I need to prove $(-\vec{v})+\vec{v}=\vec{0}$ using only the axioms of vector space addition and multiplication, except the commutativity axiom.   I have tried using this approach $\vec{0}=\vec{v}+(-\vec{v})$ but I get stuck and can't find a way to get the $(-\vec{v})$ first.
123760
I'm trying to prove commutativity of addition for vector spaces, using the axioms for vector spaces. Apparently commutativity can be proven! Im having trouble getting a good feel for what is allowed and what is not. Here's my work so far:  $u+v+u+v = 2(u+v) = 2u + 2v = u+u+v+v = u+(u+v)+v$  Here I just wanna claim that $u+(v+u)+v = u+(u+v)+v$   $\Rightarrow -u+u+(v+u)+v+(-v) = -u+u+(u+v)+v+(-v)$ : here im just adding -u to the right, and -v to the left. Question: is this "adding to both sides" really legit in this context? Why?   Quick help proof: $-v+v = (-1)v+(1)v = (-1+1)v = 0v = 0 = v-v$  And another: $ 0+v = v+(-v) + v = (1)v + (-1)v + v = (1-1)v + v = v = v+0$  We have $0 + (u+v) + 0 = 0+(v+u)+0 \Rightarrow u+v = v+u$  This feels ugly and not at all elegant, especially the great leap "add -u to both sides" feels completely out of place. Do I need more lemmas? Is there a more elegant way?   //not homework or anything, just for my own pleasure, feel free to provide theory, as it is more insightful than solutions. :) Thanks!   EDIT: corrected notation a little.
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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?
67194
I need to prove for all real numbers $a$, $-(-a) = a$ using only the following axioms:     Thanks to many members of the Mathematics Stackexchange Community, I have the following proof worked out:  Theorem: The Additive Inverse Identity is Unique  $ ( \forall a,b,c \in \mathbb{R} )(a+b=0) \land (a+c=0) \\ ( \forall a,b,c \in \mathbb{R} )(a+b=a+c) \\ ( \forall a,b,c \in \mathbb{R} )(b=c) \\ \text{QED} \\ $  Theorem $\textbf{a} \cdot \textbf{0 = 0}$  $ \begin{align} a \cdot 0 &amp;= a \cdot 0 \\ a \cdot (0 + 0) &amp;= a \cdot 0 \\ a \cdot 0 + a \cdot 0 &amp;= a \cdot 0 \\ a \cdot 0 + a \cdot 0 + (-a) \cdot 0 &amp;= a \cdot 0 + (-a) \cdot 0 \\ a \cdot 0 &amp;= 0 \\ &amp;\text{QED} \\ \end{align} $  Theorem: $-\textbf{1} \cdot \textbf{(a) = (}-\textbf{a)}$  $ \begin{align} a \cdot 0 &amp;= 0 \\ a \cdot \left[ 1 + (-1) \right] &amp;= a + (-a) \\ 1 \cdot a + (-1) \cdot a &amp;= a + (-a) \\ -1 \cdot (a) &amp;= (-a) \\ &amp;\text{QED} \\ \end{align} $  Theorem: $-\textbf{(}-\textbf{a) = a}$  $ \begin{align} 0 &amp;= 0 \\ -a \cdot 0 &amp;= 0 \\ -a \cdot \left[ 1 + (-1) \right] &amp;= 0 \\ -a \cdot 1 + -a \cdot (-1) &amp;= 0 \\ -a + \left[ -(-a) \right] &amp;= 0 \\ -(-a) &amp;= a \\ \text{QED} \\ \end{align} $  Does everything look alright? Have I missed anything?  Also--why is it necessary to show that the additive inverse is unique?  Many thanks in advance.
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I have just started self-studying Linear Algebra after a 3 year hiatus from mathematics (in an extended neuroscience foray), and I want to ensure that I am sufficiently verifying commutativity of complex numbers by using the properties of real numbers.  My attempts...  (1) Commutativity of Addition:   Verify that $w + z = z + w$ for all $w, z \in \mathbb{C}$.  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{16mm} = (a + c) + (bi + di) \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{19mm} = (c + a) + (di + bi) \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{13mm} = c + (a + di) + bi \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = c + (di + a) + bi \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{17mm} = (c + di) + (a + bi) \hspace{11mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{13mm} = z + w \hspace{34mm}\text{by definition of }\mathbb{C}$$  $$\blacksquare$$  This seems a bit verbose to me - is it acceptable to combine steps? For example, rather than:  $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{13mm} = a + (bi + c) + di \hspace{10mm}\text{by associativity in }\mathbb{R}$$ $$\hspace{17mm} = a + (c + bi) + di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ can I instead write: $$w + z = (a + bi) + (c + di) \hspace{10mm}\text{by definition of }\mathbb{C}$$ $$\hspace{48mm} = a + (c + bi) + di \hspace{10mm}\text{by associativity and commutativity in }\mathbb{R}$$?  (2) Commutativity of Multiplication:  Verify that $wz = zw$ for all $w, z \in \mathbb{C}$.  $$\hspace{-14mm}wz = (a + bi)(c + di) \hspace{24mm}\text{by definition of }\mathbb{C}$$ $$\hspace{4mm} = a \cdot c + a \cdot di + bi \cdot c + bi \cdot di \hspace{10mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{9mm} = a \cdot c + bi \cdot c + a \cdot di + bi \cdot di \hspace{10mm}\text{by commutativity in }\mathbb{R}$$ $$\hspace{35mm} = c \cdot a + c \cdot bi + di \cdot a + di \cdot bi \hspace{10mm}\text{by multiplicative commutativity in }\mathbb{R}$$ $$\hspace{1mm} = c(a + bi) + di(a + bi) \hspace{20mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{5mm} = (c + di)(a + bi)\hspace{34mm}\text{by distribution in }\mathbb{R}$$ $$\hspace{4mm} = zw \hspace{55mm}\text{by distribution in }\mathbb{R}$$ $$\blacksquare$$  This also seems a bit wonky to me. Any and all suggestions on clearing up my verification, reasoning, etc. would be greatly appreciated!
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It is given in question that $x+a=(a.x)b$ and it is stated that a,b and x are vectors.It is also stated that $a.b$ not equal to $1$.The question ask to express $x$ in term $a$ and $b$,how do you this?
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I've read the following formula in :  Given finite dimensional vector spaces $V_i$ and an  $\cdots\rightarrow V_i\rightarrow V_{i+1}\rightarrow\cdots$, we have  $$ \sum_{n\in 2\mathbb{Z}}\dim V_n = \sum_{n\in 2\mathbb{Z}+1}\dim V_n $$  Is there a name for this theorem? Could anyone please tell me where to find a proof of this in the literature?  Thank you!
121835
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!  Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.  Claim:  If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.  Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.   Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.  Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} =  \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$  Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!  The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.  Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.
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I want to see if my proof is true or I thought very trivially?    If $H$ is a subspace of a finite dimensional vector space $V$, show there is a subspace $K$ such that $H\cap K=0$ and $H+K=V$    So far I have tried :    $H\subseteq V$ is a subspace $\Rightarrow\,\exists\;K=(V-H)\subseteq V$ $K$ is a subspace because it's the sum of two subspace $V$ and $(-H)$ Now we should prove that $H\cap K=0$ We assume $H\cap K\neq0\Rightarrow\exists\,\vec u\neq0\,,\,\vec u\in H\cap K$ $\vec u\in H\cap K\Rightarrow \begin{cases}\vec u\in H\\\vec u\in K\,\Rightarrow\vec u\in V-H\Rightarrow\vec u\in V,\,\vec u\notin H\end{cases}$ That's a contradiction so our hypothesis was wrong and $\vec u=0$
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I'm looking for the shortest and the clearest proof for this following theorem:  For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , I need to show $V= \ker T^n \oplus  $ Im $T^n$.  Any hints? I don't know where to start from.  Thank you.
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So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.  The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
60912
We were ask to prove that $|A \cup B| = |A| + |B| - |A \cap B|‫‪$. It was easy to prove it using a Venn diagram, but I think we might be expected to do if more formally. Is there a formal way?
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I know that any subset of V with more than k vectors is always linearly dependent, but how do I include matrix A into the vectors given to prove the statement?   Also, for part (b), does the statement mean that Ax=0 only has the trivial solution since V intersects span{u1, u2, ..., uk}={0}?  Any help would be appreciated! Thank you!
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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 $T$.
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If A is invertible, prove that $\lambda \neq 0$, and $\vec{v}$ is also an eigenvector for $A^{-1}$, what is the corresponding eigenvalue?  I don't really know where to start with this one. I know that $p(0)=det(0*I_{n}-A)=det(-A)=(-1)^{n}*det(A)$, thus if both $p(0)$ and $det(0) = 0$ then $0$ is an eigenvalue of A and A is not invertible. If neither are $0$, then $0$ is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove $\vec{v}$ is also an eigenvector for $A^{-1}$ and how to find a corresponding eigenvalue.
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I'm not sure how to go about this proof. I just need help getting started. Is there a way to prove it algebraically?
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I was wondering why do some people use redundant axioms in definitions?  If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free.   Why do we then assume it?   Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R.
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Why isn't $V= V_1 + V_2$? $V=V_a - V_c = V_a - V_b + V_b - V_c$, $V_a - V_b= V_1$ and $V_b - V_c = V_2$ Doesn't that prove that $V = V_1 + V_2$? Regardless of $V_3$, If i'm wrong , is there a way to obtain $V$ in terms of $V_1$, $V_2$ and $V_3$