Answer to Question #248553 in Linear Algebra for chaitu

Question

Diagonalize the matrix

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10 −2 −5

−2 2 3

−5 3 5

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Answer 1

$$A=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}$$

Find the eigen values

$$A-\lambda I=\begin{pmatrix} 10-\lambda & -2 & -5 \\ -2 & 2-\lambda & 3 \\ -5 & 3 & 5-\lambda \\ \end{pmatrix}$$

$$\det(A-\lambda I)=\begin{vmatrix} 10-\lambda & -2 & -5 \\ -2 & 2-\lambda & 3 \\ -5 & 3 & 5-\lambda \\ \end{vmatrix}$$

$$=(10-\lambda\begin{vmatrix} 2-\lambda & 3 \\ 3 & 5-\lambda \end{vmatrix}+2\begin{vmatrix} -2 & 3 \\ -5 & 5-\lambda \end{vmatrix}$$

$$-5\begin{vmatrix} -2 & 2-\lambda \\ -5 & 3 \end{vmatrix}=(10-\lambda)(10-7\lambda+\lambda^2-9)$$

$$+2(-10+2\lambda+15)-5(-6+10-5\lambda)$$

$$=10-70\lambda+10\lambda^2-\lambda+7\lambda^2-\lambda^3+10+4\lambda$$

$$-20+25\lambda=-\lambda^3+17\lambda^2-42\lambda$$

$$=-\lambda(\lambda-3)(\lambda-14)$$

$$\det(A-\lambda I)=0=>-\lambda(\lambda-3)(\lambda-14)=0$$

The roots are $\lambda_1=14, \lambda_2=3, \lambda_3=0.$

These are eigenvalues.

Find the eigenvectors

$\lambda =14$

$$A-\lambda I=\begin{pmatrix} 10-14 & -2 & -5 \\ -2 & 2-14 & 3 \\ -5 & 3 & 5-14 \\ \end{pmatrix}$$

$$=\begin{pmatrix} -4 & -2 & -5 \\ -2 & -12 & 3 \\ -5 & 3 & -9 \\ \end{pmatrix}$$

$R_2=R_2-R_1/2$

$$\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ -5 & 3 & -9 \\ \end{pmatrix}$$

$R_3=R_3-5R_1/4$

$$\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ 0 & 11/2 & -11/4 \\ \end{pmatrix}$$

$R_3=R_3+R_2/2$

$$\begin{pmatrix} -4 & -2 & -5 \\ 0 & -11 & 11/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_2=R_2/(-11)$

$$\begin{pmatrix} -4 & -2 & -5 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1+2R_2$

$$\begin{pmatrix} -4 & 0 & -6 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1/(-4)$

$$\begin{pmatrix} 1 & 0 & 3/2 \\ 0 & 1 & -1/2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

If we take $v_3=t,$ then $v_1=-\dfrac{3}{2}t, v_2=\dfrac{1}{2}t.$

The eigenvector is $v=\begin{pmatrix} - 3/2 \\ 1/2 \\ 1 \end{pmatrix}$

$\lambda =3$

$$A-\lambda I=\begin{pmatrix} 10-3 & -2 & -5 \\ -2 & 2-3 & 3 \\ -5 & 3 & 5-3 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 7 & -2 & -5 \\ -2 & -1 & 3 \\ -5 & 3 & 2 \\ \end{pmatrix}$$

$R_2=R_2+2R_1/7$

$$\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ -5 & 3 & 2 \\ \end{pmatrix}$$

$R_3=R_3+5R_1/7$

$$\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ 0 & 11/7 & -11/7 \\ \end{pmatrix}$$

$R_3=R_3+R_2$

$$\begin{pmatrix} 7 & -2 & -5 \\ 0 & -11/7 & 11/7 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_2=-7R_2/11$

$$\begin{pmatrix} 7 & -2 & -5 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1+2R_2$

$$\begin{pmatrix} 7 & 0 & -7 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1/7$

$$\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 &-1 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

If we take $u_3=t,$ then $u_1=t, u_2=t.$

The eigenvector is $u=\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

$\lambda =0$

$$A-\lambda I=\begin{pmatrix} 10-0 & -2 & -5 \\ -2 & 2-0 & 3 \\ -5 & 3 & 5-0 \\ \end{pmatrix}$$

$$=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}$$

$R_2=R_2+R_1/5$

$$\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ -5 & 3 & 5 \\ \end{pmatrix}$$

$R_3=R_3+R_1/2$

$$\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ 0 & 2 & 5/2 \\ \end{pmatrix}$$

$R_3=R_3-5R_2/4$

$$\begin{pmatrix} 10 & -2 & -5 \\ 0 & 8/5 &2 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_2=5R_2/8$

$$\begin{pmatrix} 10 & -2 & -5 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1+2R_2$

$$\begin{pmatrix} 10 & 0 & -5/2 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

$R_1=R_1/10$

$$\begin{pmatrix} 1 & 0 & -1/4 \\ 0 & 1 & 5/4 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

If we take $w_3=t,$ then $w_1=\dfrac{1}{4}t, w_2=-\dfrac{5}{4}t.$

The eigenvector is $u=\begin{pmatrix} 1/4\\ -5/4\\ 1 \end{pmatrix}$

Form the matrix $P$

$$P=\begin{pmatrix} -3/2 & 1 & 1/4 \\ 1/2 & 1 & -5/4 \\ 1 & 1 & 1 \end{pmatrix}$$

Form the diagonal matrix $D$

$$D=\begin{pmatrix} 14 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$

The matrices $P$ and $D$ are such that the initial matrix

$$A=\begin{pmatrix} 10 & -2 & -5 \\ -2 & 2 & 3 \\ -5 & 3 & 5 \\ \end{pmatrix}=PDP^{-1}$$

$$P=\begin{pmatrix} -3/2 & 1 & 1/4 \\ 1/2 & 1 & -5/4 \\ 1 & 1 & 1 \end{pmatrix}$$

$$D=\begin{pmatrix} 14 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$