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I have the following two tasks before me:

Let $\psi:V\to W$ be a linear Transformation and $V, W$ are finite-dimensional over $K$.
$u_1,\dots,u_m$ is a basis of ker $\psi$ and $w_1,\dots,w_n$ is a basis of Im $\psi$.
The Vectors $v_1,\dots,v_n \in V$ are choosen such that: $\psi(v_j)=w_j$ for $ j=1,\dots,n$

  1. Show that $u_1,\dots,u_m,v_1,\dots,v_n$ are linear independent in $V$

  2. Show that $u_1,\dots,u_m,v_1,\dots,v_n$ is a generating system of $V$.

What I have so far:

  1. Since {$w_1,\dots,w_n$} is a basis of Im $\psi$ and thus in particular linear independent in $W$$\implies$ $v_1,\dots,v_n$ are l.i. in $V$.
    Also, $u_1,\dots,u_m$ are linear independent in $V$, since ker $\psi$ is a subspace of $V$. We now need to show that $u_1,\dots,u_m,v_1,\dots,v_n$ is still linear independent: Assume $u_1,\dots,u_m,v_1,\dots,v_n$ was linear dependent, it then follows that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n)$ was also linear dependent. That however, can't be true since $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_n) \leftrightarrow 0,\dots,0,w_1,\dots,w_n \leftrightarrow w_1,\dots,w_n$ (Here I am not so sure if I can just remove all the zero vectors like I did.)

  2. Again we can use the fact that ker $\psi$ is a subspace of $V$ and expand its base into a base for $V$: {$u_1,\dots,u_m,v_1,\dots,v_k$} If we manage to show that $k=n$ we have proven that $u_1,\dots,u_m,v_1,\dots,v_n$ is indeed a generating system. We can achieve this by observing that $\psi(u_1),\dots,\psi(u_m),\psi(v_1),\dots,\psi(v_k) \leftrightarrow w_1,\dots,w_k$ Since ${w_1,\dots,w_k}$ must be a generating system of Im $\psi$ we simply need to prove linear independence to conclude that $k=n$.

The more I think about it the more confused I am. As pointed out I am especially confused about just removing the zero vectors to show linear independence. Also if ${w_1,\dots,w_k}$ is indeed linear independent, that means every base of $V$ gets mapped to a base of Im $\psi$?

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2 Answers 2

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For part I start like this,

  1. Suppose $\{u_1, \cdots, u_m, v_1,\cdots, v_n\}$ is linearly dependent.
  2. Then $a_1u_1 + \cdots +a_m u_m + b_1v_1 + \cdots + b_n v_n = \bf{0}$ and at least one of these scalars is nonzero.
  3. Now take $T$ of the statement above and you'll get $b_1 w_1 + \cdots + b_n w_n = T(\bf{0})$

What can you now conclude?

For part II, first employ rank - nullity, i.e dim$(V)$ = rk$(A)$ + nullity$(A) = n+m$. Now use the fact that the dimension of $V$ is the same as the number of vectors in the lin indep. set - we have a result that tells you automatically the lin indep. set is also spans, hence it's a basis.

Comment : I know the second part is not by brute force, but we can pair two theorems and get it done pretty quickly.

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Your problem is that there is no prompt to say that the dimension of $V$and $W$ is same.So you can not think that if $e$ is the basis of $V$ so the $\psi(e)$ is also the basis of $W$.

I will try to give a detailed explanation:

1.$u_1,u_2,\dots,u_m$ is the bais of $\ker{\psi}$, so you can find $\forall k_1,K_2,\dots,k_m \in K$, that $\psi(k_1u_1+k_2u_2+\dots+k_mu_m)=0$.But we know that in $V$,only $0=0$,but through $\psi$, many $u=0$ in the $V$ will become $\psi(u)=0$,if as you think, only $\psi(0)=0$ in the W, obviously it's wrong.

2.The similar as 1....

Ok,I know my explanation is terrible...

The key point is that if $e$ is the basis of $V$, but $\psi(e)$ is not the basis of $W$,cause you don't know that they have a same dimension, so your inference in 1. 2. is not true fundamentally...

We can have a extreme example,we let $V$to be $K=\mathbb{R}^2$ and $W$ to be $\{0\}$, so the $\psi$,$\forall x \in K, \psi(x)=0$

We can figure that the on of the basis of $V$ is $((0,1),(1,0))$,but the basis of $W$...

That's why you fell confused

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  • $\begingroup$ Thanks I will have to go through that again, but I think I get what you are saying.. I was wrong in assuming that $\psi$ was mapping a standardbasis onto a standardbasis? $\endgroup$ Jan 10, 2021 at 13:51
  • $\begingroup$ Yes,I just mean that $\endgroup$
    – Hovard
    Jan 10, 2021 at 13:56
  • $\begingroup$ @ghupftwieghatscht $\endgroup$
    – Hovard
    Jan 10, 2021 at 14:02
  • $\begingroup$ what would a transformation of an n dimensional vectorspace into an n dimensional vectorspace that maps a standardbasis to a standardbasis even be? this seems like just renaming the elements? $\endgroup$ Jan 10, 2021 at 14:20
  • $\begingroup$ Ah....Most like that,but also exists a special case that zero mapping,and the else,may be like you say(I'm not very sure.....you can try to prove it@ghupftwieghatscht $\endgroup$
    – Hovard
    Jan 11, 2021 at 11:31

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