Let for the linear operator $A:E\to E$, $A^k=0$, then $A^n=0$.  Here $n=dim E$ and $n &lt;k.$  I tried to find the characteristic oolynomial that I think it is the clue, however did nit succeed.   Any suggestion?
39380
$A$ is a $n\times n$ matrix such that $ A^m = 0 $ for some positive integer $m$.   Show that $A^n = 0$.   My attempt: For $n &gt; m$, it's obvious since matrix multiplication is associative.  For $n &lt; m$, $A^n\times A^{m-n} = 0$; not sure what to do next. Also I know that $\det A = 0$.
35967
This is exercise 10 in section 3.2 of Kreyszig's Introductory Functional Analysis with Applications:     (Zero Operator): Let $T:X \to X$ be a bounded linear operator on complex inner product space $X$. If $\langle Tx,x\rangle =0$ for all $x \in X$, show that $T=0$.    I had solved little kindly guide me further       Solution:   Let $\langle Tx,x\rangle =0$ for all $x \in X$. Let $x=u+av$ where $v,u$ belong to $X$ and $a$ be the scalar then  $$ \langle Tx,x\rangle =\langle T(u+av), u+av\rangle $$ since $T$ is linear then  \begin{align} \langle Tx,x\rangle &amp;=\langle Tu+aTv, u+av\rangle \\     &amp; =\langle Tu,u\rangle +\overline{a}\langle Tu,v\rangle +a\langle Tv,u\rangle +a\overline{a}\langle Tv,v\rangle \\     &amp; =\overline{a} \langle Tu,v\rangle +a\langle Tv,u\rangle  \end{align}
103343
$f:V \rightarrow V$, $\dim(V) = n$  a) If $n$ is odd, prove that there does not exist a linear transformation with $\operatorname{im}(f) = \ker(f)$  b) Is it false to affirm the previous statement if $n$ is even?  My answer: $$ i = \dim(\operatorname{im}(f))\\ k = \dim(\ker(f)) \begin{cases} n=2i                           &amp; \text {if $i=k$} \\ n=i+k   &amp; \text{if $n\neq k$} \\ \end{cases}  $$  b) Yes.  Is that a valid proof?
85184
I feel kind of dumb but I needed this little claim as a part of a proof I'm writing, and I figured out that I'd better just ask, since I could not find the correct algebraic manipulation needed in order to prove it.  So I want to prove this: Suppose $H$ is an Hilbert space, and $A: H\to H$ is a linear operator (if that matters, which I believe it does not, we can assume that $A$ is a projection). Suppose that for every $x\in H$, $(Ax,x)=0$. Then $A=0$.  What I tried to do is a little bit of Cauchy-Schwarz and some algebra but nothing worked out.  Thanks!
21788
Does anyone have a reference for the fact that if A is a normal operator on a Hilbert space the $\|A^n\|=\|A\|^n$? I have managed to prove this fact using a hint on another question, but I need an actual bibliographical reference, please help.
1217
Seems like I , I think I am missing something important.     Let $V$ be an $n$ dimensional inner product space ($n \geq 1$), and $T\colon\mathbf{V}\to\mathbf{V}$ be a linear transformation such that:         $T^2 = T$   $||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;         Prove that a subspace $U \subseteq V$ exist, such that $T$ is the orthogonal projection on $U$.   Now, I know these things:   The fact hat $T^2 = T$ guarantees that $T$ is indeed a projection, so I need to prove that T is an orthogonal projection (I guess this is where $||T(a)|| \leq ||a||$ kicks in). To do this I can prove that:   For every $v$ in $ImT^{\perp}$, $T(v) = 0$ Alternatively, I can prove that for every $v$ in $ImT$ and $u$ in $KerT$, $(v,u)=0$. $T$ is  (according to Wikipedia) The matrix $A = [T]_{E}$ when $E$ is an orthonormal basis, is hermitian (this is equivalent to the previous point). What else?    I've been thinking about it for quite some time now, and I'm pretty sure there is something big I'm missing, again. I just don't know how to use the data to prove any of these things.  Thanks!
12185
Let $f : U \to V$ and $g : V \to W$ be linear transformations on the vector spaces $U$, $V$, and $W$. Supposedly,  $$ \dim(\ker(g \circ f)) = \dim(\ker(f)) + \dim(\ker(g) \cap \operatorname{im}(f)). $$  How might I go about proving that?    (Attempt:) The $\dim(\ker(g)\cap\operatorname{im}(f))$ term suggests to me that I should define a vector space $V' = \ker(g) + \operatorname{im}(f)$ to invoke the theorem that  $$ \dim(V') = \dim(\ker(g)) + \dim(\operatorname{im}(f)) - \dim(\ker(g) \cap \operatorname{im}(f)), $$  but I don't really see where to go from there.
108527
Let $A$ be an $n \times n$ matrix with real enteries. Define $\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n$ , where $\langle,\rangle$ is a standard inner product on $\mathbb R^n$. Then  $\langle x,y\rangle_A$ is an inner product iff  1)  $\ker A = \{0\}$.  2) $\operatorname{rank} A= n$.  3) all eigenvalues of $A$ are positive.  4) all eigen value of $A$ are non negative.  Clearly if $\ker A \neq \{0\}$, then there exist $x \neq 0$ such that $Ax =0$, then $\langle Ax, Ax\rangle_A = 0$, then $\langle ,\rangle_A$ is not inner product, so $\ker A = \{0\}$, thus rank of $A$ is $n$.  I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
43879
Given that $A_{m \times n}$ has real entries, I want to prove that $\operatorname{trace}(A^TA) = 0$ if and only if $A = 0$. In other words, I want to show that the only way for the trace of $(A^TA)$ to be zero is if $A$ is a zero matrix, and that if $A$ is a zero matrix then $A^TA$ has a trace of zero. Intuitively this makes sense to me. My idea is that in order to get zeros on the main diagonal of any product of matrices I'd need at least one of the matrices to have zeros on its main diagonal. In this case, because the product is between $A$ and its transpose, I figure it $A$ does indeed need to be a zero matrix. However, I'm having difficulty turning my intuition into an actual proof.
68911
This is an old exam problem: Let $V$ and $W$ be finite dimensional vector spaces over a field $F$ and let $T: V \to W$ be a linear transformation. Define $\hat{T}: W^* \to V^*$ by $(\hat{T}(f))(v)=f(T(v))$. Here, $U^*$ denotes the dual vector space $U$. Since $\hat{T}$ is a linear transformation prove $\ker \hat{T} = \text{ann}(\text{range } T)$ where $\text{ann}(Y) = \{g \in U^* : g(y)=0 \text{ for all } y \in Y \}$ for a subspace $Y$ of vector space $U$.  I'm really not sure how to start this problem. Any advice would be great!
35004
Let $M_n$ denote the vector space of all $n\times n$ real valued matrices and for any matrix $A\in M_n$ let $A^t$ denote its transpose. Define the linear map $T:M_n\rightarrow M_n$ as $$T(A)=A+A^t.$$ What is the minimal polynomial of $T$? What are the eigenvalues and the corresponding eigenspaces of $T$? What is the characteristic polynomial of $T$?  I learned about minimal polynomials of operators from $\textit{Linear Algebra Done Right}$ by Axler. In his description he explains how in order to find the minimal polynomial you must find the smallest positive integer $m$ such that $T^m$ is a linear combination of $I, T, T^2, \ldots, T^{m-1}$. That is, the smallest integer $m$ such that $$T^m = a_0I+a_1T+a_2T^2+ \cdots +a_{m-1}T^{m-1}.$$  As for the eigenvalues...since we are working with real-valued matrices, then $A$ and $A^t$ have the same eigenvalues, so if $\lambda \in \mathbb{R}$ is an eigenvalue of $A$ (and hence of $A^t$) wouldn't $2\lambda$ be an eigenvalue of $A+A^t$? I apologize if I am saying utter nonsense...
79831
Let $T_1,T_2$ be two linear transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$. Let $\{x_1,\cdots , x_n\}$ be a basis of $\mathbb{R}^n$. Suppose that $T_1x_i \neq 0$ for every $i=1,2,\cdots, n$ and that $x_i \bot \ker T_2$ for every $i=1,2,\cdots, n$. Which of the following is/are necessarily true?   $T_1$ is invertible. $T_2$ is invertible.  both $T_1,T_2$ are invertible.  neither $T_1$ nor $T_2$ is invertible.   My Approch:  Given dimension of $\mathbb{R}=n$ and  $x_i\perp Ker(T_2)$  So, $\text{Ker}T_2=0$  Row space and null space are perpendicular to each other.  From Rank Nullity Theorem:  let T : V → W be a linear map. Then the rank of T is the dimension of the image of T and the nullity of T is the dimension of the kernel of T, so we have  $\text{dim}(\text{Im}(T))+ \text{dim}(\text{Ker}(T))=\text{dim}(V)$  Applying above result we get,  $\text{dim}(\text{Im}(T))+ 0=n$  Therefore, $T_2$ is one-one.  I could not solve this problem completely. what to do next?
107768
I think I understand the proof for square matrices, such that $(AB)^{-1}=B^{-1}A^{-1}$, but I'm not sure if I can just say the same for the bounded linear operators A and B.  Does the existence of $(AB)^{-1}:X\rightarrow X$ imply the existence of $B^{-1}$ and $A^{-1}$?
33276
I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise $4$ from page $115$. It is in the section of The transpose of a Linear Transformation.     Let $V$ be a finite-dimensional vector space over the field   $\mathbb{F}$ and let $T$ be a linear operator on $V$. Let $c$  be a   scalar and suppose there is a non-zero vector $\alpha$ in $V$ such   that $T\alpha=c\alpha.$ Prove that there is a non-zero linear   functional $f$ on $V$ such that $T^{t}f=cf.$ ($T^{t}$ is the transpose of $T$.)   I tried to solve this question by induction on $\dim V$. I was able to show the base case, that is, when $\dim V=1$, but I got stuck in the  inductive step. If $\alpha $ is a non-zero vector, then we can find a base $\mathcal{B}=\{\alpha,\alpha_{1},\ldots\alpha_{m}\}$ of $V$. We can write $V=W_{1}\oplus W_{2}$, where $W_{1}=\langle \alpha \rangle$ and $W_{2}=\langle \alpha_{1},\ldots,\alpha_{m}\rangle.$ I can not show that $T(W_{2})\subset W_{2}.$ Anyway, $\alpha \notin W_{2}.$  EDIT: If $T$ is a linear transformation from $V$ to $W$, then $T^{t}$ is the linear transformation from $W^{\star}$ into $V^{\star}$ such that $$(T^{t}f)(\alpha)=g(T\alpha)$$ for every $f\in W^{\star}$ and $\alpha \in V$.
23262
Let $E$ be  a complex Hilbert space. Let $T\in \mathcal{L}(E)$.  I have two questions:     Why it is not true that for an arbitrary operator $T\in \mathcal{L}(E)$, we have $\langle Tu\;|\;u\rangle=0,\;\forall u\in E \Longrightarrow T=0$? And is  this property true for normal operators?   I think it is true for self adjoint operators because the norm of a self adjoint operators is given by  $$\left\|T\right\|= \sup\big\{\;\left|\langle Tu\;|\;u\rangle \right|,\;\;u \in E\;, \left\| u \right\| = 1\;\big\}$$  Thank you.
33425
As we know, for $V$ vectoral space and a orientation $\mathcal{O}$ on $V$ and $e_{1},...,e_{n}$, the hodge star operator $\ast:\wedge V^*\rightarrow\wedge V^*$ is defined for  $\ast(e_{1}\wedge...\wedge e_{n})=\pm 1$(acoording to the orientation preserve/invert for the basis $\{e_{i}\}$)  $\ast(1)=\pm e_{1}\wedge...\wedge e_{n}$(again $+$ if $[\{e_{i}\}]=\mathcal{O}$ and $-$ if $[\{e_{i}\}]=-\mathcal{O}$)  $e_{1}\wedge...\wedge e_{k}=\pm e_{k+1}\wedge...\wedge e_{n}$  Already prove that $\ast\ast=(-1)^{n(n-k)}\cdot$ and $&lt;T,S&gt;=\ast(T\wedge\ast S)=\ast(S\wedge \ast T)$ (where $T=\eta_{1}\wedge...\wedge \eta_{1}$, $S=\omega_{1}\wedge...\wedge \omega_{1}$ and $&lt;T,S&gt;=det(&lt;\omega_i,\eta_{j}&gt;)$)  For any $\xi\in V$ denote $\gamma:\ast:\wedge^{k+1} V^*\rightarrow \wedge^{k} V^*$ the adjoint of left exterior multiplication by $\xi$ $$&lt;\gamma(T),S&gt;=&lt;T,\xi\wedge S&gt;$$ for all $T\in\wedge^{k+1} V^*$ and $S\in\wedge^{k} V^*$.  I need to prove that $\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)$ but I'm a little confused, how prove that $\gamma$ has this expression from this property?
91537
As everyone know that we can use a matrix $A$ to represent an operator $T$.  The adjoint of a matrix $A$ is denoted as $A^*$, which takes complex conjugate of $A$ and then transpose.      My problem is, what is the relationship between $A^*$ and $A^{-1}$?    I mean as we know $Ax = b$, $x = A^{-1}b$. (If not emphasize on the dimension of $x$ and $b$.