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Let-A-2-2-1-3-Find-a-non-singular-matrix-P-such-that-P-1-AP-is-a-diagonal-matrix-




Question Number 98250 by bobhans last updated on 12/Jun/20
Let A= (((2    2)),((1    3)) ) . Find a non singular matrix  P such that P^(−1) AP is a diagonal matrix.
$$\mathrm{Let}\:\mathrm{A}=\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\mathrm{3}}\end{pmatrix}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{non}\:\mathrm{singular}\:\mathrm{matrix} \\ $$$$\mathrm{P}\:\mathrm{such}\:\mathrm{that}\:\mathrm{P}^{−\mathrm{1}} \mathrm{AP}\:\mathrm{is}\:\mathrm{a}\:\mathrm{diagonal}\:\mathrm{matrix}. \\ $$
Commented by john santu last updated on 12/Jun/20
find eigen vector det(A−λI)=0   determinant (((2−λ      2)),((     1       3−λ)))= 0 ⇒λ = 1; 4  Eigen vector for λ = 1   (A−I) ((x),(y) ) =  ((0),(0) ) ⇒ (((1    2)),((1    2)) )  ((x),(y) ) = ((0),(0) )  eigen−vector  [((−2)),((   1)) ]  for λ = 4 ⇒ (((−2     2)),((   1     −1)) )  [(x),(y) ]=  ((0),(0) )  → x−y = 0 ; eigen−vector =  [(1),(1) ]  ∴ P =  [((−2      1)),((    1      1)) ] such that   P^(−1) AP =  [((1    0)),((0    4)) ]■
$$\mathrm{find}\:\mathrm{eigen}\:\mathrm{vector}\:\mathrm{det}\left(\mathrm{A}−\lambda\mathrm{I}\right)=\mathrm{0} \\ $$$$\begin{vmatrix}{\mathrm{2}−\lambda\:\:\:\:\:\:\mathrm{2}}\\{\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{3}−\lambda}\end{vmatrix}=\:\mathrm{0}\:\Rightarrow\lambda\:=\:\mathrm{1};\:\mathrm{4} \\ $$$$\mathrm{Eigen}\:\mathrm{vector}\:\mathrm{for}\:\lambda\:=\:\mathrm{1}\: \\ $$$$\left(\mathrm{A}−\mathrm{I}\right)\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\Rightarrow\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{1}\:\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{eigen}−\mathrm{vector}\:\begin{bmatrix}{−\mathrm{2}}\\{\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\mathrm{for}\:\lambda\:=\:\mathrm{4}\:\Rightarrow\begin{pmatrix}{−\mathrm{2}\:\:\:\:\:\mathrm{2}}\\{\:\:\:\mathrm{1}\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:\begin{bmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{bmatrix}=\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\rightarrow\:\mathrm{x}−\mathrm{y}\:=\:\mathrm{0}\:;\:\mathrm{eigen}−\mathrm{vector}\:=\:\begin{bmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{bmatrix} \\ $$$$\therefore\:\mathrm{P}\:=\:\begin{bmatrix}{−\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\\{\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\end{bmatrix}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{P}^{−\mathrm{1}} \mathrm{AP}\:=\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{4}}\end{bmatrix}\blacksquare \\ $$

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