# Thread: Simple eigenvector/value problem

1. ## Simple eigenvector/value problem

Attatched is the document with my question... i think the problem is that i really dont understand the steps to finding and testing eigenvalues and vectors. i use the characteristic equations but i seem to be getting alot of eigen-vectors, that, when tested are only true for the trivial case... ie. multiple of zero.

thanks.

**Edit**

Just realised that if i let k be equal to -3, the eigenvector i got will be equal to the one in the solution.... does this mean that we can have k as anyy value except for zero?

Also... if we have matrix A, eigenvector = u and eigenvalue = b, when we test to see if we have obtained the correct solution by checking if Au = bu, does it prove true if we have b = 0?

2. Originally Posted by Schniz2
Attatched is the document with my question... i think the problem is that i really dont understand the steps to finding and testing eigenvalues and vectors. i use the characteristic equations but i seem to be getting alot of eigen-vectors, that, when tested are only true for the trivial case... ie. multiple of zero.
The eigenvector you found, $\left[
\begin{array}{c}
-\frac34\\
1\end{array}\right]$
, is not correct. Could you show us how you arrived at that answer?

If we call the matrix $A$, we have $\left(\lambda I - A\right)\textbf{x} = \textbf{0}$, which, for $\lambda_1 = 2$ gives us the system:

$\left[\begin{array}{cc|c}
2 + 1 & 4 & 0\\
-3 & 2 - 6 & 0\\
\end{array}\right] = \left[\begin{array}{cc|c}
3 & 4 & 0\\
-3 & -4 & 0\\
\end{array}\right]$

which gives us

$\textbf{x} = t\left[\begin{array}{c}
4\\-3
\end{array}\right]$
.

You can see that $\left[\begin{array}{c}
4\\-3
\end{array}\right]$
is an eigenvector of $A$ because

$\left[\begin{array}{cc}
-1 & -4\\
3 & 6
\end{array}\right]
\left[\begin{array}{c}
4\\-3
\end{array}\right]
=\left[\begin{array}{c}
8\\-6
\end{array}\right]
=2\left[\begin{array}{c}
4\\-3
\end{array}\right]
=\lambda_1\left[\begin{array}{c}
4\\-3
\end{array}\right]$

But be sure to see that any nonzero multiple of $\left[\begin{array}{c}
4\\-3
\end{array}\right]$
is also an eigenvector of $A$. So, if you had, for example, $\left[\begin{array}{c}
-\frac4{3}\\1
\end{array}\right]$
, you would be okay.

Originally Posted by Schniz2
Just realised that if i let k be equal to -3, the eigenvector i got will be equal to the one in the solution.... does this mean that we can have k as anyy value except for zero?
I'm not sure what you are asking here. Please clarify.

3. Originally Posted by Reckoner
But be sure to see that any nonzero multiple of $\left[\begin{array}{c}
4\\-3
\end{array}\right]$
is also an eigenvector of $A$. So, if you had, for example, $\left[\begin{array}{c}
-\frac4{3}\\1
\end{array}\right]$
, you would be okay.

I'm not sure what you are asking here. Please clarify.
sorry, the one i had was $\left[\begin{array}{c}
-\frac4{3}\\1
\end{array}\right]$

At the end of an example in my notes they prove an eigenvector is correct by showing that the eigenvector multiplied by zero proves it. Is this allowed? *see zero.jpg*

Also, the equation we were given to use was *see equation.jpg*... is this equivalent to the one you used?

thanks.

4. Originally Posted by Schniz2
sorry, the one i had was $\left[\begin{array}{c}
-\frac4{3}\\1
\end{array}\right]$

At the end of an example in my notes they prove an eigenvector is correct by showing that the eigenvector multiplied by zero proves it. Is this allowed? *see zero.jpg*
That's because 0 is an eigenvalue for that matrix. Its eigenvectors, however, cannot include the zero vector, by definition.

Originally Posted by Schniz2
Also, the equation we were given to use was *see equation.jpg*... is this equivalent to the one you used?

thanks.
Multiply both sides by -1.

5. ahh ok, i think i understand now....

so the eigenvalue you get at the beginning, should be the eigenvalue you get when you check?

thanks heaps for all your help