A scale drawing of a square has a side length of 9 inches. The drawing has a scale of 1 in. : 5 mi. Find the actual perimeter an
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(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
See the figure for the graph:
(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )
∴ x → -x hence '-1' is the eigen value.
∴ y → y hence '1' is the eigen value.
also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.
( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.
(b) ∵ T(x, y) = (-x, y)
T(x) = -x = (-1)(x) + 0(y)
T(y) = y = 0(x) + 1(y)
Matrix Representation of
now, eigen value of 'T'
after solving the determinant,
we get two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
Hence,
(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
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