# Base of Kernel and pre-image of base of Image are linear independent and span V.

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 $$\psi$$?

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.

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

• 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? Jan 10, 2021 at 13:51
• Yes,I just mean that Jan 10, 2021 at 13:56
• @ghupftwieghatscht Jan 10, 2021 at 14:02
• 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? Jan 10, 2021 at 14:20
• 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 Jan 11, 2021 at 11:31