Constraints :  Basis and dimensions are not introduced yet Matrix and system of equations can't be used  I am aware that this might possibly be a duplicate of the , but I wanted to take this question further and see if any of you have come across an alternative proof than the one mentioned in the first answer. Btw, I am new to this platform, so in case I am making a mistake by posting this, please let me know.
121057
So I am in an introductory linear algebra(with proofs) class, and I came across the following theorem:  If $S$ is a subspace of a vector space $V(F)$ and is generated by $\{a_1,...,a_n\}$ then any subset $\{b_1,...,b_m\}$ of $S$ with $m&gt;n$ is linearly dependent.   The proof in the textbook is done through induction (as most proofs regarding finite sets are) but I really do not follow it and I feel there is a simpler way of doing it. A hint would be appreciated.
27278
Some days ago, I was thinking on a problem, which states that $$AB-BA=I$$ does not have a solution in $M_{n\times n}(\mathbb R)$ and $M_{n\times n}(\mathbb C)$. (Here $M_{n\times n}(\mathbb F)$ denotes the set of all $n\times n$ matrices with entries from the field $\mathbb F$ and $I$ is the identity matrix.) Although I couldn't solve the problem, I came up with this problem:  Does there exist a field $\mathbb F$ for which that equation $AB-BA=I$ has a solution in $M_{n\times n}(\mathbb F)$?  I'd really appreciate your help.
115054
$$\begin{align}\mathrm D &amp;= \left|\begin{matrix} (b+c)^2 &amp; a^2 &amp; a^2 \\ b^2 &amp; (a+c)^2 &amp; b^2 \\ c^2 &amp; c^2 &amp; (a+b)^2 \end{matrix}\right|\\ &amp;= (a+b+c)\left|\begin{matrix} b+c - a &amp; a^2 &amp; a^2 \\ b - a -c &amp; (a+c)^2 &amp; b^2 \\ 0 &amp; c^2 &amp; (a+b)^2 \end{matrix}\right| \\ &amp;= (a+b+c)^2\left|\begin{matrix} b+c - a &amp; 0 &amp; a^2 \\ b - a -c &amp; a+c - b &amp; b^2 \\ 0 &amp; c - a-b &amp; (a+b)^2 \end{matrix}\right|\\ &amp;= (a+b+c)^2\left|\begin{matrix} b+c - a &amp; 0 &amp; a^2 \\ 0 &amp; a+c - b &amp; b^2 \\ c - a-b &amp; c - a-b &amp; (a+b)^2 \end{matrix}\right|\end{align}$$  Can $\rm D$ be further simplified without expanding ?  I feel it should be because this was competition question.
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Consider the system of simultaneous linear equations  \begin{align} a_{11} &amp;x_1 &amp;&amp;\; + \;&amp;&amp; a_{12} x_2   &amp;&amp;\; + \cdots + \;&amp;&amp; a_{1n} x_n &amp;&amp;\; = \;&amp;&amp;&amp; 0 \\ a_{21} &amp;x_1 &amp;&amp;\; + \;&amp;&amp; a_{22} x_2   &amp;&amp;\; + \cdots + \;&amp;&amp; a_{2n} x_n &amp;&amp;\; = \;&amp;&amp;&amp; 0 \\ \vdots\;&amp;\; &amp;&amp;     &amp;&amp; \;\;\;\vdots\;\;\; &amp;&amp;                &amp;&amp; \;\;\;\vdots\;\;\; &amp;&amp;     &amp;&amp;&amp; \vdots \\ a_{m1} &amp;x_1 &amp;&amp;\; + \;&amp;&amp; a_{m2} x_2   &amp;&amp;\; + \cdots + \;&amp;&amp; a_{mn} x_n &amp;&amp;\; = \;&amp;&amp;&amp; 0 \\ \end{align}  for constants $a_{ij}$ and unknowns $x_j$ , in the case where $m &lt; n$ prove that there  must exist infinitely many solutions to this system of equations.  Okay, so it's pretty easy to prove that this set of equations must have solutions because we have m equations and n values, we x_1=x_n=0 is our trivial solution and if we have more than this one solution, then out null space will be greater than {0} which implies we don't have an injective set of equations. As m is less than n we know the system is not injective, which tells us there must be non-zero solutions. But I want to show these solutions are infinite which means for x_n solutions, the solution should include a parameter. How can I go about this?
116727
I tried solving the above question but was unable to prove it. I used Descartes rule of sign, factorisation techniques, and many other things but could not figure out the solution.
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I'd like a hint rather than a full solution.  The problem I am considering is the following:     $X$ is an $n\times m$ matrix $Y$ is $m\times n$      Show that      $(I - XY)^{-1}\cdot X = X\cdot(I - YX)^{-1}$   The first $I$ is of dimension $n$ and the second $I$ is of dimension $m$.
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Got this as homework and I don't know how to tackle this. Help please!  Prove that the isometries of $\mathbb{S}^{n} \subset \mathbb{R}^{n+1}$, with the induced metric, are restrictions to $\mathbb{S}^{n}$ of the linear orthogonal transformations.
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I don't understand why this statement is wrong:     $V$ is a vector space, and $W$ is a subspace of $V$. $K$ is a basis of $V$. We can manage to find a subset of $K$ that will be a basis of $W$.   Sorry if my English is bad... and if you can show me an example of something that contradicts it, it'd be great.
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Do anybody know how matrices are built into ZFC theory ? I pretty have no idea how to build them from ZFC axioms. I would like a constructive proof of their existence if possible.  Thanks in advance.
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?
37567
A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&amp;0 \\ 0&amp;-1-\epsilon  \\   \end{pmatrix}$$ (a) exists for all $\epsilon &gt; 0$. (b) does not exist for any $\epsilon &gt; 0$. (c) exists for some $\epsilon &gt; 0$. (d) None of the above is true.  Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix. Could someone please give me a hint on how to go about this problem? Thank you very much for your help in this regard.
121890
I'd like to know if the following problem of elementary linear algebra is already solved / solvable.  For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?
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I'm just getting back into having to do linear algebra and I am having some trouble with some elementary questions, any help is much appreciated.  Suppose that $A = [a_{ij}]$ is an $m\times n$ matrix and $B = [b_{ij}]$ and is an $n\times m$ matrix and $BA = I$ the identity matrix that is $n \times n$.  Show that if for some $b \in \Bbb{R}^m$ the equation $Ax = b$ has a solution, then the solution is unique.
55046
I was wondering why it is almost impossible to find a geometrical explanation of why adding two linear equations helps us to find a solution of a system of linear equations? Am I right that adding two linear equations will result in an equation of a line which will pass through a point where two linear equations intersect?  If it is right then I completely don't understand why such a crucial point in understanding how to solve systems of equations is never properly taught even at a university level.
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One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.
102988
My question:      Is $e_1 \wedge e_2 + e_3 \wedge e_4 \in \bigwedge^2 V$ not completely decomposable if $e_1$, $e_2$, $e_3$, $e_4$ is a basis for $V$?
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So I'm trying to prove the classic dimension theorem for linear transformations. Here is what I have so far:  Given a linear transformation $T:V\rightarrow W$  Let the basis of $\mathrm{Im}T$ be $\{u_1,...,u_m\}$ where $\forall u\in V:T(u)$.  Let the basis of $\ker T$ be $\{v_1,..,v_k\}$ where $\forall v:T(v)=0$.  Lets prove $B=\{u_1,...,u_m,v_1,..,v_k\}$ is a basis of $V$:  (1) $B$ is linearly independent: Assume (falsely) that $B$ isn't linearly independent. Since $\{u_1,...,u_m\}$ and $\{v_1,..,v_k\}$ are both the basis of $\ker T$ and $\mathrm{Im}T$ repectively, they must each be linearly independent.If this is the case, $\exists b\in V:b\in \{v_1,..,v_k\}\cap \{v_1,..,v_k\}$ however this contradicts $\mathrm{Im}T\cap \ker T=\emptyset$ so $B$ must be linearly independent.  (2)$\mathrm{span}(B)=V$: Assume (falsely) that $b_2\in V| b_2\notin \mathrm{span}(B)$. If this is the case, then $b_2\notin \{\forall u\in V:T(u)\}\cap\{\forall v:T(v)=0\}$. This is impossible because $b_2\in V$ so $\mathrm{span}(B)=V$  Since (1) and (2) are true, according to basis identities, $B$ is a basis of $V$. This means that $\dim V=|B|=\dim\ker T+\dim\mathrm{Im}T$ QED