# 2 questions : Kernel/Image of a linear transformation and orthonormal basis

• Feb 10th 2009, 12:41 PM
arbolis
2 questions : Kernel/Image of a linear transformation and orthonormal basis
Hello MHF,
It is the continuation of the exercise I posted in my precedent thread.
So I recall, I have a vectorial subspace of $\mathbb{R}^4$, $W$. $W$ is defined to be $\{ (x,y,z,w)\in \mathbb{R}^4 \text{such that }2x-z+w=0, x-y-z=0 \}$.
I've found a basis of $W$ and a basis of $W^{\perp}$. They now ask me to find an orthonormal basis of $\mathbb{R}^4$ using the basis I found in the past items.
For $W$ I have that a basis is $\{ (0,1,-1,-1), (1,0,1,-1) \}$. While for $W^{\perp}$ I have that a basis is $\{ (2,0,-1,1), (1,-1,0,-1) \}$.
Oh wait... doesn't that mean that a basis of $\mathbb{R}^4$ is the union of the 2 basis I found? But it wouldn't necessarily be orthonormal especially because I have a factor 2 in the basis of $W^{\perp}$. So I don't see how I can answer the question.

Now the next question seems even much harder. I must find a linear transformation $T:\mathbb{R}^4 \to \mathbb{R}^4$ such that $\ker (T)=W$ and $\Im (T)=W^{\perp}$.
My attempt : $T(x,y,z,w)=0\Leftrightarrow T(x,y,x-y,-x-y)=0$ and I'm at a loss.
• Feb 10th 2009, 06:09 PM
NonCommAlg
Quote:

Originally Posted by arbolis
Hello MHF,
It is the continuation of the exercise I posted in my precedent thread.
So I recall, I have a vectorial subspace of $\mathbb{R}^4$, $W$. $W$ is defined to be $\{ (x,y,z,w)\in \mathbb{R}^4 \text{such that }2x-z+w=0, x-y-z=0 \}$.
I've found a basis of $W$ and a basis of $W^{\perp}$. They now ask me to find an orthonormal basis of $\mathbb{R}^4$ using the basis I found in the past items.
For $W$ I have that a basis is $\{ (0,1,-1,-1), (1,0,1,-1) \}$. While for $W^{\perp}$ I have that a basis is $\{ (2,0,-1,1), (1,-1,0,-1) \}$.

Oh wait... doesn't that mean that a basis of $\mathbb{R}^4$ is the union of the 2 basis I found? yes!

let $u_1, u_2, u_3, u_4$ be as i found in here. the set $\{u_1,u_2,u_3,u_4 \}$ is an orthogonal basis for $\mathbb{R}^4.$ so if you put $v_j=\frac{u_j}{||u_j||}, \ 1 \leq j \leq 4,$ then $\{v_1,v_2,v_3,v_4 \}$ would be an orthonormal basis for $\mathbb{R}^4.$

Quote:

Now the next question seems even much harder. I must find a linear transformation $T:\mathbb{R}^4 \to \mathbb{R}^4$ such that $\ker (T)=W$ and $\Im (T)=W^{\perp}$.
My attempt : $T(x,y,z,w)=0\Leftrightarrow T(x,y,x-y,-x-y)=0$ and I'm at a loss.
let $B_1=\{u_1,u_2,u_3,u_4 \}$ and $B_2$ be the standard basis for $\mathbb{R}^4.$ the linear transformation you're looking for in basis $B_1$ simply sends $u_1,u_2$ to 0 and $u_3,u_4$ remain fixed, i.e.

$T(c_1u_1+c_2u_2+c_3u_3+c_4u_4)=c_3u_3+c_4u_4.$ thus $[T]_{B_1}=\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$ now use what you've learned about "change of basis" to find $S=[T]_{B_2}.$ then the answer to
your question would be: $T(\bold{x})=S \bold{x},$ where $\bold{x}=\begin{bmatrix}x \\ y \\ z \\ w \end{bmatrix}.$
• Feb 10th 2009, 06:48 PM
math2009
W basis $\{\vec{v}_1= (0,1,-1,-1), \vec{v}_2=(1,0,1,-1) \}$ are orthonal each other. we only compute orthnormal (unit vector),
$\vec{u}_1=\frac{\vec{v}_1}{\|\vec{v}_1 \|}=\frac{1}{\sqrt{3}}\begin{bmatrix} 0 \\ 1 \\ -1 \\ -1 \end{bmatrix}
$
, $\vec{u}_2=\frac{\vec{v}_2}{\|\vec{v}_2 \|}=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 0 \\ 1 \\ -1 \end{bmatrix}
$

**********
$W^{\perp} basis$ are not orthnormal. We can compute by formular:
$\vec{u}_1=\frac{\vec{v}_1}{\|\vec{v}_1 \|}$
$\vec{v}^{\perp}_2=\vec{v}_2 - \vec{v}_2^{\parallel}=\vec{v}_2-(\vec{u}_1.\vec{v}_2)\vec{u}_1$
$\vec{u}_2=\frac{\vec{v}^{\perp}_2}{\|\vec{v}^{\per p}_2 \|}$

$\vec{u}_1=\frac{1}{\sqrt{6}}\begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix}$ , $\vec{u}_2=\frac{1}{\sqrt{6}}\begin{bmatrix} 0 \\ 2 \\ 1 \\ 1 \end{bmatrix}$

**********
In previous question "Give a basis of a subspace", $Dim(ker(T))+Dim(Im(T))=2+2=4 =Dim(R^4)$, it mean T exist. And it is rotation matrix. It's more complex.
• Feb 11th 2009, 12:34 AM
math2009