$\huge \begin{split}E_{\lambda} &= \setbuild{\vec x \in \R^n}{A\vec x =\lambda \vec x}\\ &= \op{Nul}(A-\lambda I) \end{split} $ The [[Eigenspace]] of a [[Matrix]] $A$ is a [[Linear Subspace]] of all [[Eigenvector|Eigenvectors]] of $A$ with an [[Eigenvalue]] of $\lambda$. The [[Eigenspace]] is equal to the [[Set]] of all solutions to the [[Homogeneous Equation]] $(A-I)\vec x = \vec 0$ / the [[Nullspace]] of $A-I$. ### Basis Given $A\in M_{n\times n}$: - Find the [[Eigenvalue|Eigenvalues]] $\lambda_{1},\dots,\lambda_{k}$ of $A$. - Find a [[Basis]] of each [[Eigenspace]] $E_{\lambda_{1}}, \dots E_{\lambda_{m}}$ - Make a [[Set]] of the [[Vector|Vectors]] from all of these bases. This [[Set]] will always be [[Linear Independence|Linearly Independent]]. $\begin{align} A\vec v_{1}&=\vec v_{1}\\ A\vec v_{2} &= 3\vec v_{2}\\ A\vec v_{3} &= 4\vec v_{3}\\ \end{align} $ $\begin{align} \let c_{1}, c_{2}, c_{3} &\in \R :\\ c_{1}\vec v_{1} + c_{2}\vec v_{2} + c_{3}\vec v_{3} &= \vec 0 \end{align} $ $ \begin{align} (A-I)(c_{1}\vec v_{1} + c_{2}\vec v_{2} + c_{3}\vec v_{3}) &= (A-I)\vec 0\\ c_{1}(A\vec v_{1}-\vec v_{1}) + c_{2}(A\vec v_{2}-\vec v_{2}) + c_{3}(A\vec v_{3}-\vec v_{3}) &= \vec 0 \\ c_{2}2\vec v_{2} + c_{3}(A\vec v_{3}-\vec v_{3}) &= \vec 0\\ (A-\underbrace{3}_{\lambda_{2}}I)(c_{2}(A\vec v_{2}-\vec v_{2}) + c_{3}(A\vec v_{3}-\vec v_{3}) ) &= \vec 0\\ c_{3}(A\vec v_{3}- 3\vec v_{3}) &= \vec 0 \end{align} $ >[!example] > Find the [[Eigenvalue|Eigenvalues]], make a [[Basis]] for each [[Eigenspace]], wite $A$ in [[Eigendecomposition|Diagonal]] form > >$\large A = \mat{3&1\\4&2} $ >>[!check]- Solution >>$\large \begin{align} >>\det(A-\lambda I) &= 0 \\ >> >>\det{\mat{ 3-\lambda&1\\4&2-\lambda }} &= (3-\lambda)(2-\lambda)-4 >>\\ >>&= \lambda^2 -5t+6-4\\ >>&= \lambda^2 -5t+2\\ >>\lambda &= \frac{5}{2} \pm \frac{\sqrt{17}}{2} >>\end{align}$ >>$\large \begin{align} >>E_{\lambda_{0}} &= \op{Nul}\pa{A-\left( \frac{5}{2}-\frac{\sqrt{ 17 }}{2} \right)I}\\ >>&\sim >>\augmented{cc|c}{ >>\frac{1}{2} + \frac{\sqrt{ 17 }}{2} & 1 & 0\\ >>4 & -\frac{1}{2}+ \frac{\sqrt{ 17 }}{2} &0 >>}\\ >>&\sim \augmented{cc|c}{ >>1 & -\frac{1}{8}+ \frac{\sqrt{ 17 }}{8} & 0\\ >>0&0&0 >>}\\ >>&\sim \begin{cases} >>x_{1} + \left( -\frac{1}{8} + \frac{\sqrt{ 17 }}{2} \right) &=0\\ >>0&=0 >>\end{cases}\\ >>&\sim \begin{cases} >>x_{1}&= \left( \frac{1}{8}+ \frac{\sqrt{ 17 }}{2} \right) t\\ >>x_{2}&= t >>\end{cases}\\ >> >>E_{\lambda_{0}} &= \op{span}\left(\mat{ 1+\sqrt{ 17 }\\ 8 }\right) \\ >>\end{align}$ >> >>Following the same logic for $\lambda_{1}$ : >>$\large >>E_{\lambda_{1}} = \op{span}\left(\mat{ 1-\sqrt{ 17 }\\ 8 }\right) >>$ >> >>$\large >>A = >>\mat{ >>\vec \lambda_{0} & \vec \lambda_{1} >>} >>\mat{ >>\lambda_{0} & 0 \\0&\lambda_{1} >>} >>\mat{ >>\vec \lambda_{0} & \vec \lambda_{1} >>}^{-1} \\ >>$ >>$\tiny >>A=\mat{ >>1+\frac{\sqrt{ 17 }}{8} & 1- \frac{\sqrt{ 17 }}{8}\\ >>8 & 8 >>} >>\mat{ >>\frac{5}{2} - \frac{\sqrt{ 17 }}{2} & 0 \\ >>0&\frac{5}{2} + \frac{\sqrt{ 17 }}{2} \\ >>} >>\mat{ >>1+\frac{\sqrt{ 17 }}{8} & 1- \frac{\sqrt{ 17 }}{8}\\ >>8 & 8 >>}^{-1} >>$ >[!example] >$ >A= \mat{ 2 & -1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 1} >$ >>[!check]- Solution >>$ >>\begin{align} >> >>0&= \det(A- \lambda I) \\ >>&=\mat{ >>2-\lambda&-1&1\\ >>1& -1-\lambda & 0 \\ >>0 & 1 & 1- \lambda >>}\\ >>&= >> >>0\left|\array{ -1&1& \\-1-\lambda & 0 } \right| >>-\left|\array{ >>2-\lambda&1\\1&0 >>}\right| >>+ >>(1-t) >>\left|\array{ >>2-t & -1 \\ >>1 & -1-t >>} \right| \\ >>&= >> >>-\pa{(2-t)0-1} + (1-t)((2-t)(-1-t)-(-1)) \\ >>&= \cdots \text{ Trust me bro } \cdots \\ >>&= (\lambda-0)(\lambda-(-1+\sqrt{ 2 }))( \lambda - (-1- \sqrt{ 2 }) ) \\ >> \lambda &= \set{ >> 1 + \sqrt{ 2 } , >> 1 - \sqrt{ 2 } , >> 0 >> } >> >>\end{align} >>$ >> >>$ >>\begin{align} >>E_{o} &= \op{Nul}(A-0I) \\ >>&= \cdots \\ >>&= \op{span}\pa{\mat{-1\\-1\\1}} >>\end{align} >>$