Eigenvalues?

Algebra Level pending

Consider the usual R 2 \mathbb{R} ^2 dot product.

Let γ ∈ [ a b c d ] ; a , b , c , d ∈ ( R ) \gamma \in \begin{bmatrix}a & b \\c & d \end{bmatrix};a,b,c,d \in \mathbb(R) be a symetric matrix. It's known that:

d e t ( γ ) = − 1 det(\gamma)=-1

N ( γ + I ) = L ( ( 1 , 1 ) ) . N( \gamma + I)=L({(1,1)}).

If γ 2 × [ 2 0 ] = [ x y ] \gamma ^2 \times \begin{bmatrix}2 \\0 \end{bmatrix} = \begin{bmatrix}x \\y \end{bmatrix} enter the value of x + y x+y .

Clarifications:

  • N ( A ) N(A) denotes the Null Space of the matrix A;

  • I I denotes the identity matrix;

  • U = L ( v ) U=L(v) denotes the span.


The answer is 2.

Joao Miguel Coelho by Joao Miguel Coelho , 24, Portugal

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1 solution

γ \gamma is a 2 × 2 2\times2 matrix then it has maximum 2 2 eigenvalues.

Since d e t ( γ ) = − 1 det(\gamma)=-1 and N ( γ + I ) ≠ 0 N( \gamma + I)\neq 0 , 1 1 and − 1 -1 are the only eigenvalues of γ \gamma .

γ \gamma is symetric then N ( γ + I ) N( \gamma + I) and N ( γ − I ) N( \gamma - I) are orthogonal, giving us that N ( γ − I ) = L ( ( 1 , − 1 ) ) . N( \gamma - I)=L({(1,-1)}).

Then:

γ 2 × [ 2 0 ] = γ γ ( [ 1 1 ] + [ 1 − 1 ] ) = γ ( [ 1 1 ] − [ 1 − 1 ] ) = ( [ 1 1 ] + [ 1 − 1 ] ) = [ 2 0 ] \gamma ^2 \times \begin{bmatrix}2 \\0 \end{bmatrix} = \gamma \gamma (\begin{bmatrix}1 \\1 \end{bmatrix}+\begin{bmatrix}1 \\-1 \end{bmatrix})=\gamma (\begin{bmatrix}1 \\1 \end{bmatrix}-\begin{bmatrix}1 \\-1 \end{bmatrix})=(\begin{bmatrix}1 \\1 \end{bmatrix}+\begin{bmatrix}1 \\-1 \end{bmatrix})=\begin{bmatrix}2 \\0 \end{bmatrix}

So the answer is 2 + 0 = 2 2+0=2

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