MHB 051 how they got the eigenspaces ?

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The discussion focuses on finding eigenspaces for the matrix A = [[-1, 2], [-6, 6]]. The correct eigenvalues are identified as 2 and 3, with corresponding equations derived for eigenvectors. For eigenvalue 2, the equations simplify to 3x = 2y, leading to an eigenvector (2, 3) and its multiples. The participants clarify that the eigenvectors span the eigenspace associated with each eigenvalue. The conversation emphasizes the importance of correctly identifying eigenvalues to determine the corresponding eigenspaces.
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ok I didn't understand how they got the eigenspaces

the original matrix was
$A=\left[\begin{array}{rrr}−1&2\\−6&6\end{array} \right]$
so think I got values correct $\lambda=2,3$

https://dl.orangedox.com/wlKD7eKSWiQ79alYD6
 
Last edited:
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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".
 
Last edited:
Country Boy said:
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".
That last is wrong because, of course, the eigenvalue was 2, not 4!
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.
 
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