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Find a basis for the eigenspace corresponding to each listed eigenvalue of A given below:

\[ \boldsymbol{ A = \left[ \begin{array}{cc} 1 & 0 \\ -1 & 2 \end{array} \right], \lambda = 2, 1 } \]

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:

\[ \left[ \begin{array}{c} 7 \\ 1 \end{array} \right] \]

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