MHB Kernel of Linear Map: Show $\ker \phi$ Equation

mathmari
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Hey! :o

Let $1\leq n,m\in \mathbb{N}$, $V:=\mathbb{R}^n$ and $(b_1, \ldots , b_n)$ a basis of $V$. Let $W:=\mathbb{R}^m$ and let $\phi:V\rightarrow W$ be a linear map.
Show that $$\ker \phi =\left \{\sum_{i=1}^n\lambda_ib_i\mid \begin{pmatrix}\lambda_1\\ \vdots \\ \lambda_n\end{pmatrix}\in \textbf{L}(\phi (b_1), \ldots , \phi (b_n))\right \}$$

I have done the following:

Let $v\in V$. Since $(b_1, \ldots , b_n)$ is a basis of $V$, we have that $\displaystyle{v=\sum_{i=1}^n\lambda_ib_i}$.

Then we have that $$v\in \ker \phi \iff \phi (v)=0_W \iff \phi \left (\sum_{i=1}^n\lambda_ib_i\right )=0_W \iff \sum_{i=1}^n\lambda_i\phi (b_i)=0_W$$

Is this correct so far? (Wondering)
 
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mathmari said:
Is this correct so far?

Hey mathmari!

Yep. Correct. (Nod)

Btw, what is $\mathbf L$? (Wondering)
 
Klaas van Aarsen said:
Yep. Correct. (Nod)

Btw, what is $\mathbf L$? (Wondering)

The definition is: $$\mathbf L=\left \{(\lambda_1, \ldots , \lambda_k)^T\in \mathbb{R}^k\mid \sum_{i=1}^k\lambda_iv_i=0\right \}$$

So we get $$v\in \ker \phi \iff \phi (v)=0_W \iff \phi \left (\sum_{i=1}^n\lambda_ib_i\right )=0_W \iff \sum_{i=1}^n\lambda_i\phi (b_i)=0_W\iff (\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$$

But to get the desired result it has to be $v=(\lambda_1, \ldots , \lambda_n)^T$, or not? So did we have to take at the beginning this assumption? (Wondering)
 
mathmari said:
The definition is: $$\mathbf L=\left \{(\lambda_1, \ldots , \lambda_k)^T\in \mathbb{R}^k\mid \sum_{i=1}^k\lambda_iv_i=0\right \}$$

So we get $$v\in \ker \phi \iff \phi (v)=0_W \iff \phi \left (\sum_{i=1}^n\lambda_ib_i\right )=0_W \iff \sum_{i=1}^n\lambda_i\phi (b_i)=0_W\iff (\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$$

Yep. (Nod)

mathmari said:
But to get the desired result it has to be $v=(\lambda_1, \ldots , \lambda_n)^T$, or not? So did we have to take at the beginning this assumption?

No. $(\lambda_1, \ldots , \lambda_n)^T$ is not an element of $V$, is it? And it shouldn't be. (Shake)
It's not an element of the kernel either.
Don't we already have the desired result? (Wondering)
What do you think is missing?
 
Klaas van Aarsen said:
Yep. (Nod)
No. $(\lambda_1, \ldots , \lambda_n)^T$ is not an element of $V$, is it? And it shouldn't be. (Shake)
It's not an element of the kernel either.
Don't we already have the desired result? (Wondering)
What do you think is missing?

Ohh now I think I got it. I thought we have to show that $v\in \ker \phi \iff v\in L$, but $(\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$ is just the condition that $v$ is in $\left \{\sum_{i=1}^n\lambda_ib_i\mid \begin{pmatrix}\lambda_1\\ \vdots \\ \lambda_n\end{pmatrix}\in \textbf{L}(\phi (b_1), \ldots , \phi (b_n))\right \}$, right? (Wondering)

So from $$v\in \ker \phi \iff \phi (v)=0_W \iff \phi \left (\sum_{i=1}^n\lambda_ib_i\right )=0_W \iff \sum_{i=1}^n\lambda_i\phi (b_i)=0_W\iff (\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$$ we have that $v=\sum_{i=1}^n\lambda_ib_i$ is in the kernel iff $(\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$ which means that $v=\sum_{i=1}^n\lambda_ib_i$ is contained in $\left \{\sum_{i=1}^n\lambda_ib_i\mid \begin{pmatrix}\lambda_1\\ \vdots \\ \lambda_n\end{pmatrix}\in \textbf{L}(\phi (b_1), \ldots , \phi (b_n))\right \}$.

Is this correct? (Wondering)
 
mathmari said:
Ohh now I think I got it. I thought we have to show that $v\in \ker \phi \iff v\in L$, but $(\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$ is just the condition that $v$ is in $\left \{\sum_{i=1}^n\lambda_ib_i\mid \begin{pmatrix}\lambda_1\\ \vdots \\ \lambda_n\end{pmatrix}\in \textbf{L}(\phi (b_1), \ldots , \phi (b_n))\right \}$, right?

So from $$v\in \ker \phi \iff \phi (v)=0_W \iff \phi \left (\sum_{i=1}^n\lambda_ib_i\right )=0_W \iff \sum_{i=1}^n\lambda_i\phi (b_i)=0_W\iff (\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$$ we have that $v=\sum_{i=1}^n\lambda_ib_i$ is in the kernel iff $(\lambda_1, \ldots , \lambda_n)^T\in \mathbf L$ which means that $v=\sum_{i=1}^n\lambda_ib_i$ is contained in $\left \{\sum_{i=1}^n\lambda_ib_i\mid \begin{pmatrix}\lambda_1\\ \vdots \\ \lambda_n\end{pmatrix}\in \textbf{L}(\phi (b_1), \ldots , \phi (b_n))\right \}$.

Is this correct?

Yep. All correct. (Nod)
 
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