For each of the following symmetric matrices A, find an...

For each of the following symmetric matrices $A,$ find an orthogonal matrix $P$ and a diagonal matrix $D$ such that $P^{\prime} A P$ is diagonal: (a) $A=\left[\begin{array}{rr}1 & 2 \\ 2 & -2\end{array}\right]$ (b) $A=\left[\begin{array}{rr}5 & 4 \\ 4 & -1\end{array}\right]$ (c) $A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]$

Transcript

00:02 At a the matrix 1 2 2 minus 2 so the characteristic polynomial of a is given by p lambda of a is lambda square minus the trace of a times lambda plus the determinant of a and this is equal to square minus plus lambda minus 6 so the angle vectors is given by lambda e equal to minus 1 plus o minus square 1 plus the 2 and this is equal to 2 over 2 so the angle vectors are lambda 1 equal 2 and the lambda 2 equal minus 2.

01:21 So the egg vectors of a is given by the product of the matrix a with the vector xy is equal to the angle value 2 times the vector xy.

01:43 And we obtain a linear system x plus 2y equal to x and 2x minus 2y equal to y so this is equal to minus x plus 2y equal 0 and 2y minus 4 y equal 0 multiply the line 1 by 2 and add in the line 2 then imply that x is equal to 2 y so the angle vector associated to the angle value lambda lambda 1 is equal to x y by replacing is equal to 2 y and y and this is a vector to 1 times 1 2.

03:09 The second egg vector is given by the product of the matrix a with the vector xy which 1 minus 3 and you between a linear system x plus 2y minus 2 x minus 2 x minus 2 y equal to minus 3 x to minus 3 y he is so this is equivalent for x plus 2 y equal 0 and minus 2 x minus y equals 0 and the solution of this system is y equal to x.

04:32 So the angle vector associated to the angle value number 2 is equal to the vector is given by the vector 1, 2 and the vector 2, sorry, the vector 2 1 and the vector 1 minus 2.

05:17 So the diagonal matrix d is the angle values is equal to the inverse of the matrix of the angle vector.

05:34 The matrix a minus the matrix 2, 1 minus 2.

05:52 For the item b, let a, the matrix 5, 4 minus 1.

06:10 The characteristic pollen of a is given by lambda square minus 4 times lambda minus 21...

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