Find the eigenvectors in the given problem

[chwala]

Homework Statement

See attached.

Relevant Equations

Diagonalization of matrices.

Consider this attachment,

In my understanding we have these equations, using P={x,y,z,m], For A.P=0

$24x-12y+4z=0$

$4m=0$

I can see how they got the first one, by letting $y=0, 24x+4z=0$ ⇒$x=1,y=-6$ that is clear,

My question is, we can as well have other possibilities other than what is given, correct?

I may as well make $z=0$ to give, $v_1=\begin{vmatrix} 1 \\ 2 \\ 0\\ 0 \end{vmatrix}$

 

[fresh_42]

$(1,2,0,0)^\tau = v_1+2v_2,$ so yes. If $\{v_1,v_2\}$ is a basis, so is $\{v_1+2v_2,v_2\}.$
There is never only one basis (except one-dimensional $\mathbb{F}_2$ spaces).

 

[Gavran]

There are some typos in the original post.

using P={x,y,z,m]

using $\textbf{P}=\begin{bmatrix}x&y&z&m\end{bmatrix}^T$

For A.P=0

For $\textbf{A}\cdot\textbf{P}=-2\cdot\textbf{P}$

by letting $y=0, 24x+4z=0$ ⇒$x=1,y=-6$ that is clear,

by letting $y=0$, $24x+4z=0$ $\implies$ $x=1$, $z=-6$ that is clear,

 

[WWGD]

Well, given this is a lower triangular matrix its determinant is the product of the diagonal elements and nonzero, so its kernel is {0}. Or did I miss something?

 

[WWGD]

Given a basis $B_n; |B_n|=n$ and an invertible $n \times n$ matrix M, then $MB_n$ is a basis too.