find the basis

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February 7th 2011, 07:00 PM
Taurus3

find the basis

Let L be the line spanned by v_1=[1, 1, 1]. Find a basis {v_2, v_3} for the plane perpendicular to L, and verify that B= {v_1,, v_2, v_3} is a basis for R3.

Let ProjL denote the projection onto the line L. Find the matrix B for ProjL with respect to the basis B.

Can you just help me get started with the problem. I'll do the rest on my own.
February 7th 2011, 07:48 PM
TheEmptySet

Quote:

Originally Posted by Taurus3

Let L be the line spanned by v_1=[1, 1, 1]. Find a basis {v_2, v_3} for the plane perpendicular to L, and verify that B= {v_1,, v_2, v_3} is a basis for R3.

Let ProjL denote the projection onto the line L. Find the matrix B for ProjL with respect to the basis B.

Can you just help me get started with the problem. I'll do the rest on my own.

Since $v_1$ is perpendicular to the plane you want to span it is the planes normal vector.

Since the equation of the plane must pass though the origin it will have the form
$x+y+z=0$

Can you finish from here?
February 7th 2011, 08:07 PM
Taurus3

ummm......actually no. Sorry. I mean I know how to find the eigenvalues and then the basis. But this question is weird.
February 7th 2011, 08:19 PM
TheEmptySet

Quote:

Originally Posted by Taurus3

ummm......actually no. Sorry. I mean I know how to find the eigenvalues and then the basis. But this question is weird.

We don't need eigenvectors or eigenvalues.

We need to find two vectors that span the above plane. e.g we need to find the basis for the null space of this matrix.

$\begin{bmatrix} 1 & 1 & 1& 0\end{bmatrix}$

If it helps you can think of it as this matrix with added rows of zeros.

$\begin{bmatrix} 1 & 1 & 1& 0 \\ 0 & 0 & 0& 0 \\ 0 & 0 & 0& 0 \\ \end{bmatrix}$

Since this is already in reduced row form We know that we have two free parameters. Let $z=t,y=s$ then

$x+t+s=0 \iff x =-t-s$

So the basis of the nullspace is

$\begin{pmatrix} x \\ y \\ z\end{pmatrix} = \begin{pmatrix} -t-s \\ s \\ t\end{pmatrix} = \begin{pmatrix} -t \\ 0 \\ t\end{pmatrix}+ \begin{pmatrix} -s \\ s \\ 0\end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1\end{pmatrix}t+\begin{pmatrix} 0\\ -1 \\ 1\end{pmatrix}s$

You can verify that the two above vectors are perpendicular to the first.

Can you finish from here?
February 7th 2011, 09:03 PM
Taurus3

So for my 2nd question, would it just be B= [(1, 1, 1), (0, -1, 1), (-1, 0, 1)]?
February 8th 2011, 05:42 AM
TheEmptySet

Quote:

So for my 2nd question, would it just be B= [(1, 1, 1), (0, -1, 1), (-1, 0, 1)]?

No that is the matrix of the linear transformation from the standard basis to your new basis $B$ .

Hint: In your new basis $T(v_1)=v_1$ and $T(v_2)=T(v_2)=0$ So what is the matrix of this transformation?
February 8th 2011, 02:40 PM
Taurus3

this is what I don't get. How do you know T(v1)=v1 and that T(v2)=0?
February 8th 2011, 04:39 PM
TheEmptySet

Quote:

Originally Posted by Taurus3

this is what I don't get. How do you know T(v1)=v1 and that T(v2)=0?

Since $\displatstyle T(\vec{x})=\text{proj}_{v_1}\vec{x}=\frac{\vec{v_1 }\cdot \vec{x}}{||v_1||^2}\vec{v_1}$

This is the definition of projecting one vector onto another. This will verify the above "claim".