Confirm by multiplication that x is an eigenvector of A, and find the corresponding eigenvalue.

Given: \begin{align} A = \begin{pmatrix} 1&2\\3&2\\\end{pmatrix}, && x = \begin{pmatrix} 1\\-1\\\end{pmatrix} \end{align} I know: $Ax = \lambda x$

My work:

I know $\lambda I - A$

\begin{pmatrix} \lambda - 1&-2\\-3&\lambda - 2\\\end{pmatrix}

From there I know the characteristic polynomial is $\lambda^2 - 3\lambda - 4 = 0$ through ad-bc (since this is a 2 x 2 matrix)

I can successively trying out each factor of c (which is 4) : positive and negative of 1,2,4.

4 turns out to be the only one that works. So $\lambda - 4 = 0$. So the $\lambda$ = 4.

I also know I can divide the characteristic polynomial by $\lambda - 4$ and get $\lambda + 1$. Setting $\lambda + 1 = 0$. $\lambda$ is $-1$.

Answer: So I got two eigenvalues which is $-1$ and $4$.

Dilemma I am having with eigenvector:

The problem is I am not sure if the given eigenvector applies for both the left and right side of the equation Ax = $\lambda$x. Or is it just the left side?

Work I have done using the given eigenvector x:

I know Ax = $\lambda$x

\begin{align} \begin{pmatrix} 1&2\\3&2\\\end{pmatrix} \cdot \begin{pmatrix} 1\\-1\\\end{pmatrix} = \begin{pmatrix} 1*(1) + &2 (-1)\\3*(1)&2(-1)\\\end{pmatrix} = \begin{pmatrix} -1\\1\\\end{pmatrix} = Ax. \end{align} Problem I am facing: What do I do after this step? Do I use the given value of the eigenvector $x$ on the right side of the equation $Ax = \lambda x$ along with the eigenvalue I find to see if the equation satisfies itself? How do I know if the given eigenvector is actually correct?


The directions are confirm by multiplication. All you need do is compute $Ax$ for the given $A$ and $x$ then compare that result to the given $x$.


Now it is clear that $\lambda=-1$. Because we have $Av=-v$ thus we must have $\lambda=-1$


You are given the matrix A and the possible eigenvector x1.

You correctly find the eigenvalues, λ1 = -1 and λ2 = 4.

By the way, the characteristic equation gives both eigenvalues: characteristic polynomial = λ^2 - 3λ - 4 = (λ +1)(λ - 4) = 0, implying λ1=-1 and λ2=4.

You'll need to find the second eigenvector, x2.

Find x2 so that (A−λ2*I)*x2=0

Then, show that these are in fact eigenvectors and eigenvalues of A.

You have the defining relationship, Ax = λx, which says that the eigenvalue scales the eigenvector in the exact same way the matrix does!

Just do the multiplications to demonstrate this.

Ax1 = λ1 x1

Ax2 = λ2 x2

  • $\begingroup$ Am I suppose to put both value of the given eigenvector on the left side or (left and right side) of the equation? $\endgroup$ – Nicholas May 12 '14 at 18:34
  • $\begingroup$ @Nicholas just multiply $Ax$ and see what the outcome is...you basically get a multiple of $x$, and that factor of multiplication is the eigenvalue $\endgroup$ – Christiaan Hattingh May 12 '14 at 18:36
  • $\begingroup$ I get \begin{align} Ax = \begin{pmatrix} -1\\1\\\end{pmatrix} \end{align} = 4 \begin{pmatrix} 1\\-1\\\end{pmatrix} and Ax = \begin{pmatrix} -1\\1\\\end{pmatrix} = -1 \begin{pmatrix} 1\\-1\\\end{pmatrix} It looks like the eigenvalue of 4 will satisfied the equation but the eigenvalue of -1 will. Correct me if I am wrong but that is how I am seeing the math. $\endgroup$ – Nicholas May 12 '14 at 18:39

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