Is a Shifted Solution Still Valid?

[Yankel]

Hello all

A simple question.

It is known that

\[\left ( \begin{matrix} x_{1}\\ x_{2}\\ x_{3} \end{matrix} \right )\]

is a solution of a homogenous system of equations.

I need to determine if

\[\left ( \begin{matrix} x_{1}+1\\ x_{2}+1\\ x_{3}+1 \end{matrix} \right )\]

is also a solution, and why ?

thank you !

 

[I like Serena]

Hello all

A simple question.

It is known that

\[\left ( \begin{matrix} x_{1}\\ x_{2}\\ x_{3} \end{matrix} \right )\]

is a solution of a homogenous system of equations.

I need to determine if

\[\left ( \begin{matrix} x_{1}+1\\ x_{2}+1\\ x_{3}+1 \end{matrix} \right )\]

is also a solution, and why ?

thank you !

Hi Yankel,

That depends.

A homogenous system means that $A\mathbf x=\mathbf 0$.

So when would $A(\mathbf x + \mathbf 1)=\mathbf 0$ also be a solution? (Wondering)

 

[Yankel]

I am not sure. I will guess. Maybe when the solution is (-1,-1,-1) and then x+1 is the trivial solution ?

 

[I like Serena]

I am not sure. I will guess. Maybe when the solution is (-1,-1,-1) and then x+1 is the trivial solution ?

Can you simplify:
$$\mathbf A(\mathbf x + \mathbf 1) = \mathbf 0$$
using $\mathbf A \mathbf x = \mathbf 0$ and the properties of linear transformations, which are:
$$\begin{cases}\mathbf A(\mathbf a + \mathbf b) = \mathbf A \mathbf a + \mathbf A \mathbf b \\ \mathbf A (\lambda \mathbf a) = \lambda (\mathbf A \mathbf a)\end{cases}$$

 

[Yankel]

Oh...

A=0 ?

 

[I like Serena]

Oh...

A=0 ?

Huh? No, I'm afraid not. At least not necessarily.

What I mean is if you can get rid of the parentheses in $\mathbf A(\mathbf x+ \mathbf 1) = \mathbf 0$? Then we'll see what we can deduce from that...

 

[Yankel]

you get:

Ax+A=0

but isn't Ax=0 ?

 

[I like Serena]

you get:

Ax+A=0

but isn't Ax=0 ?

$A \mathbf x$ is indeed $\mathbf 0$, but your equation should be:
$$A \mathbf x + A \mathbf 1 = \mathbf 0$$
Note that $A \mathbf 1$ actually means:
$$A \mathbf 1 = A \begin{bmatrix}1\\1\\1\end{bmatrix}$$
which is different from A.

 

[Yankel]

this is the answer ? it's a solution only if

A1=0 ? if (1,1,1) is a solution ?

 

[I like Serena]

this is the answer ? it's a solution only if

A1=0 ? if (1,1,1) is a solution ?

Yep! (Nod)

So we can say that $\mathbf x + \mathbf 1$ will generally not be a solution, unless (1,1,1) is a solution.

I guess that should suffice to answer the question, although we can say a little more.

It means that (1,1,1) must be an eigenvector with eigenvalue 0.
And we know that $\mathbf x$ is also an eigenvector with eigenvalue 0.
However, the one does not have to be a multiple of the other, since a 3-dimensional matrix can have more than 1 eigenvector with eigenvalue 0. (Nerd)

 

[Yankel]

Thanks !