\[ \boldsymbol{ A = \left[ \begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array} \right], \lambda = 2, 1 } \]
The aim of this question is to find the basis vectors that form the eigenspace of given eigenvalues against a specific matrix.
To find the basis vector, one only needs to solve the following system for $ x $:
\[ A x = \lambda x \]
Here, $ A $ is the given matrix, $ \lambda $ is the given eigenvalue, and $ x $ is the corresponding basis vector. The no. of basis vectors is equal to the no. of eigenvalues.
Expert Answer
Given matrix A:
\[ A = \left[ \begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array} \right] \]
Finding eigen vector for $ \boldsymbol{ \lambda = 2 }$ using the following defining equation of eigen values:
\[ A x = \lambda x \]
Substituting values:
\[ \left[ \begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] = ( 2 ) \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] \]
\[ \Bigg \{ \begin{array}{l} (1)(x_1) + (0)(x_2) = 2(x_1) \\ (-1)(x_1) + (2)(x_2) = 2(x_2) \end{array} \]
\[ \Bigg \{ \begin{array}{l} x_1 = 2x_1 \\ -x_1 + 2x_2 = 2x_2 \end{array} \]
\[ \Bigg \{ \begin{array}{l} x_1 – 2x_1 = 0\\ -x_1 + 2x_2 – 2x_2 = 0 \end{array} \]
\[ \Bigg \{ \begin{array}{l} – x_1 = 0\\ -x_1 = 0 \end{array} \]
Since $ \boldsymbol{ x_2 } $ is unconstrained, it can have any value (lets assume $1$). So the basis vector corresponding to eigen value $ \lambda = 2 $ is:
\[ \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \]
Finding eigen vector for $ \boldsymbol{ \lambda = 1 } $ using the following defining equation of eigen values:
\[ A x = \lambda x \]
Substituting values:
\[ \left[ \begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] = ( 1 ) \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] \]
\[ \Bigg \{ \begin{array}{l} (1)(x_1) + (0)(x_2) = x_1 \\ (-1)(x_1) + (2)(x_2) = x_2 \end{array} \]
\[ \Bigg \{ \begin{array}{l} x_1 = x_1 \\ -x_1 + 2x_2 = x_2 \end{array} \]
First equation gives no meaningful constraint, so it can be discarded and we only have one equation:
\[ -x_1 + 2x_2 = x_2 \]
\[ 2x_2 – x_2 = x_1\]
\[ x_2 = x_1\]
Since this is the only constraint, if we assume $ \boldsymbol{ x_1 = 1 } $ then $ \boldsymbol{ x_2 = 1 } $. So the basis vector corresponding to eigen value $ \lambda = 2 $ is:
\[ \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \]
Numerical Result
The following basis vectors define the given eigen space:
\[ \boldsymbol{ Span \Bigg \{ \left[ \begin{array}{c} 0 \\ 1 \end{array} \right] \ , \ \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] \Bigg \} } \]
Example
Find a basis for the eigenspace corresponding to $ \lambda = 5 $ eigenvalue of $A$ given below:
\[ \boldsymbol{ B = \left[ \begin{array}{cc} 1 & 0 \\ 2 & 7 \end{array} \right] } \]
The eigen vector equation:
\[ B x = \lambda x \]
Substituting values:
\[ \left[ \begin{array}{cc} -1 & 0 \\ 2 & -7 \end{array} \right] \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] = ( 7 ) \left[ \begin{array}{c} x_1 \\ x_2 \end{array} \right] \]
\[ \Bigg \{ \begin{array}{l} (-1)(x_1) + (0)(x_2) = 7(x_1) \\ (2)(x_1) + (-7)(x_2) = 7(x_2) \end{array} \]
\[ \Bigg \{ \begin{array}{l} x_1 = x_1 \\ 7x_2 = x_1 \end{array} \]
First equation is meaningless, so we only have one equation:
\[ 7x_2 = x_1 \]
If $ x_2 = 1 $ then $ x_1 = 7 $. So the basis vector corresponding to eigen value $ \lambda = 7 $ is:
\[ \left[ \begin{array}{c} 7 \\ 1 \end{array} \right] \]