# Simple eigenvector/value problem

• May 23rd 2008, 08:37 PM
Schniz2
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?
• May 23rd 2008, 09:54 PM
Reckoner
Quote:

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, $\displaystyle \left[ \begin{array}{c} -\frac34\\ 1\end{array}\right]$, is not correct. Could you show us how you arrived at that answer?

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

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

which gives us

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

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

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

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

Quote:

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.
• May 23rd 2008, 10:37 PM
Schniz2
Quote:

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

I'm not sure what you are asking here. Please clarify.

sorry, the one i had was $\displaystyle \left[\begin{array}{c} -\frac4{3}\\1 \end{array}\right]$

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.
• May 23rd 2008, 10:45 PM
Reckoner
Quote:

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

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.

Quote:

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.
• May 23rd 2008, 11:00 PM
Schniz2
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 :)