\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)$ and as they span whole space we would have   $\mathbb{R}^n\perp Ker(T_2)$ and thus $Ker(T_2)=0$ so it is injective so it is invertible...  Thus both $T_1$ and $T_2$ are invertible?  I would be so thankful if someone can assure what it has been done here is sufficient/clear.  THank you
86514
I am asked to prove this without the rank nullity theorem   My Attempt at a Proof  For the $\implies$; If $g:V \rightarrow V$  is injective then the dimension of the kernel is 0, and so as ($im$) and ($ker$) are both subspaces of $V$, $dim(im)=dim(v)$ which implies $g$ is surjective.    For the $\impliedby$ ; If $g:V \rightarrow V$  is surjective, then $dim(v)=dim(im)$ and $dim(ker)$ must be 0, so $g$ is injective.    It seems to me that i have indirectly used the  rank nullity theorem, would there be a more suitable proof that doesn't incorporate it?
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.
31407
Question  Let $T:V \rightarrow V$ be a linear map. How do I prove that $T \circ T = T_0$ ( the zero linear map) iff $R(T) \subset N(T)$?  Attempt  \begin{eqnarray} T\circ T=T(T(v))&amp;=&amp;T(T(v-0))\\ &amp;=&amp;T(T(v)+T(-0))\\ &amp;=&amp;T(T(v))+T(T(-0))\\ &amp;=&amp;T(T(v))+T(-T(0))\\ &amp;=&amp;T(T(v))-T(T(0))&amp;=&amp;0, \end{eqnarray} so this last bit gives \begin{eqnarray} T(T(v))=T(T(0)), \end{eqnarray} which means $T(v)=T(0)=0~\forall~v\in V$, so $R(T)=0$, this zero of which is in the null space for all linear maps $T$.
52020
Let $V$ be a finite-dimensional vector space over $F$ with $\dim(V) = n$. Let $ T:V \rightarrow V$ be a linear transformation.  Assuming that $T$ has $n$ different eigenvalues. prove that :$$ TS = ST \iff \textrm{each eigenvector of} \ T \ \textrm{is an  eigenvector for }S\textrm{ (maybe with a different eigenvalue)}$$   I managed to prove the right to left direction.   Thanks for helping.
53717
Let $V$ be a vector space. Determine all linear transformations $T:V \rightarrow V$ such that $T=T^2$. I am given      $x=T(x)+(x-T(x))$ for every x in V, and show that $V={y:T(y)=y} \oplus N(T)$ as a hint   I don't have a first thought on this problem, what can identity mapping tell us? And what's the general intuition for this problem?
91187
Let $V$ be a finite-dimensional $K$-vector space and let $T$ be a linear operator from $V$ to $V$.  I already proved that for every polynomial $p(x) \in K[x]$, $\ker p(T)$ and $Im p(T)$ are $T$-invariant. Is it true that for every $W$ proper $T$-invariant subspace, $W$ is the kernel of $p(T)$ for a polynomial $p$?  My guess is yes. If $p$ is the characteristic polynomial of the restriction of $T$ to $W$, then $p(T)v=0$ for every $v$ and then $W\subset \ker p(T)$. How can I finish the converse?
92890
Let $C, C', C''$ be three categories.  Let $C \xrightarrow{F_1, F_2} C'$ and $C' \xrightarrow{G_1, G_2} C''$, be four functors, and let $\theta : F_1 \Rightarrow F_2, \ \lambda : G_1 \Rightarrow G_2$, be two natural transformations.  Then I've already shown algebraically that if we defined $(\lambda * \theta)_X \equiv G_2(\theta_X) \circ \lambda_{F_1 X}$, then $\lambda * \theta$ is a natural transformation $G_1 \circ F_1 \Rightarrow G_2 \circ F_2$.  Now I want to show that for all $X \in C$, the following diagram commutes (or the two formulas for $\lambda * \theta$ are the same.  A book says by naturality of $\theta$ and functoriality of $G_1$ the square commutes.  But I'm having trouble seeing this.  Please explain using diagrams or give a hint.