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- Thread starter karush
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- Jan 30, 2018

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Yes, the eigenvalues are 3 and 4.

Any eigenvector corresponding to eigenvalue 3 must satisfy

$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y \end{bmatrix}$

so -x+ 2y= 3x and -6x+6y= 3y. What must x and y be? (Those two equations reduce to the same thing so there are an infinite number of solutions- a one dimensional subspace.)

Any eigenvector corresponding to eigenvalue 4 must satisfy

$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}4x \\ 4y \end{bmatrix}$

so -x+ 2y= 4x and -6x+6y= 4y. What must x and y be?

Those eigenvectors span the "eigenspace".

Any eigenvector corresponding to eigenvalue 3 must satisfy

$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y \end{bmatrix}$

so -x+ 2y= 3x and -6x+6y= 3y. What must x and y be? (Those two equations reduce to the same thing so there are an infinite number of solutions- a one dimensional subspace.)

Any eigenvector corresponding to eigenvalue 4 must satisfy

$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}4x \\ 4y \end{bmatrix}$

so -x+ 2y= 4x and -6x+6y= 4y. What must x and y be?

Those eigenvectors span the "eigenspace".

Last edited:

- Jan 30, 2018

- 821

That last is wrong because, of course, the eigenvalue was 2, not 4!Any eigenvector corresponding to eigenvalue 4 must satisfy

$\begin{bmatrix}-1 & 2 \\ -6 & 6\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}4x \\ 4y \end{bmatrix}$

so -x+ 2y= 4x and -6x+6y= 4y. What must x and y be?

Those eigenvectors span the "eigenspace".

Instead -x+ 2y= 2x and -6x+ 6y= 2y.

2y= 3x and -6x= -4y both reduce to 3x= 2y. In particular, if we take x= 2, then 3x= 6= 2y, y= 3. One eigenvector is (2, 3) but any multiple is also an eigenvector.