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Considering the space $\mathbb{R}^3$$\ $, let D be a diagonal matrix similar to the matrix T. Find the matrix P that satisfy:

$P^{-1}TP=D$

The matrix T is:

$$T = \begin{bmatrix} 4 & -2 &2 \\ -2 & 2 & 0\\ 2 & 0 & 2\\ \end{bmatrix} $$

So far, I've found all eigenvalues and eigenvectors, which are:

$λ_1$ = 0 -> $V_1$=($1/\sqrt{3}$, $1/\sqrt{3}$, -$1/\sqrt{3}$)

$λ_2$ = 2 -> $V_2$=(0, $1/\sqrt{2}$, $1/\sqrt{2}$)

$λ_3$ = 6 -> $V_2$=($2/\sqrt{6}$, -$1/\sqrt{6}$, $1/\sqrt{6}$)

But from now on, I don't know how to proceed. Any suggestions?

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  • $\begingroup$ Thanks! Stupid question: why are the eigenvectors the columns of P? $\endgroup$ – woz Jun 14 '15 at 19:08
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    $\begingroup$ @woz $P$ is a change of basis matrix, which consists of the vectors of the "new" basis (in this case the eigenvectors) as columns, written with respect to the "old" basis (in this case the standard basis). $\endgroup$ – ptrsinclair Jun 14 '15 at 19:15
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We can see what's going on here by using block-matrix multiplication: let $v_1,v_2,v_3$ denote the columns of $P$. We then have $$ P^{-1}TP = D \implies\\ TP = PD \implies\\ T(v_1 \quad v_2 \quad v_3) = (v_1 \quad v_2 \quad v_3) \pmatrix{d_1 \\ & d_2 \\ && d_3} \implies\\ (T v_1\quad Tv_2\quad T v_3) = (d_1 v_1 \quad d_1 v_2\quad d_1 v_3) $$ Because $P$ is invertible, its columns are linearly independent. What we can take away from this, then, is that if $P$ diagonalizes $T$, then its columns should be linearly independent eigenvectors of $T$.

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