Question

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing anything), find the eigenvalues and eigenvectors of t. b) find the matrix representation of t to compute the eigenvalues and eigenvectors of t to confirm your answer in (a)

Answer 1

(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 $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

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 $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

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 $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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