Theorem: Any two norms on a finite-dimensional vector space $V$ are equivalent.   My book begins the proof by noting that we can identify $V$ with $\mathbb R^n$ for some $n\in\mathbb N$, and therefore we only need to prove the theorem for $\mathbb R^n$. However, I don't see why this is true. I know that there is a bijection between $V$ and $\mathbb R^n$ for some $n\in\mathbb N$, or equivalently, there is a bijection between a basis in $\mathbb R^n$ and a basis in $V$. But how to proceed from there on?  I've read somewhere that an isomorphism is basically just the relabelling of the elements... but how do I know that things like inequalities and the like stay preserved between two identified elements? Things like "the vector space structure is preserved" sounds slightly vague to me, so is there something more rigorous to say?
121385
Let $V$ be a finite dimensional vector space (of dimension n) over a field $F$. I need to show that $V$ is isomorphic to $F^n$ as abelian groups. However, I don't really understand what does "isomorphic as abelian groups" mean, my (poor) attempt to solve this was:  Let $f:V\to F^n$, $f((v_1,v_2,\ldots,v_n)):=v_1+\ldots+v_n$, clearly $f$ is a group homomorphism, but it is not biyective since, for example, in the case where $dim(V)=2$ and $F=\mathbb{R}$ we have $f((2,-2))=0=f((3,-3))$, I've also tried it by defining $f$ to be the product of a vector's entries but obviously it didn't work so I'm stuck!   I think this should be an easy problem but since I don't quite understand it, it's giving me problems. Thank you for your help.
58822
I am trying to understand more about the Bidualspace (or double dual space). The whole idea is that $V$ and $V^{**}$ are canonically isomorphic to one another, which means that they are isomorphic without the choice of a basis, which means there exists an isomorphism between them which does not depend on choosing some basis (on either of the spaces). (suggestion by @DonAntonio)    Let $V$ be a finite dimensional Vectorspace with Basis $\beta = \lbrace v_1, \dots , v_n \rbrace $ (the infinite dimensional case is discussed at ). I do understand that although we make a choice of Basis here, it will later on somehow be obsolete (which I don't see why)  Define: \begin{align} \begin{matrix} V &amp; \overset{\Phi}{\longrightarrow}&amp; V^* &amp; \overset{\Phi^*}{\longrightarrow}&amp;V^{**} \\  \sum_{i=1}^n \lambda_i v_i &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^* &amp; \longmapsto &amp; \sum_{i=1}^n \lambda_i v_i^{**} \end{matrix} \end{align} Discussion: I know and have already shown that $V$ and $V^*$ are isomorphic (not canonically isomorphic!) to one another, meaning that $\lbrace v_1^*, \dots , v_n^* \rbrace$ defines a Basis for $V^*$. Although I did not show it I am at peace with the statement that the mapping $\Phi^*$ introduces another isomorphism between $V^*$ and $V^{**}$. It is the canonical isomorphism between $V$ and $V^{**}$ that bothers me.  My tutors reasoning: In the following I will use $\checkmark$ to highlight whether or not I understand something.   $v_i^{**} \in V^{**}=\hom(V^{*},k)  \checkmark$, sure nothing to add here $v_i^{**} (\sum_{j=1}^n \lambda_j v_j^{*})=\lambda_i \checkmark$, should be completely analogous to $v_i^*(v_j)=\lambda_i$ $\Phi^* \circ \Phi (v) =: \iota_v$ $$ \iota: \begin{cases} V &amp; \longrightarrow V^{**} \\ v &amp; \longmapsto \iota_v \end{cases}$$ where $\iota_v( \varphi)= \varphi(v)$ and $\varphi \in V^*$ is a linear functional. My tutor said that $\iota$ is 'suddenly' a canonical isomorphism, independent of the choice of the Basis $\beta$ because $\Phi$ and $\Phi^*$ are isomorphisms. This is the step where I understand nothing about.   What happened next (two more equations): I told my tutor that I don't see why $\iota$ is independent of a basis choice, because at $\Phi^* \circ \Phi= \iota$ we still make a choice for a basis at $\Phi$ namely $\beta$. My tutor said that this seems to be the case at first sight and made two more calculations which I in fact "understand": $$ \Phi^* \circ \Phi \left( \sum_{i=1}^n \lambda_i v_i \right) ( \varphi) = \Phi^* \left( \sum_{i=1}^n \lambda_i v_i^* \right) ( \varphi) = \left( \sum_{i=1}^n \lambda_i v_i^{**} \right)(\varphi) \\ = \sum_{i=1}^n \sum_{j=1}^n \lambda_i \mu_i \underbrace{v_i^{**}(v_j^*)}_{= \delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ I understand this calculation, it's mainly an application of the above introduced definitions and the Kronecker-Delta. Apparently I am supposed to have an "aha" moment here, which unfortunately didn't occur yet.  The next calculation he did was $$ \varphi (v) = \left(\sum_{j=1}^n \mu_j v_j^{*} \right) \left( \sum_{i=1}^n \lambda_i v_i \right)= \sum_{j=1}^n \sum_{i=1}^n \mu_j \lambda_i \underbrace{v_j^*( v_i)}_{\delta_{ij}} = \sum_{i=1}^n \lambda_i \mu_i $$ which evidently gives the same result as above. I would appreciate some wording on how those two equations (which calculations I understand) help me to see why there appears to be a canonical isomorphism between $V$ and $V^{**}$
115553
I would appreciate a hint on the following problem:     Let $V$ be a finite dimensional vector space over $F$. There are two scalar products such that: $$ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=0 \implies \langle v,w\rangle_2=0\Big) $$ Show that $$ \exists \ c \in F \ \ \forall \ w,v \in V \ \Big(\langle v,w\rangle_1=c \langle v,w\rangle_2\Big) $$   I have tried to define an orthonormal basis with respect to $\langle \cdot,\cdot\rangle_1$ hoping that the transformation matrices would be different by a constant, yet it seems to lead nowhere.
85458
So the question asks: Let A be a $2\times 2$ matrix, and suppose that there exist scalars $λ_1,λ_2$ in$\mathbb{R}$ and nonzero vectors $v_1,v_2$ in $\mathbb{R}$ such that $Av_1=λ_1v_1, Av_2=λ_2v_2$, and $λ_1$ does not equal to $λ_2$.Prove that the vectors $v_1,v_2$ form a basis of $R^2$.   So so far I got:   As the eigenvector $v_1,v_2$ are corresponding to two different eigen values$λ_1,λ_2$ respectfully.   Therefore the eigenvectors $v_1,v_2$ are linearly independent.   And since $v_1,v_2$ spans A in $\mathbb{R}$.   Hence,  $v_1,v_2$ forms a basis of $\mathbb{R}$.   Does this proof look right? Do I need to write a linear combination of $v_1,v_2$ to show they span A?
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.
25950
On an infinite dimensional, do there exist two norms which are not equivalent?  I actually know that on a infinite dimensional space, the number of inequivalent norms are $ 2^{dimX} $ However, I want to know if we can explicitly construct two inequivalent norms on any given space.
120705
Consider a finite-dimensional vector space $V$ over the field $\mathbb{F}$ with a linear operator $T.$ Prove that we have $\operatorname{rank}(T^3) + \operatorname{rank}(T) \geq 2 \operatorname{rank}(T^2).$   First, let us assume that $\dim_{\mathbb{F}}(V) = n.$ Our immediate observations include that $T^3(V) \subseteq T^2(V) \subseteq T(V)$ so that $\operatorname{rank}(T^3) \leq \operatorname{rank}(T^2) \leq \operatorname{rank}(T)$ and $\operatorname{rank}(T^3) \geq \operatorname{rank}(T) + \operatorname{rank}(T^2) - n$ by Sylvester's Rank Inequality. Can we do some manipulations with what we have to get the desired inequality, or do we require something else?
40319
Let $V,W$ be normed vector spaces, and $L(V,W)$ be the space of bounded linear operators. Usually I would only see the statement "If $W$ is Banach, then $L(V,W)$ is Banach.". But  writes that there is a converse: "If $L(V,W)$ is Banach, and if $V$ is non-trivial, then $W$ is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof.  I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in $W$, and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy)
123394
Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Prove that $V$ has a subspace of dimension $r$ for each $0 \le r \le n$.  Is this as simple as saying that $V$ has a basis of $n$ elements, and then take $r$ elements from this basis and this will generate an $r$-dimensional subspace? Or am I missing something? It seems too simple.  Thanks
32156
I want to prove the fact that  If $W\subseteq V$ is a vector subspace of $V$ than   $dim(W)=dim(V)=n \implies W=V$   Without using the fact that if $dim(V)=n$ any set of $n$ indipendent vectors in $V$ is necessarily a basis of $V$, neither the fact that a basis of $\mathscr{L}=(\vec{v_1},...,\vec{v_n})$ is necessarily the set $\big\{ \vec{v_1},...,\vec{v_n} \big\}$. (Since these two facts are consequences of the proof).  By hypotesis I have that $W\subseteq V$ which means that all the vectors in $W$ are in $V$. I must prove the opposite inclusion $V\subseteq W$: all the vectors in $V$ are a linear combination of $n$ vector of a basis, but I don't know how to prove that all the vectors of that basis are contained in $W$.  Can I say that they must belong to $W$ since $W$ has $n$ vectors in its basis too? (But I think in this way I'm assuming the facts listed above).  Can you help me to finish the proof?   Thanks for your help
31280
My question is:  Let $ V $ be a vector space (over $ \mathbb K\in\{\mathbb{R}, \mathbb{C}\} $), $ X,Y\subseteq V $ two subspaces equipped with norms $ \|\cdot\|_X, \|\cdot\|_Y $ such that $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$ are Banach spaces and $ D\subseteq X\cap Y$. If $ D $ is dense in $ (X,\|\cdot\|_X) $ and $(Y,\|\cdot\|_Y)$, is $ D $ also dense in $ X\cap Y $ equipped with $ \|\cdot\|:=\|\cdot\|_X + \|\cdot\|_Y $?  At first sight, it seemed very clear to me that this should be true. But I even fail to answer the following (possibly) easier question:  Let $ X $ be a vector space over $ \mathbb K $ equipped with two norms $ \|\cdot\|_1, \|\cdot\|_2 $ such that $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_1)$ are Banach spaces and $ D\subseteq X$. If $ D $ is dense in $ (X,\|\cdot\|_1) $ and $(X,\|\cdot\|_2)$, is $ D $ also dense in $ X $ equipped with $ \|\cdot\|:=\|\cdot\|_1 + \|\cdot\|_2 $?  The answer is yes, if $ \|\cdot\|_1, \|\cdot\|_2 $ are equivalent, so I tried to think about a counterexample using nonequivalent norms on a specific space and I also found a nice paper about nonisomorphic complete norms () but it didn't helped me so far to construct anything useful for my question.  Thanks for your help!
64073
There is a theorem in the course notes but the lecturer didn't given the proof:   If $A,B\in M_n(K)$, where $K$ is a field. Then $A\sim B$ if and only if they represent the same linear transformation $T$ of an $n$-dimensional vector space $V$ over $K$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$. i.e.,  $$ A\sim B\iff A=[T]_\mathcal{B},\ B=[T]_{\mathcal{C}}. $$  I know how to prove the '$\Leftarrow$' but I dont think I can prove the other direction '$\Rightarrow$'.   Could please somebody help me?
4487
Can you read my proof and tell me if it's correct? Thanks.  Let $V$ be a vector space over a complete topological field say $\mathbb R$ (or $\mathbb C$) with $\dim(V) = n$, base $e_i$ and norm $\|\cdot\|$. Let $v_k$ be a Cauchy sequence w.r.t. $\|\cdot\|$. Since any two norms on a finite dimensional space are equivalent, $\|\cdot\|$ is equivalent to the $l^1$-norm $\|\cdot\|_1$ which means that for some constant $C$, $\varepsilon &gt; 0$, $k,j$ large enough, $$ \varepsilon &gt; \|v_j - v_k\| \geq C \|v_j - v_k\|_1 e_i= C \sum_{i=1}^n |v_{ji} - v_{ki}| \geq |v_{ji} - v_{ki}|$$ for each $1 \leq i \leq n$. Hence $v_{ki}$ is a Cauchy sequence in $\mathbb R$ (or $\mathbb C$) for each $i$. $\mathbb R$ (or $\mathbb C$) is complete hence $v_i = \lim_{k \to \infty} v_{ki} $ is in $\mathbb R$ (or $\Bbb C$) for each $i$. Let $v = (v_1, \dots , v_n) = \sum_i v_i e_i$. Then $v$ is in $V$ and $\|v_k - v\| \to 0$:  Let $\varepsilon &gt; 0$. Then $$ \|v_k - v\| \leq C \|v_k - v\|_1 = C \sum_{i=1}^n |v_{ki} - v_i| \leq C^{'}n \varepsilon$$  for $k$ large enough.
102663
Suppose $A: n \times n$ and $B: n \times m$ and that $A$ is   invertible.      Prove that $\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}$ is   consistent is equivalent to $\forall \vec{b} \in \mathbb{R}^n, (AB)  \vec{x} = \vec{b}$ is consistent.   I'm assuming this boils down to proving that $B$ and $AB$ are equivalent expressions, but I can't see why this would be the case. Why does $A$ being invertible make the two statements equivalent (that is, why is $A$ invertible required)?
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.
44833
The $p$-norm on $\mathbb R^n$ is given by $\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$. For $0 &lt; p &lt; q$ it can be shown that $\|x\|_p\geq\|x\|_q$ (, ). It appears that in $\mathbb{R}^n$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$(),  $\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$ (). Geometrically, it is easy to see that opposite inequalities must hold in $\mathbb R^n$. For instance, for $n=2$ and $n=3$ one can see that for $0 &lt; p &lt; q$, the spheres with radius $\sqrt n$ with $\|\cdot\|_p$ inscribe spheres with radius $1$ with $\|\cdot\|_q$. It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $n=2$ it is easily proven (see below), but not for $n&gt;2$. So my questions are:  How can relation (3) be proven for arbitrary $n\,$? Can this be generalized into something of the form $\|x\|_{p} \leq C \|x\|_{q}$ for arbitrary $0&lt;p &lt; q\,$? Do any of the relations also hold for infinite-dimensional spaces, i.e. in $l^p$ spaces?   Notes: $\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$, hence  $=2\|x\|_{2}^{2}$ $\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$. This works because $|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$, but only because $(|x_{1}| - |x_{2}|)^2 \geq 0$, while for more than two terms $\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$ gives an inequality that never gives the right signs for the cross terms.
90442
I am trying to prove the matrix norm equivalence for norms 1, 2, $\infty$ and Frobenius.  I have managed to solve find the constants for $||.||_{1}$ and  $||.||_{2}$ but I cannot see how to continue if I want to prove the following:  $||A||_{\infty} \leq \sqrt{n}||A||_{2}$   and the ones for Frobenius:  $||A||_{F} \leq \sqrt{n}||A||_{1}$  $||A||_{F} \leq \sqrt{n} ||A||_{2}$  Any help is appreciated. Thank you for your time