information:classification
Eigenvectors

Once the (exact) value of an eigenvalue is known, the corresponding eigenvectors can be found by finding nonzero solutions of the eigenvalue equation, that becomes a system of linear equations with known coefficients. For example, once it is known that 6 is an eigenvalue of the matrix A = {\displaystyle A={\begin{bmatrix}4&1\\6&3\end{bmatrix}}}

we can find its eigenvectors by solving the equation A v = 6 v {\displaystyle Av=6v}, that is = 6 ⋅ {\displaystyle {\begin{bmatrix}4&1\\6&3\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}=6\cdot {\begin{bmatrix}x\\y\end{bmatrix}}}

This matrix equation is equivalent to two linear equations { 4 x + y = 6 x 6 x + 3 y = 6 y {\displaystyle \left\{{\begin{aligned}4x+y&=6x\\6x+3y&=6y\end{aligned}}\right.} that is { − 2 x + y = 0 6 x − 3 y = 0 {\displaystyle \left\{{\begin{aligned}-2x+y&=0\\6x-3y&=0\end{aligned}}\right.}

Both equations reduce to the single linear equation y = 2 x {\displaystyle y=2x}. Therefore, any vector of the form T {\displaystyle {\begin{bmatrix}a&2a\end{bmatrix}}^{\textsf {T}}}, for any nonzero real number a {\displaystyle a}, is an eigenvector of A {\displaystyle A} with eigenvalue λ = 6 {\displaystyle \lambda =6}.

The matrix A {\displaystyle A} above has another eigenvalue λ = 1 {\displaystyle \lambda =1}. A similar calculation shows that the corresponding eigenvectors are the nonzero solutions of 3 x + y = 0 {\displaystyle 3x+y=0}, that is, any vector of the form T {\displaystyle {\begin{bmatrix}b&-3b\end{bmatrix}}^{\textsf {T}}}, for any nonzero real number b {\displaystyle b}.