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The problem I am working on is:

Consider the Matrix $A= \begin{bmatrix}2&-12\\-3&2\\ \end{bmatrix}$.

Is $\lambda=3$ and eigenvalue of $A$?

What I tried:

  1. I found a similar question here: How do I show that a given value is an eigenvalue of matrix A?

  2. One of the comments said: "Just solve the linear problem $Ax = vx$, where the unknown is the vector $x \in \mathbb{R}^3$. If you find such an $x$, it is an eigenvector."

  3. So I set up the equation $A\overrightarrow{x}=\lambda \overrightarrow{x}$, substituted $\lambda = 3$ and the matrix $A$ into it, and made it into the augmented matrix:

$$ \left[ \begin{array}{cc|c} 2&-12&3\\ -3&2&3 \end{array} \right] $$

  1. This row reduces to:

$$ \left[ \begin{array}{cc|c} 1&0&\frac{-21}{16}\\ 0&1&\frac{-15}{32} \end{array} \right] $$

Since this system has a solution, I'd want to say that yes, $3$ is an eigenvalue of $A$.

But when I tried to check my work on Wolfram Alpha: https://www.wolframalpha.com/input/?i=%7B%7B2,-12%7D,%7B-3,2%7D%7D

It says the only eigenvalues of $A= \begin{bmatrix}2&-12\\-3&2\\ \end{bmatrix}$ are $\lambda_1=8$ and $\lambda_2=-4$?

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  • $\begingroup$ Why don't you use Characteristic polynomial to check it? $\endgroup$ – archangel89 Dec 7 '17 at 23:45
  • $\begingroup$ Solve det(A-\lambda.I)=0 $\endgroup$ – archangel89 Dec 7 '17 at 23:46
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    $\begingroup$ You have set up the augmented matrix incorrectly. The augmented matrix you have is for the system of equations 2x- 12y= 3 and -3x+ 2y= 3. t $\endgroup$ – user247327 Dec 7 '17 at 23:59
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As another answer states, the problem is in step 3. Your augmented matrix corresponds to the system $$ Ax = \begin{pmatrix}3\\ 3\end{pmatrix} = 3 \begin{pmatrix} 1\\ 1\end{pmatrix} $$ not to $Ax = 3x$, as you had hoped. (Unless by some stroke of luck $x$ happens to equal $\begin{pmatrix} 1\\ 1\end{pmatrix}$.)

Instead, note that if $Ax = 3x$ for a nonzero vector $x$, then $(A - 3I)x = 0$ so $A - 3I$ has a nontrivial kernel. So does the equation $(A - 3I)x = 0$ have a nontrivial solution?

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The problem is on the third step. The equality $A\vec x=3\vec x$ is equivalent to$$\left\{\begin{array}{l}2x-12y=3x\\-3x+2y=3y\end{array}\right.$$and this, in turn, is equivalent to$$\left\{\begin{array}{l}-x-12y=0\\-3x-y=0.\end{array}\right.$$

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You want to find $\lambda$ such that

$Av = \lambda v\\ Av = \lambda Iv\\ (A - \lambda I)v = \mathbf 0$

Which implies that $(A - \lambda I)$ has a non-trivial kernel, is singular, has $\det (A - \lambda I) = 0$

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