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I was reading my lecture notes on using matrices to solve ODEs and came across this and am having trouble understanding it:

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(Please note that the two eigen values that are calculated are 2 and 0 respectively).

I have tried checking the solution and all I am coming up with is a=b (I did this by solving simultaneously. a+b=2a and a+b=2b. I don't understand how the solution v=(1,1)^T comes from. In addition I also do not understand where the solution for the second eigenvalue (0) comes from either.

Please could some one kindly help me understand this. Thank You

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5 Answers 5

up vote 2 down vote accepted

Since $a=b$, an eigenvector is of the form $\pmatrix {b\cr b\cr}$. And it's easy to check that (for $b \ne 0$) all those are in fact eigenvectors. The ``solution'' just arbitrarily took $b$ to be $1$, but there's no reason to do so: all nonzero scalar multiples of an eigenvector are also eigenvectors.

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You are right, the most general eigenvector for the eigenvalue $2$ is of the form $\begin{pmatrix} a \\a \end{pmatrix}$ for nonzero $a$. It seems that the author was only interested in one, specific, eigenvector for this eigenvalue. The simplest choice that comes to mind is taking $a=1$. I assume he'd done the same for the other eigenvalue ($0$).

I think using the word "the" as in "the solution" is confusing.

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If we solve the system of equations $Av=2v$ where $v=(a,b)^T$ we find that $a=b$ so there's infinitely many solutions which form the eigenspace $\ker(A-2I)$ so we need just to choose one eigenvector i.e. an arbitrary non zero vector in this eigenspace so we can choose $v=(1,1)^T$ or $(2,2)^T$ or even $(-1000,-1000)^T$.

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You are almost there. Eigenvectors are unique only up to an overall scaling factor. If $(1,1)$ is an eigenvector $(2,2)$ is an eigenvector too, and so is $(3,3)$, etc. You have shown that $a=b$. This implies that the eigenvector is $(a,a)$. This is just a general case of $(1,1)$.

Often, the vector presented as "the solution" will be a vector normalized to have unit length. In this case, it would be $(1/\sqrt{2},1/\sqrt{2})$.

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Once you have calculated the eigenvalues, e.g., by calculating the roots ${\lambda_1, \lambda_2 }$of the characteristic polynomial Det($A-\lambda I $), you then find the solutions (meaning a basis for the nullspace of) to each of $A-\lambda I v=0.

Here, $\lambda_1,\lambda_2$={0,2}

For the value $0$, we need to find a basis for the nullspace of the matrix $(A-0.I)=A$ of , i.e., we must find a basis for the nullspace of A itself. We do this by setting Av=0, from which we get that a+b=0, i.e., a=-b , so that , e.g., the vector (1,-1) spans the nullspace of $A=(A-0.I)$.

We then do the same thing for the second $\lambda$ value 2,i.e., we find the a basis for the nullspace of $ A-2.I $ , the matrix with entries (-1,1,1,-1) ( I will Latex the matrix soon). So we need to find a basis for this nullspace. We get the two equations a-b=0 and b-a=0 that must be satisfied by the nullspace vectors. This means that we can use, e.g., the vector $(1,1)^T$ as a basis.

So your eigenvectors here are {(1,-1),(1,1)}

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