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

• MHB
• mathmari

#### mathmari

Gold Member
MHB
Hey!

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)

Is this correct so far?

Hey mathmari!

Yep. Correct. (Nod)

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

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)

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)

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?

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)

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)