# [SOLVED]051 how they got the eigenspaces ?

#### karush

##### Well-known member

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 $\lambda=2,3$

307 hw

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#### Country Boy

##### Well-known member
MHB Math Helper
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".

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#### Country Boy

##### Well-known member
MHB Math Helper
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.