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.

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:

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