Question

If T:R^3 tends to R^3 is defined by T(x, y, z)=(2x+y-2z, 2x+3y-4z, x+y-z) Find all eigen value of T and find a basis of each eigen space. Is T is diagonalizable?

Answer 1

Solution: Given that "T(x, y, z)=(2x+y-2z, 2x+3y-4z, x + y-z)"

First we find matrix representation of T relative to the standard basis B={(1,0,0),(0,1,0),(0,0,1)}

So matrix obtained is $A=\left[\begin{array}{ccc}2&1&-2\\2&3&-4\\1&1&-1\\\end{array}\right]$

To find eigenvalues , we consider characteristic equation,

$\left|\begin{array}{ccc}2-\lambda&1&-2\\2&3-\lambda&-4\\1&1&-1-\lambda\\\end{array}\right|=0$

we know that characteristic equation is given by

$\lambda^3-(Trace~of~A )\lambda^2+(Sum ~of~minors~along~diagonal)\lambda-det(A)=0$

Trace of A=sum of diagonal elements=2+3-1=4

Sum of minors along diagonal=1+0+4=5

det(A)=2

$\therefore\lambda^3-(Trace~of~A )\lambda^2+(Sum ~of~minors~along~diagonal)\lambda-det(A)=0$ is written as

$\lambda^3-(4 )\lambda^2+(5)\lambda-2=0$

$\lambda^3-4 \lambda^2+5\lambda-2=0$ is called the characteristic equation.

Put $\lambda=1$ the characteristic equation satisfies. Therefore \lambda=1 is one of the eigen value of the characteristic equation.

Now we find other eigen values by synthetic division method.

1|1 -4 5 -2

| 1 -3 2

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| 1 -3 2 0

$\lambda^2-3\lambda+2=0$

$\lambda^2-2\lambda-\lambda+2=0$

$\lambda(\lambda-2)-1(\lambda-2)=0$

$(\lambda-1)(\lambda-2)=0$

$\lambda=1,\lambda=2$

Therefore the eigenvalues are $\lambda=1,1,2$

To find eigenspace corresponding to $\lambda=1:$

Put $\lambda=1$ in the equation $(A-\lambda I)X=0~ where ~X ~is vector ~X=\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]$

$\therefore\left[\begin{array}{ccc}2-\lambda&1&-2\\2&3-\lambda&-4\\1&1&-1-\lambda\\\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\\\end{array}\right]$

$\therefore\left[\begin{array}{ccc}1&1&-2\\2&2&-4\\1&1&-2\\\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\\\end{array}\right]$

Solving this by by Gauss elimination method, we get $y~and~z$ as free variable.

$\therefore X=\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]=\left[\begin{array}{ccc}-y+2z\\y\\z\\\end{array}\right]=y\left[\begin{array}{ccc}-1\\1\\0\\\end{array}\right] +z\left[\begin{array}{ccc}2\\0\\1\\\end{array}\right] =y[v1]+z[v2]$

$Where~v1=\left[\begin{array}{ccc}-1\\1\\0\\\end{array}\right] ~and~v2=\left[\begin{array}{ccc}2\\0\\1\\\end{array}\right]$ $~are~two~eigen~vectors~corresponding~ to~\lambda=1$

To find eigenspace corresponding to $\lambda=2:$

Put $\lambda=2$ in the equation $(A-\lambda I)X=0~ where ~X ~is vector ~X=\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]$

$\therefore\left[\begin{array}{ccc}2-\lambda&1&-2\\2&3-\lambda&-4\\1&1&-1-\lambda\\\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\\\end{array}\right]$

$\therefore\left[\begin{array}{ccc}0&1&-2\\2&1&-4\\1&1&-3\\\end{array}\right]\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\\\end{array}\right]$

Solving this by by Gauss elimination method, we get $z$ as free variable.

$\therefore X=\left[\begin{array}{ccc}x\\y\\z\\\end{array}\right]=\left[\begin{array}{ccc}z\\2z\\z\\\end{array}\right]=z\left[\begin{array}{ccc}1\\2\\1\\\end{array}\right] =z[v3]$

$Where~v3=\left[\begin{array}{ccc}1\\2\\1\\\end{array}\right] ~is~the~eigen~vectors~corresponding~ to~\lambda=2$ .

Now, to check T is diagonalizable :

To check T is diagonalizable or not, it is sufficient to prove that $AP=PD$

$where~A=\left[\begin{array}{ccc}2&1&-2\\2&3&-4\\1&1&-1\\\end{array}\right],~P=[v1~v2~v3]=\left[\begin{array}{ccc}-1&2&1\\1&0&2\\0&1&1\\\end{array}\right]$

$~and~D=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&2\\\end{array}\right]=diagonal ~elements~ of~eigen~values.$

Now, $AP=\left[\begin{array}{ccc}2&1&-2\\2&3&-4\\1&1&-1\\\end{array}\right]\left[\begin{array}{ccc}-1&2&1\\1&0&2\\0&1&1\\\end{array}\right] ~~~~~~~~~~~~~~~~~[matrix~ multiplication]$

$=\left[\begin{array}{ccc}-1&2&2\\1&0&4\\0&1&2\\\end{array}\right]..................................(1)$

$PD=\left[\begin{array}{ccc}-1&2&1\\1&0&2\\0&1&1\\\end{array}\right]\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&2\\\end{array}\right]$

$=\left[\begin{array}{ccc}-1&2&2\\1&0&4\\0&1&2\\\end{array}\right]..................................(2)$

$from~(1)~and~(2)~,AP=PD$

Hence T is diagonalizable.