Question

18. Suppose R^2 has weighted inner product given as <u, v> =(3u base 1 v base 1 + 2u base 2 V base 2 for u = (u base 1, u base 2), v = (V base 1, v base 2). Let u = (1, 2), v = ( 2, - 1) and K = 3. Then the valued of <u, kv> is.....

(i) 4

(ii) 6

(iii) 18

(iv) None

19. Suppose that u, v $\isin$ V are such that ||u|| = 2, ||u +v|| = 3 and ||u - v|| = 4. Then ||v|| is?

(i) 17/2

(ii) √17

(iii) Does not exist

(iv) None

20. For a given matrix A=$\begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}$, the matrix P that is orthogonally diagonalizes A is of the following matrices are diagonalisable

(i)P= $\begin{bmatrix} 1/√2 & 1/√2 \\ 1/√2 & - 1/√2 \end{bmatrix}$

(ii)P= $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

(iii)P=$\begin{bmatrix} -1/√2 & 1/√2 \\ - 1/√2 & 1/√2 \end{bmatrix}$

Answer 1

18.

$<u, v> =3u_ 1 v _ 1 + 2u _ 2 v _ 2$

$<u, kv>=3\cdot3\cdot2-2\cdot2\cdot3=18-12=6$

Answer: (ii) 6

19.

$||u \pm v|||^2=||u||^2\pm 2 <u, v>+||v||^2$

then:

$||u + v|||^2+||u - v|||^2=2(||u||^2+||v||^2)=9+16=25$

$||v||=\sqrt{25/2-4}=\sqrt{17/2}$

Answer: (iii) Does not exist

20.

$\begin{vmatrix} 3-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix}=0$

$\lambda^2-6\lambda+8=0$

$\lambda_1=2,\lambda_2=4$

$x+y=0\implies x=-y$

first eigenvector:

$\begin{pmatrix} x_1 \\ -x_1 \end{pmatrix}$

$-x+y=0\implies x=y$

second eigenvector:

$\begin{pmatrix} x_2 \\ x_2 \end{pmatrix}$

$P=\begin{pmatrix} x_2 &x_1\\ x_2&-x_1 \end{pmatrix}$

Answer:

(i)P= $\begin{bmatrix} 1/√2 & 1/√2 \\ 1/√2 & - 1/√2 \end{bmatrix}$