U, V are subspaces of a finite dimensional vector space W, Let $\ P_U$ and $\ P_V$ be the orthogonal projections onto U and V respectively. When is it true that $\ P_UP_V = P_VP_U$?
120513
Let $\mathcal{S}$ and $\mathcal{T}$ be two subspaces of $\mathbb{R}^n$, let $P$ be the orthogonal projection of $\mathbb{R}^n$ on $\mathcal{S}$ and let $Q$ be the orthogonal projection of $\mathbb{R}^n$ onto $\mathcal{T}$.    Show that if $P$ and $Q$ commute, then $PQ$ is a projection and $PQ$ is the projection onto $\mathcal{S}\cap \mathcal{T}$.  Is the converse assertion true? Suppose $PQ$ is the orthogonal projection of $\mathbb{R}^n$ onto the intersection $\mathcal{S}\cap \mathcal{T}$. Must $P$ commute with $Q$.   Anybody has advice on how i should start proving this assertion?
43478
Given the following problem     Let $U$ and $V$ be finite dimensional vector spaces, such that $\dim(U) = \dim(V)$. Prove that $U$ and $V$ are isomorphic.   I was just wondering if somebody could critique and validate my proof.  Proof  Two vector spaces $U$ and $V$ are isomorphic if and only if there exists a bijection $\phi: U \to V$. That is, an invertible linear map between the two. Let $\dim(U) = \dim(V) = n$ and let $\left\{u_1, \dots, u_n\right\}$ and $\left\{v_1, \dots, v_n\right\}$ be bases for $U$ and $V$, respectively. It follows that any vector $u \in U$ can be written as  $$ u = c_1u_1 + \cdots + c_nu_n $$  We define $\phi: U \to V$ as  $$ \phi(c_1u_1 + \cdots + c_nu_n) = c_1v_1 + \cdots + c_nv_n $$  Observe that $\phi$ is invertible. That is, given a vector  $$ v = d_1v_1 + \cdots + d_nv_n $$  in $V$, we define $\phi^{-1}: V \to U$ as  $$ \phi^{-1}(d_1v_1 + \cdots + d_nv_n) = d_1u_1 + \cdots + d_nu_n $$  Since there exists a bijection between $U$ and $V$, they are isomorphic.
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.
79183
Suppose that we have a Hilbert Space $H$ and $M$ is a closed subspace of $H$. Let $T\colon H\rightarrow M$ be the orthogonal projection onto $M$.   I have to show that $T$ is compact iff $M$ is finite dimensional.  So if we assume that $M$ is finite dimensional then $\overline{T(B(0,1))}$ is a closed bounded set in a finite dim vector normed space and so it is compact. Which gives that $T$ is compact.  But I am unsure how to prove that if $T$ is compact then $M$ is finite dimensional?  Thanks for any help
13996
Suppose $W$ is a subspace of the finite-dimensional vector space $V$ and $\dim W = \dim V=n$  Let $\{\overrightarrow w_1,\ldots,\overrightarrow w_n\}$ be a basis for $W$ and $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ be a basis for $V$.  The set $\{\overrightarrow w_1,\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ still spans $W$. Since $W$ is a subspace of $V$, $W\subset V$ and all the $w_i$'s can be written in terms of $\{ \overrightarrow v_{1},\ldots,\overrightarrow v_n\}$.  $\therefore$ We can remove one by one all the $w_i$'s from the set $\{\overrightarrow w_{1},\ldots,\overrightarrow w_n,\overrightarrow v_1,\ldots,\overrightarrow v_n\}$ while still having it span $W$, which leaves us with the set $\{\overrightarrow v_1,\ldots,\overrightarrow v_n\}$. We know it to be linearly independent since it is a basis for $V$. Therefore the basis of $V$ is also a basis of $W$. From there on, it is trivial to prove double inclusion $W\subset V$ and $V\subset W$ which gives us $W=V$.  Is my approach correct?
108971
How to prove (if it is true ... I suppose it is) that orthogonal projection from N-dimensional space onto 2-dimensional plane and rotation in this plane (I assume components of vectors out of the plane are not affected by the rotation) are commutative operations independently from dimension of N-dimensional vector space ?
32901
Given a subspace $U \subset V$ of a vector space $V$, show that there exists a subspace $U' \subset V$, such that $U+U'=V$ and $U \cap U'=\{0_V\}$.  I am aware that $U'$ is the complement of $U$ in $V$.   Would it be correct to prove this over the basis extension theorem?  For every subspace $U$ there exists a basis of $U$, let it be $B_U$. It is true that $B_U \subset V$. According to the basis extension theorem one can extend any set of linearly independant vectors in $V$ to a basis of $V$. Thus one can extend $B_U$ to a basis of $V$. Let this basis be $B_V$. It is true, since the vectors in $B_V$ are linearly independant, that $\langle B_U \rangle \cap \langle B_V \setminus B_U \rangle = \{0_V\}$. It is also obviously true that $\langle B_U \rangle + \langle B_V \setminus B_U \rangle =V $. Let $\langle B_V \setminus B_U \rangle=U'$.   Is this correct or am I missing something?
114608
Let $T$ be a linear operator on a vector space $V$, which is a $T$-cyclic subspace of itself. How do I prove that $UT=TU$ iff $U=g(T)$ for some polynomial $g$?
109591
If $P$ is an $n\times{n}$ orthogonal matrix prove that the columns of $P$ form an orthonormal set in $R^n$. So I know that a matrix is orthogonal if $AA^T$=$I$ but I'm not sure how that would help me here. Any help is appreciated.
121915
Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following:     Suppose $V$ is finite-dimensional and $\Gamma$ is a subspace of $V'$. Show that $$\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0$$   where $V'$ is the dual space of $V$ and, for any $S\subset V$, $S^0$ is the annihilator of $S$.  Attempt:  Let $S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}$. Clearly, $\Gamma\subset S^0$.I tried to show that $S^0\subset\Gamma$ using some bases of $V$ and $V'$, but I failed.  I also tried to show that $\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}$. If $s_1,\dots,s_n$ is a basis of $S$ and $s_1',\dots,s_n'$ its dual, and if $\psi_1,\dots,\psi_p$ is a basis of $\Gamma$, it's easy to see that $s_1',\dots,s_n',\psi_1,\dots,\psi_p$ is a linearly independent list of $V'$. Also, if you extend $s_1,\dots,s_n$ to a basis $s_1,\dots,s_n,v_1,\dots,v_m$ of $V$, then its dual $s_1',\dots,s_n',v_1',\dots,v_m'$ is a basis of $V'$; in fact $S^0=\text{span}\{v_1',\dots,v_m'\}$. Two remarkable facts:$$\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0$$ because $v_j\notin S$, and $$\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0$$ because $\psi_i\neq 0$; in other words the matrix of the inclusion map from $\Gamma$ to $V'$ with respect to the basis of $\Gamma$ and the dual base of the chosen basis of $V$ has no $0$ row nor $0$ column. But this doesn't seem to provide a way to prove that any linear comination of the $(v_i')_{1\le i\le m}$ is a linear combination of the elements of the basis of $\Gamma$. I believe that $v_1,\dots, v_m$ should be chosen more carefuly but I fail to.  Could you please help me? Thank you in advance!
5902
I am trying to prove (or disprove) the converse of Schur's Lemma in finite dimensional vector spaces. I am not sure if it holds in this case, but I have tried to apply the idea that proves it in representation theory (see for example Theorem 4.3  that uses Maschke's theorem or in questions  and ).  The converse of Schur's Lemma in finite dimensional vector spaces is:     Let $V$ be a finite dimensional vector space over the complex numbers. Let $S$ be a set of endomorphisms of $V$ and assume that every endomorphism $A$ of $V$ such that $$AB=BA\text{ for all }B\in S$$ is of the form $\lambda I, \ \lambda\in\mathbb{C}$. Then $V$ is a simple $S$-space.   (*an endomorphism of $V$ is a linear operator from $V$ to $V$  *$V$ is a simple $S$-space if the only $S$-invariant subspaces of $V$ are $V$ itself and the zero subspace)  What I've tried so far:  Assume to the contrary that $V$ is not a simple $S-$space. Then, there exists a subspace $W$ of $V$ such that $W\not =\{0\}$, $W\not= V$ and $W$ is $S$-invariant. Let $W'=V\setminus W$. Then, since $V$ is finite, we can easily prove that $V=W\oplus W'$ (this is my attempt to translate Maschke's theorem in vector spaces). Consequently, for every $v\in V$, there exists unique $w\in W$ and $w'\in W'$, such that $v=w+w'$. Define the projection $P:V\rightarrow V$ by $Pv=w$ for every $v\in V$.   What is left to prove is that $PB=BP$ for all $B\in S$. Then $P$ is clearly not a scalar and thus we have the contradiction we are looking for.  There is a difficulty in showing that $PB=BP$ for all $B\in S$, because $W'$ might be $S$-invariant or not. More precisely:   Let $v\in V=W\oplus W'$ and $v\not =0_V$.  If $v\in W$ then $Bv\in W$ because $W$ is $S$-invariant, therefore: $PBv=Bv$ and $BPv=Bv$.  If $v\in W'$ then $Pv=0$, hence $BPv=B\cdot 0=0$, and  (i) if $Bv\in W'$ then $PBv=0$.  (ii) if $Bv\in W$ then $PBv=Bv$ and this is where the problem occurs.  Any hints, ideas or counterexamples would be very helpful.
36881
Show that $P_n = \{$Polynomials with real coefficients of degree $≤ n\}$ is an $(n+1)$-dimensional subspace of the infinite-dimensional vector space of all real polynomials.
87087
Let $V$, a vector space over $\mathbb{F}$ and $W_1,W_2 \subseteq V$ such that $W_1 \oplus W_2 = V$. For $i=1,2$, let $\langle , \rangle $ on $W_i$. Prove that there is a unique inner product on $V$ such that:  $W_2 = W^\perp_1$ For $i=1,2$ and for all $v,u\in W_i$: $\langle v, u \rangle = \langle v, u \rangle _i$   Uniqueness: Let $v = w_1 + w_2$ and $u = w_1' + w_2'$ $$\langle u, v \rangle_V = \langle w_1,w_1' \rangle + \langle w_1,w_2' \rangle + \langle w_2,w_1' \rangle + \langle w_2,w_2' \rangle = \langle w_1,w_1' \rangle  + \langle w_2,w_2' \rangle $$ So the inner product is determined uniquely by $\langle, \rangle_i $ for $i=1,2$. How do I show the existence of such inner product?
80091
Let $V$ be a finite dimensional vector space and $U^0$ and $W^0$ denote the annihilators of subspaces $U$ and $W$, respectively.  Then I need to show that $(U\cap W)^0 = U^0 + W^0$.  I can show that $U^0 + W^0 \subseteq (U\cap W)^0$ like so: Let $f\in U^0$ and $g\in W^0$.  Then, because $U^0, W^0 \subseteq (U\cap W)^0$, we see that if $f\in U^0$ and $g\in W^0$, then for any $v\in U\cap W$ and $a,b \in \Bbb F$, $(af+bg)(v) = af(v) + bg(v) = 0$.  Thus $af+bg\in (U\cap W)^0$.  But because $f$ and $g$ were arbitrary, this implies that any linear combination of elements of $U^0$ and $W^0$ is an element of $(U\cap W)^0$.  I.e. $U^0+W^0 \subseteq (U\cap W)^0$.  But I haven't had any luck proving the other direction.  Obviously I should start with $f\in (U\cap W)^0$.  But then it looks like I need to somehow break that function into two to show its in $U^0 + W^0$.  That's throwing me off.
38632
Prove that if $V$ and $W$ are three-dimensional subspaces of $\Bbb R^{5}$, then $V$ and $W$ must have a non-zero vector in common.  So far I got $V = \{v_{1}, v_{2}, v_{3}\}$ and $W = \{ w_{1}, w_{2}, w_{3}\}$ because they are 3-dimensional subspaces. Then $V \cup W = \{v_{1}, v_{2}, v_{3}, w_{1}, w_{2}, w_{3}\}$  My prof said that now I need to prove that they are linearly independent so this is a contradiction and the statement is wrong because we are in $\Bbb R^{5}$ and we have 6 elements on it.  I don't know how to express them as a linearly independent proof, so I hope anyone can give me a good explanation on it.
41180
I got the following question as my homework.     Given $V$ is a vector space with $P \in \operatorname{End} V$. $P \circ P = P$ ("P is idempotent"). Show that $V = \operatorname{Ker} P \oplus \operatorname{Im} P$.   One $P$ I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example $\begin{pmatrix} x \\ y \\ z\end{pmatrix} \mapsto \begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. Then $\operatorname{Ker} P$ would give the line $\begin{pmatrix} 0 \\ 0 \\ z\end{pmatrix}$ and $\operatorname{Im} P$ would contain all $\begin{pmatrix}x \\ y \\ 0\end{pmatrix}$. So the result of $\operatorname{Ker} P \oplus \operatorname{Im} P$ is of course $V$.  But how do I prove that in a mathematical way?
59318
$W$ and $U$ are subspaces of vector space $V$. Show that $U\cup W$ is not necessarily a subspace of $V$.   I know that one counter-example is enough to show that. What example could fit here?