Question

Number 40 and 46 please;

PD of squares of the four sides of a parallel- ogram is the sum of squares of its diago- nals ) 6 In each of the following cases, find all the eigenvalues and associated eigenvectors of A. Diagonalize A, if possible. [-4 5 6]. Find the projection of v onto u. If we denote this projection by w, verify that v -w is orthogonal to u. Ilustrate this by drawing a picture. -1 2-3 (a) A 2 2 -6 0-2 0 (b) A-2 1 -2 0-2 2 41. Recall that a square matrix Q of ordern is said to be an orthogonal matrix if and only if QTQ QQI. Prove that the following statements are equivalent: (c) A 7-2 1-4 2 -4 1 (a) Q is orthogonal; (b) columns of Q form an orthonormal 47. For each of the following matrices set of column vectors in Rn 2 3 -1 6 (c) rows of Q for an orthonormal set of

Answer 1

40) Given, u = (1,-2,3) and v = (-4,5,6).

Now, w = proj_uv

i.e., w = [<v,u>/<u,u>]u

i.e., $w=\frac{(-4)*1+5*(-2)+6*3}{1*1+(-2)*(-2)+3*3}\begin{bmatrix} 1\\ -2\\ 3 \end{bmatrix}$

i.e., $w=\frac{-4-10+18}{1+4+9}\begin{bmatrix} 1\\ -2\\ 3 \end{bmatrix}$

i.e., $w=\frac{4}{14}\begin{bmatrix} 1\\ -2\\ 3 \end{bmatrix}$

i.e., $w=\frac{2}{7}\begin{bmatrix} 1\\ -2\\ 3 \end{bmatrix}$

i.e., $w=\begin{bmatrix} 2/7\\ -4/7\\ 6/7 \end{bmatrix}$

Now, $v-w=\begin{bmatrix} -4\\ 5\\ 6 \end{bmatrix}-\begin{bmatrix} 2/7\\ -4/7\\ 6/7 \end{bmatrix}$

i.e., $v-w=\begin{bmatrix} -30/7\\ 39/7\\ 36/7 \end{bmatrix}$

Then, <v-w,u> = (-30/7)*1+(-2)*(39/7)+3*(36/7)

i.e., <v-w,u> = -(30/7)-(78/7)+(108/7)

i.e., <v-w,u> = (-30-78+108)/7

i.e., <v-w,u> = 0

Therefore, v-w is orthogonal to u.

46) Given matrix is : A = $\begin{bmatrix} 4 & 2 & 2\\ 2 & 1 & -4\\ 2 & -4 & 1 \end{bmatrix}$

Now, the characteristic equation of A is $\begin{vmatrix} 4-x & 2 & 2\\ 2 & 1-x & -4\\ 2 & -4 & 1-x \end{vmatrix}$ = 0

i.e., (4-x)[(1-x)²-16]-2[(2-2x)+8]+2[-8-(2-2x)] = 0

i.e., (4-x)(-3-x)(5-x)+8(x-5) = 0

i.e., (5-x)[(4-x)(-3-x)-8] = 0

i.e., (5-x)(x²-x-20) = 0

i.e., (5-x)(x-5)(x+4) = 0

i.e., x = -4, 5, 5

Therefore, eigenvalues of A are -4, 5, 5.

For $\lambda$ = -4 : (A+4I)X = 0

i.e., $\begin{bmatrix} 8 & 2 & 2\\ 2 & 5 & -4\\ 2 & -4 & 5 \end{bmatrix}$ $\begin{bmatrix} x\\ y\\ z \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

i.e., 8x+2y+2z = 0

2x+5y-4z = 0

2x-4y+5z = 0

Solving we get, y = z = -2x

Let us take x = k. Then, y = -2k and z = -2k.

Therefore, the eigenvectors corresponding to the eigenvalue -4 are $\begin{bmatrix} k\\ -2k\\ -2k \end{bmatrix}$ ,i.e., $k\begin{bmatrix} 1\\ -2\\ -2 \end{bmatrix}$ , where k is real number.

For $\lambda$ = 5 : (A-5I)X = 0

i.e., $\begin{bmatrix} -1& 2 & 2\\ 2 & -4 & -4\\ 2 & -4 & -4 \end{bmatrix}$ $\begin{bmatrix} x\\ y\\ z \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

i.e., -x+2y+2z = 0

2x-4y-4z = 0

i.e., x = 2y+2z

Let us take y = k and z = m. Then, x = 2k+2m.

Therefore, the eigenvectors corresponding to the eigenvalue 5 are $\begin{bmatrix} 2k+2m\\ k\\ m \end{bmatrix}$ ,i.e., $k\begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix}+m\begin{bmatrix} 2\\ 0\\ 1 \end{bmatrix}$ , where k is a real number.

Let X = $\begin{bmatrix} 1 & 2 & 2\\ -2 & 1 & 0\\ -2 & 0 & 1 \end{bmatrix}$ . Then X is non singular matrix.

Here, X^-1 = $\begin{bmatrix} 1/9 & -2/9 & -2/9\\ 2/9 & 5/9 & -4/9\\ 2/9 & -4/9 & 5/9 \end{bmatrix}$ .

Now, X^-1AX = $\begin{bmatrix} 1/9 & -2/9 & -2/9\\ 2/9 & 5/9 & -4/9\\ 2/9 & -4/9 & 5/9 \end{bmatrix}$ $\begin{bmatrix} 4 & 2 & 2\\ 2 & 1 & -4\\ 2 & -4 & 1 \end{bmatrix}$ $\begin{bmatrix} 1 & 2 & 2\\ -2 & 1 & 0\\ -2 & 0 & 1 \end{bmatrix}$

= $\begin{bmatrix} 1/9 & -2/9 & -2/9\\ 2/9 & 5/9 & -4/9\\ 2/9 & -4/9 & 5/9 \end{bmatrix}$ $\begin{bmatrix} -4 & 10 & 10\\ 8 & 5 & 0\\ 8 & 0 & 5 \end{bmatrix}$

= $\begin{bmatrix} -4 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 5 \end{bmatrix}$

Which is the diagonalisation of A.