Prove $T$ is diagonalizable and find a basis $B$ of $V$ such that the matrix associated to $T$ be diagonal.

Question

Let $V=\mathbb{R}^3$ and $T:V\rightarrow V$ the linear operator defined by $T(x,y,z)=(3x-z,2x+4y+2z,-x+3z)$

Prove $T$ is diagonalizable and find a basis $B$ of $V$ such that the matrix associated to $T$ be diagonal.

Let $B=\{e_1,e_2,e_3\}$ the canonical basis of $V$ this implies $T_{BB}=\begin{bmatrix} 3 &2 &-1 \\ 0&4 &0 \\ 1&2 &3 \end{bmatrix}$ then by definition $P_T(\lambda)=det(xI-T)=det(xI-T_{BB})=det\begin{bmatrix} -3+x &-2 &1 \\ 0&-4+x &0 \\ -1&-2 &-3+x \end{bmatrix}=-40+34x-10x^2+x^3=(x-4)(x^2-6x+10)$

in consequence, the characteristic polynomial of $T$ is $(x-4)(x^2-6x+10)$

Then, $x_1=4$ is an eigenvalue. My question, exists other eigenvalue?

If not, then $T$ is diagonalizable, because we have an unique eigenvalue an that eigenvalue go to produce an unique eigenvector and that vector is linearly independent.

Find the basis $B$ can be a little difficult to me, can someone help me?

Answer 1

if a n * n matrix has n different eigenvalues then it is diagonalizable. here you have to solve $$det(xI−T)=0$$ for x . so you have $$ (-4+x)[(-3+x)(-3+x)-1]=0$$ $$ (-4+x)(X^2+9-6x-1)=0$$ $$(-4+x)(x^2-6x+8)=0$$ $$x=4,-2,4$$ now you have to find eigenspace, you have to show that T-4I span 2 dimension subspaces.

Answer 2

$4$ is the only real eigenvalue. The remaining two are $3\pm i$.

Regarding the basis for which $T$ is diagonal, think of the set of eigenvctors associated with these eigenvalues.

Hint: After change to basis $V$, $T$ becomes $V^{-1}TV$.

Answer 3

This is just to elaborate on shere's answer:

Clearly your matrix is wrong. The transformation T is represented as:

$$T_{BB}=\begin{bmatrix} 3 & 2 & -1 \\ 0 & 4 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$

Now, this means that the characteristic equation is got by:

$$ det \begin{bmatrix} x-3 & -2 & 1 \\ 0 & x-4 & 0 \\ 1 & -2 & x-3 \end{bmatrix} = 0 $$

This gives us the equation:

$$x^3-10x^2+32x-32=0$$

$$(x-4)(x^2-6x+8)=0$$

$$(x-4)(x-2)(x-4)=0$$

Therefore, the eigenvalues of $T$ are $4$ and $2$. (shere got it wrong here)

This gives us three linearly independent eigenvectors for $T$:

$$[1\quad 0\quad -1],\; [0\quad 1\quad 0],\; and\; [1\quad -2\quad 1]$$

which together form a basis and thus, $T$ is diagonalizable.